Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter

You can Download Properties of Matter Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter

Properties of Matter Questions and Answers

Properties of matter

The matter commonly seen in three states, solid, liquid and gas. Matter has volume and mass. Matter is anything which occupies space and has volume. The solid has a definite shape. But liquid occupy the shape of the container. Gas has no definite shape.

Properties Of Matter Class 8 Questions And Answers Question 1.
explain the peculiarities of materials.
Answer:
Tiny particles of matter

The substances are made up of many tiny particles which cannot be seen by naked eye. Even though we cannot see the tiny particles of sugar in sugar solution, we can understand that tiny particles of sugar is dissolved in the solution while we taste it.

All the substances are made up of tiny particles which cannot be seen by naked eye and these tiny particles bear all the properties of the substances.

Properties Of Matter Class 8 Notes Pdf Kerala Syllabus Question 2.
identify and picturise the arrangement of particles in different states of matter.
Answer:
Arrangements of particles in different states of matter

The particles of solids are very close to each other. The freedom of movement of particles is limited. The attractive force between particles is very high.
Properties Of Matter Class 8 Questions And Answers

The particles of liquid are relatively further apart and have more freedom for movement than in the solid state. The attractive force is less than that of solid state. The particles in gaseous state remains far away from one another and the attractive force is very low.

Change of State

When ice is heated, it changes into water and when water is further heated it boils and changes into steam.
Properties Of Matter Class 8 Notes Pdf Kerala Syllabus
Properties Of Matter Questions And Answers Kerala Syllabus

Properties Of Matter Questions And Answers Kerala Syllabus Question 3.
Which form of energy is responsible for the change of state here?
Answer:
Here the form of energy is responsible for the change of state is heat.
Hsslive Guru 8th Class Chemistry Kerala Syllabus

When heat is absorbed energy of the particles, distance between the particles, movement of the particles are increased. The attractive force between particles is decreased.
Basic Science Class 8 Solutions Chapter 4 Kerala Syllabus

Some substances, when heated, change directly into gas without forming liquid. This phenomenon is sublimation.

Diffusion of substances in different states

The fragrance of incense stick fills the room when the stick is lit. The fragrance spreads because its particles are spread in air. If a drop of ink is added carefully into the water in a beaker, the ink spread in the water.

The spontaneous mixing of different particles having freedom of movement is Diffusion.

Hsslive Guru 8th Class Chemistry Kerala Syllabus Question 4.
Distinguish pure substances and mixtures and tabulate them?
Answer:
Pure Substances and Mixtures
Depending on their nature, substances can be classified into two.
1. Pure Substances
2. Mixtures Materials made of particles of identical nature are called pure substances
Eg: Water, common salt, sugar

The substances made of particles of different nature are called mixtures.
Eg: Salt solution, sugar solution

Separating the Components of a Mixture.

Most of the substances found in nature are mixtures. We use different methods of separation of components in rice mixed with stones, tea mixed with tea dreg, mixture of methanol and ethanol.

Distillation

Distillation is used for separation of components in salt solution.
8th Class Biology Notes Pdf Kerala Syllabus

When one component of the mixture is volatile and the others do not vapourise under the same condition, they can be separated by distillation. More over components of a mixture possess a large difference in their boiling points, they can be separated by distillation. Eg:Mixture of water (boiling point ioo°C)and acetone(56°C)

Fractional Distillation

If the boiling points of components have very small differences, fractional distillation is to be used to separate them.
Kerala Syllabus 8th Standard Chemistry Notes

Eg: Mixture of Ethanol (boiling point 78°C) and methanol (boiling point 65°C).

When vapours of the mixture pass through the fractionating column, repeated. liquefaction and vapourisation take place. Subsequently, the vapours of low boiling methanol enter the condenser from the fractionating column, condense to liquid and get collected in the round bottomed flask first. Similarly, ethanol with higher boiling point can be collected later in another round bottomed flask.

Separating Funnel

Separating funnel is an apparatus used for separating immiscible liquids from their mixture.
Basic Science For Class 8 Chapter 4 Kerala Syllabus
We can use separating funnel to separate the mixture of water and kerosene which are having difference in density.

Sublimation

This method can be used to separate the components which have the property of sublimation from the mixture. This method can be used for separating the components of a mixture of ammonium chloride and sand.
Kerala Syllabus 8th Standard Physics Notes

Centrifugation

This is a method for separating components from a mixture, based on the difference in the mass of particles. This method is used in clinical laboratories to separate blood cells from blood samples and also for separation of butter from curd.

In order to separate insoluble particles in a liquid mixture on the basis of their mass difference, centrifuge is used. The process is known as centrifugation.

Chromatography

Chromatography is the method used to separate more than one solute dissolved in the same solvent. This method was first employed for separating coloured substances and hence this process came to be known as Chromatography. Chromatography is employed to separate components from dyes and to separate poisonous substances mixed with blood.

Properties of Matter Textbook Questions and Answers

Basic Science Class 8 Solutions Chapter 4 Kerala Syllabus Question 1.
A few mixtures are given below. Tabulate the methods to separate their components and give the reasons for selecting the method.

MixtureMethodReason
Common salt and ammoni­um chloride
Sugar solution
Petrol and ker­osene
Camphor and glass powder
Iron powder and sand

Answer:

MixtureMethodReason
Common salt and ammonium Chlo-SublimationComponents having the prope-
Sugar solutionDistillationComponents having the property of evaporation
Petrol and ker­oseneFractional dist­illationComponents have small differanee in boiling po­int
Camphor and glass powderSublimationComponents having the property of sublimation
Iron powder and sandMagnetic separationComponents ha­ving magnetic properties

8th Class Biology Notes Pdf Kerala Syllabus  Question 2.
Given below are certain changes taking place to the particles during change of state. From this, find out and tabulate the changes in the particles when water boils to form steam and also when steam condenses to form water.

  • distance increases
  • attractive force decreases
  • energy increases
  • rate of movement increases
  • distance decreases
  • energy decreases
  • attractive force increases
  • rate of movement decreases

Answer:

Water changes to Steamsteam to water
Distance increasesDistance decreases
Energy increasesEnergy decreases
Attractive force decreasesAttractive force increases
Movement of particles increasesMovement of particles decreases

Kerala Syllabus 8th Standard Chemistry Notes Question 3.
Spirit kept open in a.watch glass disappears after some time. Which among the following phenomena are responsible for this?
a) sublimation
b) distillation
c) evaporation
d) diffusion
Answer:
a) sublimation

Basic Science For Class 8 Chapter 4 Kerala Syllabus Question 4.
Which are the methods that can be used to separate the components of a mixture made of common salt, ammonium chloride and sand? Write the methods in the order in which they are applied.
Answer:
sublimation → filtration → distillation

Kerala Syllabus 8th Standard Physics Notes  Question 5.
Many minerals are present in ordinary water.
a. Which is the method that can used to remove the mineals and obtain pure water?
h.In which type of mixtures is this method employed?
c. Water purified by this method is distilled water .Write two instances of its use.
Answer:
a. distillation
b. When one component of the mixture is volatile and the others do not vapourise under the same condition, they canbe separated by distillation
c. For vaccination, In storage batteries

Basic Science Class 8 Pdf Notes Kerala Syllabus Question 6.
From the following statements, tick (V) those which apply to solid substances alone.
□ Particles have little freedom of movement.
□ Distance between particles is very high
□ Particles remain very close to each other
□ Energy of particles is very high
Answer:
✓ Particles have little freedom of movement.
✓ Particles remain very close to each other

Question 7.
Take a small wooden rectangular block and find its volume (volume = length x breadth x height).
Take a big measuring jar and fill three-fourth of it with water and mark the‘water level. Then dip the block in water in the jar. (To prevent floating, nails can be inserted in the block). Mark the difference in the water level.
a. Is there any relation between the difference in the water level and the volume of the block?
b. Which property of matter is revealed by this experiment?
Answer:
a.Yes. the volume of water rises is the volume of wooden block.
b. The substance that occupies a space.

Question 8.
Electronic balances are very popular now. On an electronic balance, find the weight of an empty balloon. Again, find its weight after filling air. Now, can you find the weight of the air in the balloon?
Repeat the experiment using balloons of different size by filling them with varying quantity of air.
Answer:
Reduce the weight of balloon from the weight of balloon with air

Question 9.
Take water mixed with chalk powder in a bottle. Tie a string to the bottle and swirl it at high speed along a circular path. Observe after sometime.
Repeat the activity using other mixtures which are suspensions. To which method of separation of components of a mixture can this be connected? Are there instances where this principle is made use of. Prepare a note.
Answer:
Centrifugation is the method used. This is a method for separating components from a mixture, based on the difference in the mass of particles. This method is used in clinical laboratories to separate blood cells from blood samples and also for separation of butter from curd.

Question 10.
Take a long white chalk piece and put a mark with black ink slightly above the bottom. Keep the chalk piece dipped perpendicularly in water in a watch glass. After sometime observe the changes. Repeat the experiment using different chalk pieces marked with sketch pens of different colours. To which of the methods of separation you have studied is this related?
Answer:
Chromatography

Properties of Matter Additional Questions and Answers

Question 1.
Find the odd one out
a. Evaporation, Fermentation, Filtration, Chromatography.
b. Sea water, Muddy water, Table salt, Curd, Air
c. Distillation, Separating funnel, Filtration, Evaporation
Answer:
a. Fermentation
b. Table salt
c. Separating funnel

Question 2.
What is the use of chromatography?
Answer:
Solutes dissolved in the same solvent can be separated using chromatography.

Question 3.
What is the importance of evaporation? Where is this technique used on a large scale?
Answer:
Evaporation is used to separate solids dissolved in liquids, mainly water. It is used on a large scale to obtain common salt from sea water. Sea water is left in the open in large containers on sunny days. After a few days salt remains as residue and water evaporates. Common salt is then further purified.

Question 4. What is the advantage of distillation over evaporation?
Answer:
Distillation is more advantageous than evaporation, because liquid can be recover in distillation while in evaporation liquid is lost.

Question 5.
Which process is used to dry clothes in a washing machine?
Answer:
Centrifugation.

Question 6.
Method for separating drugs (medicine) which are miscible in blood Centrifugation, Evaporation, Filtration, Chromatography.
Answer:
Chromatography

Question 7.
From the below statements select correct one?
a. Every substance has a definite mass
b. Every substance has a definite shape
c. Every substance has a definite volume
Answer:
a. Every substance has a definite mass

Question 8.
Melting: Converting solid into liquid Sublimation:
a. Observe the first pair and fill up. the blanks in the second pair suitable.
b. Write two examples of sublimation
Answer:
a. Sublimation. A solid directly changes into gas
b. Camphor, iodine, naphthalene.

Question 9.
From Gopu’s hand the bottle containing coconut oil fall into a bucket containing water. Is it possible to separate oil and water?
Answer:
Using a separating funnel, they can be separated. The denser water will be at the bottom and the lighter oil will be at the top. Water is removed through the tap.

Question 10.
Classify the following substances according to its state and tabulate.
Water, pencil, book, air, kerosene, oxygen
Answer:

solidliquidgas
Pencil
Book
Water
kerosene
Air
Oxygen

Question 11.
Will water fill in tumbler when a glass tumbler is immersed perpendicularly into the water in a turf ? Give reason.
Answer:
No because there is air in the tum-bler. Air need space to occupy so water cannot enter into the tumbler.

Question 12.
A balloon filled with air and an empty balloon weigled in a digital balance found difference in weight. What are the properties of matter understood by this experiment?
Answer:
Matter needs space to occupy and has mass.

Question 13.
Complete the table.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 10
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 11

Question 14.
Which energy causes the change of state when ice is heated to water and water to water vapour? The three figures shows the arrangements of particles. Recoga- nise and write the state of matter.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 12
Answer:
Heat energy figure
(1) solid, figure
(2) liquid, figure
(3) gas

Question 15.
Mention three examples from daily life for diffusion.
Answer:
Spreads the smell of perfumes, smell of fruits and food item. Ink spreads in water.

Question 16.
Complete the table
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 13
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 14

Question 17.
When naphthalene balls are kept somewhere after some days its size diminishes and disappears. Name the phenomenon caused for this process.
Answer:
Sublimation

Question 18.
Mention the reason behind the smell spreads everywhere in the room while in scent stick is lighted. Sublimation, Evaporation, Diffusion
Answer:
Diffusion

Question 19.
Tabulate following as pure substances and mixtures.
Gold, soda water, steam, sugar, soil, water, common salt, sugar solution, carbon dioxide, salt solution.
Answer:

Pure substancesMixtures
GoldSoda water
SugarSugarsolution
SaltSoil
WaterSalt solution
Steam
Carbon dioxide

Question 20.
Find the method of separation of components following mixt¬ures.
a. husk from paddy
b. butter from curd
c. tea dreg from tea
d. salt from sea water
e. blood cells from blood sample
f. mixture of petrol and kerosene
g. salt and ammonia
h. poison from blood
i. water and kerosene
j. water and acetone
Answer:
a. by picking
b. centrifugation
c. filtration
d. distillation
e. centrifugation
f. fractional distillation
g. sublimation
h. chromatography
i. using separating funnel
j. distillation

Question 21.
Write two examples of sublimation from daily life.
Answer:
1.camphor
2. naphthalene balls becomes smaller

Question 22.
Salt, rava and mustard are mixed while Suresh has brought it from shop. Say methods to separate these.
Answer:
Salt dissolves in water. Then filter it. Rava and mustard are separating by picking, salt is collected by distillation.

Question 23.
What will be seen when a filter paper is marked by sketch pen and one end of the paper is dipped in water? what is this process called?
Answer:
different colours are separated. Chromatography

Question 24.
Using given methods write one example for separating components of mixture.
Distillation, magnetic separation, centrifugation, fractional distillation, filtration, sublimation.
Answer:
• Separating common salt from sea water – Distillation
• Mixture if iron powder and sulphur – Magnetic separation
• Butter from curd – Centrifugation
• Separation of kerosene, petrol, die sel – Fractional distillation
• Separating tea dreg from tea – Fil-tration
• Ammonium chloride and common salt – Sublimation

Question 25.
Solid, Liquid, Gas are the three states of matter
a. Which of this having definite volume hut not definite shape?
b. Write any two changes occurs when we change a substance from liquid state to gaseous state?
c. How to convert a gasecfas substances into liquid?
Answer:
a. Liquid
b. Increase the distance between molecules.
• Decrease the attractive force between molecules.
• Increase the speed of molecules.
• Increase the energy of molecules.
c. to be cooled (or) Reduce temperature

Question 26.
Name of the mixtures in column A, properties of components are given in column B, method of separating of component are given in the column C also. Find out the match items from B & C to A.

MixtureCharacter of componentsMethod of Separation
1. Alcohol and waterdifference in boiling pointCentrifugation
2. Soil and mentordifference in eva­porating CharacterDistillation
3. Curd and butterdifference in weight of inso­luble precipitateChromatography
Sublimation

Answer:

MixtureCharacter of componentsMethod of Separation
1. Alcohol and waterdifference in boiling pointDistillation
2. Soil and mentordifference in eva­porating characterSublimation
3. Curd and butterdifference in weight of inso­luble precipitateCentrifugation

Question 27.
a. Which device given in the picture?
b. Why should this device doesn’t use for separating petrol and kerosene mixture?
c. Which method is used to separate the mixture containing petrol and kerosene?
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter 8
Answer:
a. Separating funnel
b. Petrol and kerosene are miscible liquids.
c. Fractional distillation

Question 28.
Diffusion is a phenomenon which
causes the spread of smell.
a. What is the relation between the rate of diffusion and states of substance?
b. What is the relation between diffusion and temperature?
c. Given an example from everyday life which shows relation between diffusion and temperature.
Answer:
a.Rate of diffusion is more in gaseous state, less in solid state.
b. Rate of diffusion increases with increase in temperature.
c. The smell of hot food material spread quickly.

Question 29.
Find out the relation and fill in the blanks.
a. Separate salt from saline water :: Evaporation; Separation of components from black coloured ink :: ………..
b. Distillation :: Condenser; Fractional distillation :: ………..
c. Write down some situations for which the fractional distillation used?
d. When we use fractional distillation for separating components of mixture?
Answer:
a. Chromatography
b. Fractionating column
c. • Separating components to petroleum
• Separating components of air.
d. Used when slight differences in their boiling points.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units

You can Download Measurements and Units Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 8 Measurements and Units

In olden days different scales were used for measuring length in different countries. Vaara, One forearm (muzham) etc are some of such units. These units were not accurate. For solving this problem a standard scale of a definite length has been recognised world widw for measuring length. This standard scale is the unut of length. The basic unit of length is metre (m).

Smaller units of length

Examine the metre scale. You can see small measurements on it. These are centimetre (cm) and millimetre(mm)
1m = 1oocm 1cm=1omm
The smallest length that can be accurately measured using a metre scale is its least count.

Thickness of paper

Measurement And Units In Physics Class 8 Questions And Answers

We cannot measure the thickness of the paper even with these small units. Take too sheets of paper and pile it up to make a bundle. Measure the height of this bundle using a scale. If it falls in between two markings, make suitable changes in the number of sheets. Then measure its height. By dividing this height by the number of sheets, we will get the thickness of a sheet of paper.

Measurement And Units In Physics Class 8 Kerala Syllabus

Similarly to find out the thickness of thin wire roll it on a pencil and mea-sure the length of ten rolls Then divide it by 10.

Length of a curved line

To measure the length of the curved line put a thread on it and measure its length using a meter scale

Measurements And Units Class 8 Notes Kerala Syllabus

Distance of a sphere

A ball kept in between two wooden blocks, whose diameter is to be measured. Then we will get its diameter.

Measurement And Units In Physics Class 8 Notes

Various measurements of length

The basic unit of length is metre. Smaller units like centimetre (cm), millimetre (mm), micrometre (micron – ™ nano-metre (nm) are also used in certain situa-tions for convenience. We use a bigger unit like kilometre (km) to measure the distance between two places. Units like astronomical unit (AU), light year (ly) are also used to measuring distance to planets or stars.

Units And Measurements Class 8 Kerala Syllabus

Measurement And Units In Physics Class 8 Questions And Answers Mass

Mass is the measure of quantity of matter contained in it. The basic unit of mass is kilogram(kg).

Except kilogram, milligram(mg), gram(g), quintal, tonne are also the units of mass.

Basic Science Question Answer Chapter Wise Class 8

Measurement And Units In Physics Class 8 Kerala Syllabus  Time
The symbol is s.
60s = iminute; 6ominute = 1 hour One second is 1/ 86400 part of an average solar day.

Fundamental units

There are some quantities which are not related to one another and cannot be expressed using other quantities. Such quantities are fundamental quantities. The units of the fundamental quantities are the Fundamental Units. The system based on these fundamental units is the | International System of Units. Its short form is SI Units.

Chemistry Textbook For Class 8 Kerala Syllabus

Measurements And Units Class 8 Notes Kerala Syllabus Derived unit

Units which are expressed in terms of fundamental units or those units which are dependent on fundamental units are derived units.
Ex. Area = length x breadth

Unit of area = unit of length x unit of breadth = m xm = m2.

HSSLive.Guru

Salient features of SI units

  1. Unified units
  2.  Internationally accepted
  3.  Adequate to express all physical quantities.

Measurement And Units In Physics Class 8 Notes Density and volume

The space occupied by a body is referred to as its volume.
The unit of volume is cubic metre (m3)

Mass per unit volume of substance is its density. Density = mass/volume Unit of density = unit of mass/unit of volume = kg/m3

The rules to be followed while writing units

We have to observe certain rules according to international standards while using these units and their symbols. They are given below:

  1. The symbols of units are normally written using small letters in the English alphabet, eg. m (metre), s (second), kg (kilogram)
  2. But there are certain occasions on which capital letters of the English alphabet are used as symbols. The units named after persons are written like this.
  3. While writing the names of units never use capital letters, eg. kelvin (correct) Kelvin (wrong) newton (correct) Newton (wrong)
  4. Never use the plural form for symbols, eg. 10 kg (correct) 10 kgs (wrong) 75 cm (correct) 75 cms (wrong)
  5. Never use full stop or comma after a
    symbol except at the end of a sentence. eg. 75 cm is the length of a table, (correct) 75 cm. is the length of a table, (wrong),
  6. While writing derived units a slash (/) is used to denote division. But never use more than one slash in one derived unit. eg. m/s1 (correct) m/s/ s (wrong)
  7. When a derived unit is expressed as the product of other units use a dot or a space between them. eg. N.m or Nm
  8. 8. Do not mix the name of a unit with the symbol.
  9. While writing units along with a value, there must be single space between them.
  10. Never use more than one unit to express a physical quantity.

Measurements and Units Textbook Questions and Answers

Units And Measurements Class 8 Kerala Syllabus Question 1.
Write down the following units in the ascending order of their values
a) mm nm cm µ m.
b) 1 m 1 cm 1 km 1 mm
Answer:
a. nm → µ m → mm → cm
b. 1 mm → 1 cm → 1 m → 1 km

Basic Science Question Answer Chapter Wise Class 8 Question 2.
Which of the following units does not belong to the group?
a) kg mg g mm
b) m mm km mg
Answer:
a. mm
b. mg

HSSLive.Guru

Chemistry Textbook For Class 8 Kerala Syllabus Question 3.
Imagine that the distance from school to your friend’s house is 2250 m. State this in kilometres.
Answer:
2\(\frac{1}{4}\) km or 2.25 km

8th Class Physics Question And Answers Kerala Syllabus Question 4.
Convert the following units, into SI units without changing their values.
a. 3500 g
b. 2.5 km
c. 2 h
Answer:
a. 3.5 kg
b. 2500 m
c. 7200 s

Measurements and Units Additional Questions and Answers

Kerala Syllabus 8th Standard Notes Physics Question 1.
Take a small quantity of solution A and solution B in a jar. Density of A is 2 kg/m3 and density of B is 5 kg/m3. Which of these flot above the jar ? Give reasion.
Answer:
Solution A
Density of A is less than the density of B. So solution with less density floats above the solution that has high density.

Class 8 Chemistry Chapter 1 Kerala Syllabus Question 2.
Find out the given word pair relation; and fill in the blanks.
a. Length : m ; Volume :………
b. Displacement : m ; Acceleration:…………….
Answer:
a. m3
b. m/s²

HSSLive.Guru

Basic Science Class 8 Solutions Kerala Syllabus Question 3.
a. What is the basic unit of time?
b. What do you mean by one Solar day?
c. How much time earth will take to revolve once around the sun?
Answer:
a. Second
b. A day or a solar day is the time period from one noon to the next noon.
c. One year (365 days)

Class 8 Science Notes Pdf State Syllabus Question 4.
Find out odd one. Justify your answer?
a. meter, kilogram, kilogram/ meter3, second
b. Length, volume, mass, time
c. Volume, mass, density accelaration.
Ans:
a.kilogram/meter3 – meter, kilogram, second are basic units, kilogram/meter3 is unit of volume.
b. Volume – Volume is a derived quantity and others are basic quatities.
c. Mass – Mass is a basic quantity and others are derived quantity.

Class 8 Physics Kerala Syllabus Question 5.
What is meant by least count of a metre scale?
Answer:
The smallest length that can be accurately measured using a metre scale is its least count. The least count is 1mm when metre scale is used.

Basic Science Class 8 Pdf Notes Kerala Syllabus Question 6.
Mention a method to measure the thickness of a paper sheet.
Answer:
Take 100 sheets of paper and pile it up to make a bundle. Measure the height of this bundle using a scale. If it falls in between two markings, make suitable changes in the number of sheets. Then measure its height. By dividing this height by the number of sheets, we will get the thickness of a sheet of paper.

Hsslive Guru 8th Class Physics Kerala Syllabus Question 7.
Using following objects how can you measure the diameter of a ball?
hall, two wooden blocks, metre scale
Answer:
The ball is kept in between two wooden blocks as shown in the figure whose diameter is to be measured.
8th Class Physics Question And Answers Kerala Syllabus

Basic Science Class 8 Kerala Syllabus Question 8.
Fill in the blanks
a. 1 nm = ….. m
b. 1 km = …….cm
c. 1000g = …….kg
d. one light year = …… km
e. 60 sec = …… minute
Answer:
a. 1 nm = 10-9 m
b. 1 km = 105
c. 1000g = 1kg
d. one light year = 365 × 24 × 60 × 60 × 105 km
e. 6o sec = 1 minute

Chemistry Class 8 Kerala Syllabus Question 9.
Calculate how many seconds in a solar day.
Answer:
24 ×60 × 60s

HSSLive.Guru

Std 8 Biology Notes Malayalam Medium Kerala Syllabus Question 10.
Explain the derived units with examples.
Answer:
Units which are expressed in terms of fundamental units or those units which are dependent on fundamen¬tal units are derived units.
Ex . Area = length × breadth=m × m = m2

Question 11.
Tabulate the fundamental quantities and their units.
Answer:
Kerala Syllabus 8th Standard Notes Physics

Question 12.
What are the salient features of SI units.?
Answer:

  • Unified units
  • Internationally accepted
  • Adequate to express all physical quantities.

Question 13.
Find the volume of a box having length 2m,breadth lm and height 0.5m
Answer:
volume = length × breadth × height =2 × 1 × 0.5 = 1 m3

Question 14.
Will the mass be same if you take one thermocol and a wooden block of same volume?
Answer:
No. Mass/Volume is density. The density of thermocol is lesser than that of wooden block. The mass will not be the same.

Question 15.
Take one jar with salt water and into both jars. In which jar the egg floated over the liquid. Justify your answer.
Answer:
In the jar having salt water because it has more density.

Question 16.
Tabulate following units and symbols
a. Length
b.Mass
c. Potential difference
d. Force
e.Pressure
f. Time
g.Density
Answer:
Class 8 Chemistry Chapter 1 Kerala Syllabus

HSSLive.Guru

Question 17.
Choose the derived quantities from following.
a. Length
b. volume
c. time
d. temperature
e. density
f. velocity
g. intensity of light
Answer:
volume, density, velocity

Question 18.
Tabulate two units connected to the name of personals
Answer:
Basic Science Class 8 Solutions Kerala Syllabus

Question 19.
Find the density of an object of mass 60 kg a having length 5m breadth 3m and height 2m.
Answer:
density =mass ÷ volume
\(=\frac{60}{5 \times 3 \times 2}=\frac{60}{30}=2 \mathrm{kg} / \mathrm{m}^{3}\)

Question 20.
Find the volume of a brick which has a length, breadth and height of 3m, 2m and 4m respectively.
Answer:
Volume = Length × breadth × height = 3 × 2 × 4 = 24m3

Question 21.
How to find the area of irregular surface? Explain with example.
Answer:
The area of irregular surface can be calculated by using graph paper.

Question 22.
How do you find out the diameter of a torch cell
Answer:
Place the cell between two wooden blocks tightly. Find out the distance between the two faces of the blocks, touching the cell using a metre scale. This reading gives the diameter of the cell.

HSSLive.Guru

Question 23.
The length of a room is 8m and the breadth ia 5m. Calculate the area of the room.
Answer:
Aiea = length × breadth = 8m × 5m = 40m2

Question 24.
What is meant by density of a substance?
Answer:
Density is the mass of a substance having unit volume. If the mass of the substance is M and V is its volume,
then \(\frac{M}{V}\)

Question 25.
Find out the fundamental units from the following table and mark them. Find out which unit is used to measure each of the physical quantity.

  • Ampere
  • Metre/second
  • Kilogram
  • Kilogram/metre
  • Mole
  • Kelvin
  • Metre3
  • Kilogram metre/second

Answer:
Fundamental units

  • Ampere
  • Kilogram
  • Mole
  • Kelvin

Physical quantity

  • Ampere – Electric current
  • Kilogram – Mass
  • Mole – Amount of substance.
  • Kelvin – Temperature

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields

You can Download Let’s Regain Our Fields Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 3 Let’s Regain Our Fields

Let’s Regain our Fields Questions and Answers

Let’s Regain Our Fields Kerala Syllabus 8th Food Safety

Food Safety is a condition in which every individual are provided with food enough to lead a healthy life. It is very’ essential to ensure food safety to create a society in which there is no fear of poverty and health issues due to malnutrition. It is a challenge to ensure food safety in a condition in which the agricultural fields are declining. A culture that loves soil and agriculture has to be recreated. It is also important to regain the lost agricultural fields.

Indicators (Text Book Page No:35)

Lets Regain Our Fields Notes Kerala Syllabus 8th Question 1.
Didn’t you notice the illustration and the newspaper report?
Answer:
Population explosion, insufficient agricultural fields, negative attitude towards agriculture, unscientific vision, methods of agriculture, influence of consumer society, etc., are the causes of scarcity of food.

Lets Regain Our Fields Notes Pdf Kerala Syllabus 8th Question 2.
What is the concept indicated by the illustration?
Answer:
Modern techniques of agriculture play a major role to overcome the problems and crises mentioned. The techniques agriculture like polyhouse farming, open precision farming, integrated pest management, hydroponics. aeroponics etc are the contributions of science. The novel information and findings form remedy for food scarcity.

Let’s Regain Our Fields Pdf Kerala Syllabus 8th Question 3.
Discuss it with your friends using the given indicators. Write your inferences in the science diary.
Answer:
The major goal of food safety is to eliminate hunger and poverty from our country. Supplying food grains in cheaper rates is ultimately beneficial to the people of the lower strata.

Indicators (Text Book Page No:36)

Class 8 Biology Chapter 3 Kerala Syllabus Question 4.
What were the changes that occurred in the area of agricultural fields from the year 1971 to 2011?
Answer:
During the period 1971-2011, the area of agricultural fields considerably reduced. In 1971 the area was 8.75 lakh hectors and in 1991 it was reduced to 5.5 lakh hectors. It was reduced to 2.08 lakh hectors in the year 2011.

Let’s Regain Our Fields Notes Kerala Syllabus 8th Question 5.
What tendency could be observed in rice production and population growth during the period?
Answer:
In this period population increased from 2.13 crores to 3.34 crores. But the production of rice steeply reduced from 13.65 lakh tons to 5.69 lakh tons.

8th Class Biology Notes Pdf Kerala Syllabus Question 6.
Is this tendency desirable? Why?
Answer:
This tendency is not advisable. The increase in population and decrease in the area of crop field and the production of rice may lead the nation to poverty. The goal of food safety will not be attainable.

Hss Live Guru 8 Biology Kerala Syllabus Question 7.
What are the obstacles faced by farmers today?
Basic Science Question and Answer:
The Crises in the field of Agriculture

Farmers face a number of crises. They are mainly loss, cost of production, limited cultivable land, climate change, exploitation of brokers, environmental destruction and health issues, etc. Many of these problems can be overcome if a positive attitude towards agriculture develop. There are many possibilities to solve every problems.

i. Fertile soil

Around 20 different elements are required for the growth of plants. They are called essential elements.
eg: Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus, Potassium, Sulphur

These elements are naturally available in the soil by the decomposition .of microorganisms. The fertility of soil can be increased by proper manuring. Organisms like bacteria, fungi, algae, termite, earthworm, etc., help to increase the fertility of soil.

Indicators (Text Book Page No:38)

Hss Live Guru 8th Biology Kerala Syllabus Question 8.
What is the role of microorganisms in ensuring the natural availability of elements in the soil?
Answer:
The elements are naturally available in the soil by the decaying action of micro-organisms. When leguminous plants are grown in fields, the microorganisms that harbour in the root nodules fixes atmospheric nitrogen to the soil, bacteria, fungi, algae, termite, earthworm, etc., increase the fertility of the soil.

Hsslive Guru Biology 8th Kerala Syllabus Question 9.
What is the need of testing the soil?
Answer:
Soil analysis (soil testing) is done in order to detect the amount of elements present and their pH value of the soil.

Let’s Regain Our Fields Pdf Download Question 10.
Why does the application of fertilizers become essential for better crop yield?
Answer:
Fertility of the soil is increased by proper manuring. Manuring helps to make up the deficiency of elements in the soil that are required for plant growth. As a result better yield is obtained.

Lets Regain Our Fields Pdf Kerala Syllabus 8th Microbial Fertilizer

Microbial fertilizers are substances that contain micro-organisms that facilitate the fertility of the soil. Micro-organisms increase the amount factors, required for plant growth, in the soil. The bacteria like rhizobium, Azotobacter, azospirillum, and the aquatic plant azolla increase the amount of nitrogen in soil. Many things are to be taken care to retain the microorganisms in the soil. They are

  • Ensure the availability of organic fertilizer(biofertilizers) in the soil.
  • Sufficient water supply must be there.
  • Do not use chemical fertilizers and pesticides.

Consequences of Unscientific application of fertilizers

Indicators (Text Book Page No:39)

Kerala Syllabus 8th Standard Biology Notes Question 11.
What are the consequences of unscientific application of chemical fertilizers? Discuss on the basis of the following indicators.

  • composition of soil
  • microorganisms in soil
  • health issues
  • financial factors

Answer:
Unscientific fertigation loses the fertility and changes the natural texture of soil. Certain elements required for plant growth will exceed in the soil and certain others become insufficient. Chemical fertilizers do not provide all the elements required for plant growth in proper amount and proportion.

Excessive fertigation and the use of pesticides kill the microorganisms in the soil. Thus the natural fertility of the soil loses.

The chemical substances present in chemical fertilizers get accumulated in agricultural crops (through biological magnification). It causes many chronic diseases in organisms including man who consume this food.

Chemical fertilizers and pesticides are expensive. Fertilizers are manufactured by small scale industries as well, as multinational companies. The expenses is not affordable by the farmers. Thus unscientific application of fertilizers causes consequences of varied dimensions.

Hsslive Guru 8th Class Biology Kerala Syllabus Pest Control

Many methods are adapted today to control pests. Chemical pesticides are widely used. Chemical pesticides kill the pest as a whole. Use of chemical fertilizers cause a number environmental and health problems. High amount of chemical fertilizers are reported in ground water too.
Another possibility is to control pests using ultrasonic sound waves.
Using radiation the reproductive capacity of male pests can be lost and makes pest control effective.
Pheromone traps like devices also makes pest control effective. Pest control using the natural enemies of pest is highly effective and having no environment and health consequences.

Natural Enemies of Pest (Friendly Pests)

Organisms that eat pests, cause disease to them or parasitises on them are called their natural enemies.

8th Std Biology Notes Kerala Syllabus Integrated Pest Management

In this method, the use of chemical pesticides is highly reduced. Pest control is made possible by the combined use of biological pesticides, friendly pests, mechanical control methods, etc. It will not disturb the equilibrium of the ecosystem.
It do not destroy pests as a whole. But it prevents the multiplication of pests and the number of pest is cortrolled to prevent crop loss.

Advantages of Integrated Pest Management

  • Do not kill the pests as a whole
  • Use of chemical pesticides is highly reduced.
  • No environmental or health issues as friendly pests, mechanical control measures, Biological pesticides, etc. are used.
  • Does not disturb the equilibrium of ecosystem.

Waste Management

Indicators (Text Book Page No:41)

Hss Live Guru 8 Physics Kerala Syllabus Question 12.
Waste management and sustainable agriculture
Answer:
Live Stock management, Poultry farming, Pisciculture etc, help not only to earn income but also to the treatment of biological wastes. Composting, biogas production etc, are possible by using bio wastes. Cow dung is a very good biological manure and an essential component . for the production of biogas. By preparing cattle feed, fish food, poultry feed, etc., from bio waste more earnings can be done and biological waste management also possible.
Let's Regain Our Fields Kerala Syllabus 8th

Hsslive Guru 8th Basic Science Kerala Syllabus Certain methods of Agriculture

Many farming techniques that help to earn improved income by scientific approach are in practice.
eg: Rearing of cattle’s, Poultry farming, Sericulture, Pisciculture, Floriculture, Apiculture, Cuniculture, Mushroom culture, Horticulture, Medicinal plant cuitivation etc.
→ Live stock Management
Rearing of cattle’s for milk, meat and agricultural purposes.
→ Poultry farming
Rearing of birds for egg, meat etc.
→ Sericulture
Rearing of silk worms
→ Pisciculture
Rearing of fishes
→ Floriculture
Cultivation of flowers for commercial purposes.
→ Apiculture
Rearing of honeybees
→ Cuniculture
Rearing of rabbits
→ Mushroom culture
Cultivation of mushrooms .
→ Horti culture
Cultivation of fruits and vegetables.

Completion of Table(Text Book Page No:44)

Hss Live Guru Biology 8 Kerala Syllabus Question 13.
Complete the following table related to various agricultural sectors.
Lets Regain Our Fields Notes Kerala Syllabus 8th

Answer:

AreasProductsVarieties
Pisciculturefish, fish liver oilPearl spot, Rohu
ApicultureHoney, WaxKolan, Mellifera Njodiyan
Mushroom cultureEdible mushroomMilk mushroom, Button mushroom
Livestock managementMilk, Meatjercy(Cow) Murrah (Buffalo) Jamnapari (Goat)
CunicultureMeatGrey giant, White giant
SericultureSilk threadMulberry silkworm Tussar silkworm Muga silkworm
Poultry farmingEgg, MeatAthulyaf(hen) Muscovy (duck) Bobwhite (Quail)

Modern Techniques of Agriculture
1. Polyhouse farming
Polyhouse is a special arrangement made by completely or partially covering transparent sheet like polythene. Humidity and temperature kept constant in polyhouse. So plant growth will be fast. Nutrients are dissolved in water and sprayed. Pest attack also will be less.

2. Open Precision farming
This is a technique in which the nature of soil in the agricultural land, quantity of elements in the soil, presence of water in the soil, etc., are studied accurately with the help of modern technology and suitable crops are selected for cultivation. The attack of weeds also is less because the soil is covered using polythene sheets. Water supply or irrigation can be limited.

3. Hydroponics
Plants are grown in nutrient solution.
4. Aeroponics
Plants are grown in such a way that their roots are grown towards air and nutrients are directly sprayed to the roots.

Indicators (Text Book Page No:46)

Hss Live Guru 8th Basic Science Kerala Syllabus Question 14.
How are modern agricultural practices helpful in reducing crop loss due to climate change?
Answer:
In arrangements like polyhouses temperature and humidity are maintained constant. Plant grows fast and gets better yield. Polyhouse farming helps to reduce crop less due to climatic changes.

Basic Science Class 8 Chapter 3 Kerala Syllabus Question 15.
What are the advantages of precision farming?
Answer:
This is a technique in which the nature of soil in the agricultural land, quantity of elements in the soil, presence of water in the soil etc., are studied accurately with the help of modem technology and suitable crops are selected for cultivation. The attack of weeds also is less because the soil is covered using polythene sheets. Water supply or irrigation can be limited.

Class 8 Science Notes Kerala Syllabus Question 16.
How does cultivation become possible without depending on soil?
Answer:
• Hydroponics and aeroponics are soilless cultivation methods.
• Plants are grown in nutrient solution.
• Aeroponic plants are grown in such a way that their roots are grown towards air and nutrients are directly sprayed to the roots.

Farmers groups

Nowadays farmers groups, that provide farmers an opportunity to sell and buy their products without brokers mediators are very lively. Online gathering of farmers also is widespread as the demand for organic product raised. These online groups help to find customers for the quality organic products and to get good price for the products.

Completion of Table(Text Book Page No:51)

Question 17.
Complete the following table, adding important ideas.
Lets Regain Our Fields Notes Pdf Kerala Syllabus 8th

Answer:

ChallengesRemedies
Climatic change• Polyhouse farming
• Hydroponics
Environmental distruction and health hazards• Scientific manuring
• Integrated Pest manage ment. Biological waste Management
Cost of production• Microbial fertilizers
• Integrated farming
Crop loss• Polyhouse farming
• Open precision  farming
• Integrated Pest management
Limited space• Hydroponics
• Aeroponics
• Terrace farming Loss
Loss• Farmers societies

• Online gatherings

Let’s Regain Our Fields Let us Assess (Text Book Page No:52)

Question 1.
Cuniculture is related to
a. Keeping of honey bees
b. Rearing of rabbits
c. Cultivation of fruits and vegetables
d. Rearing of fish
Answer:
Rearing of Rabbits

Question 2.
High quality hybrid varieties provide high yield. Then, what is the need of native varieties? Record your response to this statement.
Answer:
The native varieties exist in a particular locality by acquiring natural resistance and adaptations to the climate, availability of food, nature of soil etc., though they have low productivity.
The native varieties have high resis-tance to diseases and environmental conditions. The extinction of native varieties leads to the depletion of our biodiversity. New varieties can be developed only from the native varieties.

 

Question 3.
Which is the most appropriate way to reduce crop loss due to pests?
a. Using high concentration pesticides
b. Protecting friendly pests.
c. Practicing integrated pest management
d. Applying organic pesticides only.
Answer:
C  Adopt integrated Pest control

Question 4.
‘Lower price during higher yield’. Suggest a practical solution to overcome this crisis faced by farmers.
Answer:
Collect and distribute resources through farmers societies.

Let’s Regain Our Fields Additional Questions and Answers

Question 1.
Find the odd one in each group.
Also write the common characterestic of the others.
a. Compost, Microbial fertilizer, urea, Bone-meal
b. Anthurium, Sida, Ramacham, Koovalam
c. Boer, Litchi, Rambutan, Durian
d. Phosphorus, Potassium, Nitrogen, Azetohactor
e. Rock Phosphate, Factompho- se, Muriate of Potash, Malatheon,
f. Kuthiravaly, Jyothy, Thriveni, Jaya
g. Fowl Cholera, Anthrax, Chores, Ranikatt
h. Blight disease, Root wilt, Bud rot
i. Ammonia, Urea, Compost, Factomphose.
j. APIS (njodiyan); Naaran, Melliferra.
k. Minorka, Royans, Ankona, Gramalakshmi.
Answer:
a.Urea – This is a chemical fertilizer. Others are Bio-fertilizers.
b. Anthurium – This is an ornamental plant. Others are medicinal plants.
c. Boer – It is a variety of goat. Others are fruit varieties.
d. Azetohactor – Others are essential elements.
e. Malatheon – This is a chemical pes ticide others are chemical fertilizers.
f. Kuthiravaly – This a hybrid vari ety pepper plant others are vari eties of paddy.
g. Anthrax : This is a cattle disease others are fowl diseases.
h. Blight disease, It is affect on rice plant others are diseases of coconut tree.
i. Compost – This is an organic fertiliser, others are chemical fertilisers
j. Naaran. This is prawn.Others are Honey bees used in apiculture.
k. Royans is good quality ducks, Others are fowl varieties.

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. Rearing of honey bees : Apiculture; Rearing of silkworms : …………..
b. Malabari : Goat; Vechoor : …………..
c. Cowdung: ………….. ; Azospirillum : Microbial Fertilizer
d. Jersey: Cow; Murrah: …………..
e. Mushroom culture : Growing mushroom; Cuniculture : …………..
f. Malatheon : Chemical pesticide; Neem seed kemal: …………..
g. ………….. – Virus; Anthrax Bacteria.
h. Native variety : vechoor; Hybrid variety : …………..
i. Paddy- ………….. ; Arecanut plant : Mahali
j. Cocount caterpillar – ………….. ; Gamboosia- Larvae of mosqitoes
k. …………..: Domestic hen; Pekkins : Domestic Duck.
l. Blight disease- Paddy; Quick wilt- …………..
m. Organic fertiliser – Vermiw- ash; ………….. – Factomphose.
Answer:
a. Sericulture
b. Cow
c. Organic fertilizer
d. Buffalo
e. Rabbit farming
f. Organic pesticide
g. Foot and Mouth disease
h. Jersey cross
i. Blast disease
j. Ichneumon wasp
k. Royans is good quality ducks, Others are fowl varieties.
l. pepper.
m. chemical fertilisers

Question 3.
Arrange the following items from column B ,C with column A.

A
Agricultural sectors
B
Speciality
C
Varieties
a. Horticulturei. scientific way of rea­ring of honey1. Ankora
b. Apicultureii. scientific way of rearing of rabbits2. Litchi
c. Cunicultureiii. scientific way of rearing silkworm3. Njodiyan
iv. scientific cultiva­tion of fruits and4. Naran

Answer:

A
Agricultural sectors
B
Speciality
C
Varieties
Horticulturescientific way of rea­ring of honeyLitchi
Apiculturescientific way of rearing of rabbitsNjodiyan
Cuniculturescientific way of rearing silkwormAnkora

Question 4.
Group the following statements into suitable for Polyhouse farming and Precision farming.
a. limiting the irrigation by covering soil with polyethene sheet.
b. Cover the field completely or partially by transparent polyethene sheet.
c. By regulating temperature and moisture constantly.
d. Selecting appropriate crop for agriculture only jifter understanding characters of soil, amount of elements present in soil and the presence of water.
Let's Regain Our Fields Pdf Kerala Syllabus 8th
Answer:

  • polyhouse farming – (b), (c)
  • Precision farming – (a), (d)

Question 5.
“It is essential to retain indegenious species of mango tree like Muvandan, Kilichundan even many hybrid varieties are avail-able”. Write your opinion to the farmer’s statement? Justify your answer?
Answer:

  • Farmer’s opinion is right.
  • Indigenous varieties of a locality are varieties that acquire natural immunity by adapting to the climate, the availability of food, soil texture of the place etc.
  • We can develop new high quality varieties only from indigenous varieties.

Question 6.
Find suitable term and fill the blanks;
Moovandan – Mango Tree Kasaragodu Dwaf – a. …………..
Njalippoovan – Musa Attappadi black – b. …………….
Answer:
a. Cow
b. Goat

Question 7.
Sustainable farming is an environment friendly method. Explain the reason.
Answer:
The excessive use of chemical fertilizers and pesticides may give increased profit. But this will not last long. The continuous use may spoil the natural fertility of the soil and the farmland may be changed into a barren land. By integrated cropping method, the use of outside manures, pesticides etc can be reduced. The wastes of one can be used as a manure or food for some other one. This will help to maintain the natural fertility of the soil. Moreover, the biodiversity also is conserved.

Question 8.
The practice of cultivating fruits and vegetables.
Answer:
Horticulture

Question 9.
Some statements regarding modern agricultural practices are given below. Which agricultural practice is related to this?
a. The method of farming in which nutrients are dissolved in water and are supplied on plants through dip irrigation.
b. The method plants are grown in nutrients solution
c. The method of farming by covering the rool using polythene sheets and by limiting.
Answer:
a. Polyhouse farming
b. hydroponics
c. precision farming

Question 10.
Polyhouse farming will be advantageous only in farmlands where cultivation is continuously maintained. Why?
Answer:
The cost of making a polyhouse is very high. But the yield from the crops will increase substantially.

Question 11.
Which of the following is not desirable in integrated pest control method?
a. Mechanical Pest Control
b. Excessive use of chemical pesticide
c. Friendly pests
d. Use of biopesticides
Answer:
b. Excessive use of chemical pesticide

Question 12.
What is the difference between polyhouse farming and open precision farming
Answer:
Poly house farming

  • Agricultural land is partially or completely covered by transparent polythene sheet.
  • Heat and humidity are kept constant
  • Nutrients are dissolved in water and given to plants through drip fertigation. Pest attack is compara-tively low.

Open Precision farming

  • Soil is covered by polythene sheet
  • Nature of soil, amount of elements in soil, pH of soil, presence of water etc are studied accurately using modern technology
  • Limited irrigation is needed.
  • Weed control is effective

Question 13.
What are the main characteristics of hybrid varieties?
Answer:
High yield, disease resisting capacity ability to give high yield within a short period etc. are the characteristics of hybrid varieties.

Question 14.
What are the consequences of unscientific application of chemical fertilizers?
Answer:

  • Financial loss.
  • Chemical pollution.
  • Destruction of microorganisms
  • Health issues.

Question 15.
How can grow plants without soil?
Answer:
Science has proved that cultivation is possible in the absence of soil, for example aeroponics and hydroponics. In hydroponics, plants are grow in nutrients solution. In aeroponics, plants are grow in such a way that their roots grow into air and nutrients are sprayed directly on roots.

Question 16.
What are the different varieties of buffaloes?
Answer:
Bhadawari, Jaffrabadi, Marwari, pashmina, malabari, Beetance.

Question 17.
Find out the odd one. Write the common feature of others. Rhizobium, Azetobacter, Lacto bacillus, Azospirillum
Answer:
Lactobacillus – others are nitrogen-fixing bacteria.

Question 18.
Arrange the organisms properly in the given table. Names are mentioned in the box.
Class 8 Biology Chapter 3 Kerala Syllabus
Answer:
a. Jersey, Murrah
b. Athulya, Muscovy
c. Naran, Kara
d. Grey giant, Ankora
e. Muga, Tusser
f. Mellifera, Kolan

Question 19.
What is the difference between hydroponics and aeroponics?
Answer:

  • Hydroponics is the technique by which plants are grown in nutrient solution
  • In aeroponics plants are grown in such a way that their roots are penetrating towards air and nutrients are directly sprayed to their roots.

Question 20.
What are the advantages of precision farming?
Answer:
In this method of farming the nature of soil, quality of elements in the soil, pH value of soil, presence of water etc, in the crop field are tested using modem technology and appropriate crops are selected for cultivation.

Question 21.
What are the advantage of house farming and family farming
Answer:
a. Nontoxic food
b. Maximum utilization of land/space
c. Exercise
d. Mental Pleasure
e. Collaborative work of family members

Question 22.
How do microbial fertilizers help in plant growth?
Answer:
Microbial fertilizers contains micro organisms that increase the fertility of soil. They increase the amount of growth promoting factors in soil.
Eg: Rhizobium Azetobacter raise the amount of Nitrogen in the soil.

Question 23.
Which are the foreign varieties of fowl reared in our place?
Answer:
White leghorn, Rhode Island Red, Plymouth Rock, New Hampshire

Question 24.
What are the diseases of fowls
Answer:
Ranikatt, Fowl Cholera, Salmonellosis, Diarrhoea, Chores (Bacteria) Aspergillusis (Fungus).

Kerala SSLC Biology Chapter 7 Questions Genetics for the Future

Kerala SSLC Biology Chapter 7 Questions Genetics for the Future English Medium

Sslc Biology Chapter Wise Questions And Answers Question 1.
DNA fingerprinting was discovered by
a. Watson
b. Crick
c. Mendel
d. Alec Jeffrey
Answer:
d. Alec Jeffrey

Question 2.
Fill up the blanks by observing the relationship between the first pair
Molecular scissors: Restriction endonuclease
Molecular glue: ……………………….
Answer:
Ligase

Question 3.
What is the significance of human genome project?
Answer:
Even though science has progressed a lot, we couldn’t control genetic diseases. The reason for this is that we could not identify the exact gene and its location which is responsible for each trait. In 1990, the Human Genome Project was started as an attempt to solve the issue.

As a result of experiments conducted in various laboratories of the world until 2003, the secrets of human genome were revealed. With this we can identify the location of a gene in the DNA, responsible for a particular trait.

Question 4.
The stages in the production of bacteria liar are capable of producing insulin through genetic engineering is given below. Arrange them in correct order
a. Cuttiiig the plasmid of bacteria.
b. Inserting DNA in the bacterial cell.
c. Bacteria produce in active form of insulin.
d. Cutting the gene responsible for the production of insulin from human DNA.
e. Producing active insulin.
f. Providing a favourable medium for the multiplication of bacteria.
g. Joining the gene responsible for the production of insulin from human DNA to the plasmid of bacteria.
Answer:
d, a, g, b, f, e, e

Class 10 Biology Chapter 7 Question Answer Question 5.
Genetic engineering a threat to mankind. Write your opinion about this statement.
Answer:
It is criticized that genetically modified varieties cause harm to indigenous varieties and may cause health issues to humans. Application of genetically modified pathogens and pathogens multiplied ” through biotechnology upon enemies is called Bioware. This becomes a threat to the existence of human beings.

Question 6.
Bacteria is not used as a pharm animal. Write your opinion.
Answer:
Genes responsible for the production of human insulin, growth hormones, etc., are identified and inserted in animals like cow, pig, etc., transforming them into pharm animals. Bacteria were used for such experiments in the beginning. However, the culturing of bacteria is not so easy. They should be cultured with utmost care in the laboratory, It is easy to rear animals.

Question 7.
“Genetic engineering is the branch of science that transforms the living world’’.
a. What is your opinion on the above statement?
b. Give reason to substantiate your opinion.
Answer:
a. I agree with this statement. Genetic engineering is the technology of controlling traits of organisms by bringing about desirable changes in the genetic constitution of organisms.

b. Medicine – Plants and animals that produce medicines
Gene therapy – Remedy for genetic diseases
DNA fingerprinting – Forensic test

Biology Chapter 7 Class 10 Question 8.
Arrange the statements suitably in the table given below.
a. The enzyme used to cut genes
b. Ligase
c. The enzyme used for joining genes
d. Restriction endonuclease
Answer:

Genetic scissorsGenetic glue
……………………..
……………………..
……………………..
……………………..
Genetic scissorsGenetic glue
The enzyme used to cut genesLigase
Restriction endonucleaseThe enzyme used for joining genes

Question 9.
The stages in the production of insulin from bacteria are given below. Complete the missing stages.
Kerala SSLC Biology Chapter 7 Questions Genetics for the Future 2

Answer:
a. Cutting bacterial DNA (plasmid).
b. Then insert this DNA in the bacterial cell.
c. Bacteria produce inactive form of insulin.

Question 10.
Observe the news and answer the question given below.
DNA fingerprinting helped to identify real culprit.
Kochi: Culprit can be identified through DNA testing.
a. What is the basis of DNA testing?
b. How can culprits be identified through DNA testing.
Answer:
a. The arrangement of nucleotides in the DNA of each person differs. This finding leads to the DNA testing.

b. DNA of the skin, hair, nail, blood and other body fluids obtained from the place of murder, robbery etc. is compared with the DNA of suspected persons. Thus the real culprit can be identified from among the suspected persons through this method.

Biology Chapter 7 Question 11.
Ligase is referred as molecular glue and restriction endonuclease as molecular scissors. If these are utilized, great many achievements can be produced in the fields of agriculture and medicine.
a. What is the technology mentioned above?
b. What is the use of ligase and restriction endonuclease in this technology?
Answer:
a. Genetic engineering
b. Ligase – used for joining genes Restriction endonuclease – used for cutting of genes

Question 12.
Observe the logo given below.
Kerala SSLC Biology Chapter 7 Questions Genetics for the Future 1
a. What does it indicate?
b. What are the findings of this project?
Answer:
a. Human Genome Project.
b. Human genome has about 24000 functional genes.

  • Major share of human DNA includes junk genes.
  • There is only 0.2 percent difference in DNA among humans.
  • About 200 genes in human genome are identical to those in bacteria.

Class 10 Biology Chapter 7 Question 13.
Given below is a word tree prepared by Appu for classroom presentation. Help him to complete the tree by choosing the words given in the box:
Junk genes, Ligase, Gene therapy, DNA profiling, Restriction endonuclease, Gene mapping, Plasmid, Genetic engineering
Kerala SSLC Biology Chapter 7 Questions Genetics for the Future 2
Answer:
a. Restriction endonuclease
b. Ligase
c. DNA profiling
d. Junk genes
e. Gene mapping
f. Gene therapy

Question 14.
Observe the table and form matching pairs.

a. DNA profilingi. Treatment for genetic diseases.
b. Gene mappingii. Testing the arrangement of nucleotides.
c. Gene therapyiii. The sum of genetic material present in an organism
d. Genomeiv. Locating the position of a gene in the DNA.

Answer:

a. DNA profilingi. Testing the arrangement of nucleotides.
b. Gene mappingii. Locating the position of a gene in the DNA.
c. Gene therapyiii. Treatment for genetic diseases.
d. Genomeiv. The sum of genetic material present in an organism.

Sslc Biology Chapter 7 Notes Pdf Question 15.
Complete the table.

Protein required for treatmentDisease/Symptom
………………a………………Viral diseases
Insulin………………b………………
………………c………………Pain
Somatotropin………………d………………

Answer:

Protein required for treatmentDisease/Symptom
InterferonsViral diseases
InsulinDiabetes
EndorphinPain
SomatotropinGrowth disorders

Kerala SSLC Biology Chapter Wise Questions and Answers

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters

You can Download Cell Clusters Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters

Cell Clusters Questions and Answers

Diversity among cells

Each part of human body is composed of different kinds of cells. These cells do not act independently. But they act in groups of similar cells. These cell clusters are called tissues.

Cell Clusters Class 8 Kerala Syllabus

Cell Clusters Class 8 Kerala Syllabus Tissues

The group of cells with common origin and perform a specific function are called tissues. The coordinated action of tissue, help to perform various physiological functions effectively.

Cell Differentiation

The single celled zygote divides continuously and forms the foetus that consists of cells differ in shape, size and contents. Foetal cells gradually attain change in structure and function. This process is called cell differentiation.

Indicators (Text Book Page No:24)

Cell Clusters Class 8 Notes Pdf Kerala Syllabus Question 1.
Formation of foetus.
Answer:
The single celled zygote formed by the fusion of sperm and ovum continuously divides and the foetus is formed.

8th Class Biology Notes Pdf Kerala Syllabus Question 2.
Significance of cell differentiation.
Answer:
Foetal cells gradually attain change in structure and function due to cell differentiation. Thus, different kinds of cells and tissues are formed. It helps to make the life processes effective.

Kerala Syllabus 8th Standard Biology Notes Stem cells

Stem cells are specialised cells that can be transformed into any kind of cells.

Stem cells can either transform into other cells through division or they can exist as such. When the cells in the tissues, get destructed, new cells emerge from stem cells.

It is expected that the research of stem cells may cause drastic change in the treatment of blood cancer, diabetes, Parkinson’s disease etc as desired cells are formed from stem cells.

Indicators (Text Book Page No:25)

Class 8 Science Notes Kerala Syllabus Question 3.
What are the peculiarities of stem cells when compared to other cells?
Answer:
Stem cells are specialized cells that can be transformed into any kind of cell. They can either transform into other cells or can retain their existence. When the cells in the tissues get destroyed new cells originate from stem cells.

Cell Clusters Class 8 Questions And Answers Pdf Kerala Syllabus Question 4.
How is the destruction of cells in tissues compensated?
Answer:
When the cells in the tissues get destroyed, new cells originate from cells. Stem cells are found in bone marrow, skin, digestive fact etc.

Cell Clusters Class 8 Notes Kerala Syllabus Question 5.
Why is stem cell research gaining importance?
Answer:
Nowadays desired cells can be produced from stem cells in research laboratories. The research of stem cells can cause a drastic change in the treatment of blood cancer, diabetes, Parkinson’s diseases etc, and in the production of artificial organs.

Animal Tissues

Animal tissues are mainly classified into 4
(a) Epithelial tissue
(b) Nervous tissue
(c) Muscular tissue
(d) Connective tissue

1. Epithelial Tissue:- Covers and protects the body.
Covers the inner lining of the alimentary canal Performs the functions like protection, absorption, production of secretions, etc.
2. Nervous Tissue:- Controls and Co-ordinates physiological Activities Helps to respond to particular stimuli
3. Muscular Tissue:- Makes the movement of the body possible.
4. Connective Tissue:-

  • Connects different tissues.
  • Provide support and strength to the body
  • Performs the functions like material transport, defence, etc.

Meristematic tissues Meristematic cells are specialized cells seen at the tip of the stem and root in plants. They divide rapidly and helps in the growth of plants.

Table (Text Book Page No:27)

Compare the figures of a meristematic cell and a mature cell. Find out the differences and complete the table.
Cell Clusters Class 8 Notes Pdf Kerala Syllabus
Answer:

CharacteristicsMeristematic CellsMature plant Cells
Relative size of Nucleus(a) ……………….(b) ……………
Thickness of cell wall(c) ……………….(d) ……………
Quantity of Cytoplasm(e) ……………….(f) ……………

Basic Science Class 8 Chapter 2 Kerala Syllabus Various Plant Tissues

Parenchyma, Collenchyma, and Sclerenchyma are the main plant tissues. They differ in structure and perform different functions.

The specialised tissues in plants that conduct water and salts that are absorbed by the roots to the leaves and the food prepared in the leaves to various parts, are called vascular tissues. These are called complex tissues as they are. formed by different types of cells. Xylem and Phloem are the main vascular tissues.

Work sheet (Text Book Page No:30)

8th Class Biology Notes Pdf Kerala Syllabus

Answer:
Kerala Syllabus 8th Standard Biology Notes

  • Organs like stomach and Intestine are mainly formed by nervous tissue, epithelial tissue, connective tissue, and muscular tissue.
  • Temporary storage of food materials, digestion and the secretion of digestive juices.
  • Complete digestion of food materials. Secretion of-digestive juices, reabsorption of water, absorption of nutrients.
  • Organ performs the co-ordinated functions of various tissues. The function of an organ is not the same as that of the individual tissues which constitute the organ.
  • Various organs combine to form organ system. As a result physiological functions can be performed very effectively.
    eg: teeth, tongue, glands, oesophagus, stomach, small intestine, large intestine etc constitute digestive system. It enables the complete digestion of food.

Table (Text Book Page No:31)
Complete the following table by finding out the systems to which the organs listed in the table belong to:
Class 8 Science Notes Kerala Syllabus
Answer:

OrgansOrgan System
Heart, Blood vesselsCirculatory System
Nose, Trachea, Lungs.Respiratory System
Kidney, Ureter, Urinary bladderExcretory System
Brain, NervesNervous System

An organism is composed of various organ systems. Therefore the structure of organisms is highly complicated.

Cells are the basic unit of life. Cell parts are composed of different substances.

In higher organisms diverse tissues act complementary to each other and perform various life processes.

Cell Clusters Let us assess (Text Book Page No:33)

Given below in the illustration are various tissues related to the structure of hand.
Cell Clusters Class 8 Questions And Answers Pdf Kerala Syllabus
Answer:
1. • A
• C
2. Formed from various cells
3. Only the corners of the cells are thicked

Cell Clusters Text Book Questions and Answers

Question 1.
Arrange an exhibition showing pictures and descriptions of different types of cells.

Answer:
Cell
Cell is the structural and functional unit of life.
• All organisms are formed of one or more cells.
• All vital activities of life take place inside the cells.
• Here dietary information that controls the functions of the cells are contained in the cell.
• There are three types of cells in the blood. Red blood cells, white blood cells and platelets.

Red Blood cells

Cell Clusters Class 8 Notes Kerala Syllabus

Red blood cells are the most abundant cells in the blood. They are also known as erythrocytes. Invertebrates, RBC carries oxygen to the tissues. In RBC, the cytoplasm contains an iron containing biomolecule, the haemoglobin. If in parts red colour to the blood. In human, RBCs has biconcave disc shape.

White blood cells

Basic Science Class 8 Chapter 2 Kerala Syllabus
WBC’S are also called leucocytes, they protect the body from pathogens and other antigens as a part of the immune system. They are formed from the stem cells of the bone marrow. The number of WBCs is an indicator of health. They are classified into neutrophil, eosinophil, basophil, monocyte and lymphocyte according to the diversity of cytoplasm, nucleus etc.

Platelets

Basic Science For Class 8 Chapter 2 Kerala Syllabus
They are also known as thrombocytes. They help in the coagulation of blood. They are formed from the cells called megakaryocytes.

Question 2.
Prepare a magazine specifying the importance, relevance, and scope of stem cells.
Answer:
Stem cell

Stem cell are specified cells that can transform to any kind of cell. They are also known as root cells. They transform to other cells by a long process of differentiation.

Stem cells can either transform to other cells or exist as such. When the cell in the tissues get destructed, new cells originate from stem cells. Stem cells are seen in bone marrow, skin, digestive tract etc.

Today it is possible to develop desired cells from stem cells under specific conditions of laboratories. It is expected to make tremendous change in the treatment of blood cancer, diabetes, parkinson’s diseases etc. and in the development of artificial organs through the research works of in stem cells.

Cell Clusters Additional Questions and Answers

Question 1.
Findi the odd one in each group.
Also write the common characterestic of the others.
a. Collenchyma, Sclerenchyma Xylem, Nervous tissue.
b. Blood, Muscle, ligament, Bone
c. Basophil, Platelet, Neutrophil, Lymphocytes
d. Collenchyma, Aerenchyma, Parenchyma, Chlorenchyma
e. Intercalary meristem, Primary meristem, Apical meristem, Lateral meristem
f. Xylem, Collenchyma, Scler-enchyma, Parenchyma.
g. Man, Pigeon, Duck, Amoeba
h. Large intestine, Digestive tract, Heart, Stomach
i. Bony tissue, Muscular tissue, Cartilaginous tissue, Blood
j. Parenchyma, Collenchyma, Epithelial tissue, Sclerenchyma.
Answer:
a.Nervous tissue; Others are plant tissues.
b. Muscle, Others are connective tissues.
c. Platelets, Others are white blood cells.
d. Collenchyma, Others are the different forms of parenchyma.
e. Primary meristem : Others are divide and new tissues. Primary meristem divide and from apical meristem and lateral meristem.
f. Xylem : It is conduction tissue and others are basic tissues.
g. Amoeba – Cell level organism and
others are organ system level organisms.
h. Heart : It is a circulatory organ. Others are digestive organ.
i. Muscular tissue : The others are connective tissues
j. Epithelial tissue : The others are plant tissues.

Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. Water and salts : xylem :: ………… : Phloem
b. Photosynthesis : Parenchyma : Support and Strength : ………..
c. Between bones and muscles : tendons ; Between bones and bone: ………..
d. Blood : Connective tissue : : Skin: ……….. ;
e. Control and Co-ordination : Nervous tissue : : Body Move ment:: ………..
f. Fibrous tissue : Connectother tissue; ……….. : Transport of materials
g. Storage of food : ……….. ; Flexibility to plant parts : Col-lenchyma
h. White blood cells : ……….. ; Redblood cell; O2 transportation
i. Xylem – ………..; Phloem: Transports synthsised food
j. Intercalary meristem : Inter node lengthening; Lateral meristem: ………..
k. Aerenchyma : Air cavities ; Chlorenchyma : ………..
l. Cell :Tissue; system : ………..
Answer:
a. Food
b. Sclerenchyma
c. Ligaments
d. Epithelial Tissue
e. Muscular Tissue
f. Blood
g. Parenchyma
h. Immunity
i. Transports water and minerals
j. Thickening of stem
k. Chlorophyll
l. Organ

Question 3.
Complete the flowchart which shows levels of organisation of human being.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 10
Answer:
a. cell organelle
b. Organ

Question 4.
Observe the following figure and answer the given questions.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 20
a. Find out correct one from the bracket shows the plant tissue given in the figure.
A. parenchyma
B. sclerenchyma
C. xylem
D. collenchyma
b. Write down the characteristic features of the plant cell in the given figure?
c. Give scientific explanation for the features of plant part are given below.
i. Desterity of shaft of colacasia.
ii. Hardness of shell
Answer:
a. B. sclerenchyma
b. composed of cells that are uniformly thick all over the cell wall.
c. i. presence of collenchyma cells, ii. presence of sclerenchyma cells.

Question 5.
Find out wrong items from the following statements, and also correct their underlined words.
a. Sclerenchyma can provides strength and support to plant parts.
b. Fibrous tissue can enables to respond by identifying the changes inside and outside the body.
c. Production of secretion is one of the function of Epithelial tissue.
d. muscular tissue enables the movement of the body.
Answer:

  • b, d are wrong statements.
  • b. Nervous tissue can enables to respond identifying the changes inside and outside the body.
  • d. Muscular tissue enables the movement of the body.

Question 6.
What is the difference between blood and blood circulatory system?
Answer:
Blood is a tissue in the blood circulatory system. But blood vessels, heart, nerves, etc. are the other factors of the blood circulatory system. The combined work of all these factors form the blood circulatory system.

Question 7.
All parts of the plant do not grow. Why?
Answer:
In plants, only the meristematic cells have the ability to divide and grow. Meristematic cells are mainly found in the root apex and stem apex. So the plant growth is concentrated at the root apex and stem apex.

Question 8.
What are tissues? Give examples.
Answer:
A group of similar cells is called tissue
eg : Parenchyma

Question 9.
“Body is composed of millions of cells. Are they perform the same function? How do they act independently?”
This is question raised in the class by the teacher for a discussion. What will be your answer to this question?
Answer:
Cells act in groups of similar kind. Group of similar cells that are originated from a single cell and perform a specific function are called tissues. They perform different functions.
Eg: muscular tissue, Nervous tissue, Epithelial tissue.

Question 10.
Our body is developed from a single-celled zygote. How do this much organs and organ system sare formed in our body?
Answer:
Zygote divides continuously and forms the foetus that consists of cells differ in shape, size, and content. Foetal cells gradually attain diversity in structure and function. Thus different organs and organ systems are formed.

Question 11.

Stem cells for Cancer treatment
Tvm:- Scientists say that the pr­oduction of tissues from stem cells may cause drastic change in the treatment of Cancer…

Did you read the news report? What are stem cells? How do they become useful in the treatment of Cancer?
Answer:
Stem cells are specialized cells that can be transformed to any kind of cell. Stemcells modify to other cells through prolonged differentiation. When cells destroyed, new cells originates from stem cells. In research laboratories desired cells are formed from stem cells. New cells, instead of cancer cells, can be produced from stem cells.

Question 12.
Complete the given flow chart
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 11
Answer:
a. Nervous Tissue
b. Connective Tissue
c. • Covers and protects the body
• The inner lining of alimentary ‘ canal
• Perform the function of protection, absorption, production of secretion, etc.
d. helps to respond by recognizing the changes within and out of the body
e. Makes the movement of body possible.

Question 13.
Categorise the given statements under suitable headings.
1. Provide shape to body.
2. Defence
3. Covers and protects internal organs.
4. Conduction of respiratory gases.
Answer:

BloodBone, Cartilage (Connective Tissue)
2. Defence1. Provide shape to body.
4. Conduction of respiratory gases.3. Covers and protects internal organs.

Question 14.
Identify the figures given below.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 12
a. Nervous tissue
b. Muscular tissue

Question 15.
What are the differences between the process of growth in plants and animals?
Answer:
Plant growth

  • Plants grow throughout their life.
  • In plants, growth is mainly restricted at the tips of stem and root(meristem)

Animal growth

  • Animals grow upto a particular period
  • In animals, growth is not confined to any specific regions.

Question 16.
Why plant growth is confined to specific parts?
Answer:
In plants, specific cells called meristematic cells are concentrated the tips of roots and stems. They have the ability to divide rapidly.

Question 17.
Complete the table properly

CharacteristicsMeristematic CellsMature Cells
Relative size of Nucleus(a) ……………….(b) ……………
Thickness of cell wall(c) ……………….(d) ……………
Quantity of Cytoplasm(e) ……………….(f) ……………

Answer:
a. Comparatively large nucleus
b. Small nucleus
c. Thin cell wall
d. Thick cell wall
e. More
f. Comparatively less

Question 18.
How do the water and salts absorbed by the roots reach the leaves?
Answer:
In plants, the water and salts absorbed by the roots are transported to the leaves through specialised tissues called vascular tissues.

Question 19.
Vascular tissues are called complex tissues. Why?
Answer:
Vascular tissues are formed by the union of different types of cells. Hence they are called complex tissues.

Question 20.
Picture of section of stem is given. Label the parts.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 13
Answer:
a. Collenchyma
b. Parenchyma
c. Sclerenchyma
d. Phloem
e. Xylem

Question 21.
Pair the given cells with their specific character

Parenchyma, Collenchyma,  Sclerenchyma.

a. formed of cells whose cell wall is uniformly thickened in all parts.
b. seen in tender parts of plant.
c. formed of cells whose cell wall has thickenings in the comers only
Answer:
parenchyma – b
Collenchyma –
c sclerenchyma – a

Question 22.
Write examples for vascular tissues?
Answer:
Xylem and Phloem

Question 23.
Arrange the given statements in the right column.
1. Formed of inter related cells seen as ducts.
2. Carries water and salts absorbed by the roots to the leaves.
3. Carries the food prepared in the leaves to various plant parts.
4. Provide support and strength to plant
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 15
Answer:
Xylem 2, 4 Phloem -1, 3

Question 24.
Complete the worksheet based on the hints given.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 16
Hints
A. Carries the food prepared in the leaves to various plant parts.
B. Carries water and salts absorbed by the roots to the leaves.
C. Seen in the tender parts of the leaves.
D. Thickenings seen only in the corners of the cell wall
E. Cells have uniform thickenings in the cell wall.
Answer:
A – Phloem
B – Xylem
C – Parenchyma
D – Collenchyma
E – Sclerenchyma

Question 25.
Which are the tissues that constitute the organs like stomach and intestine?
Answer:
Nervous tissue, Epithelial tissue, Connective tissue, Muscular tissue

Question 26.
Identify the organ system to which the given organs are associated?
a. Heart, Blood Vessels
b. Nose, trachea, lungs
c. Kidney, Ureter, Urinary Bladder.
d. Brain, Nerves
Answer:
a – Circulatory system
b – Respiratory system
c – Excretory system
d – Nervous system

Question 27.
Which of the statements is not related to tissues?
a. Different types of cells are seen.
b. Similar kind of cells are seen.
c. Performs specific function.
d. Formed from different cells
Answer:
d. Formed from different cells

Question 28.
Which indicator helps to identify collenchyma when tissues are observed through a microscope.
a. No thickenings in the cell wall.
b. all parts of the cell wall is thick,
c. No nucleus in the cell
d. Thickenings are seen only in the comers of the cell.
Answer:
d. Thickenings are seen only in the comers of the cell.

Question 29.
Complete the flow chart showing the levels of organization in organisms
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 2 Cell Clusters 17
Answer:
a. tissue
b. Organ system
c. Organism
d. Species

Question 30.
What is meant by voluntary muscles and involuntary muscles? Give examples:
Answer:
Tissues that work according to our wish are called voluntary muscles.
eg: muscles of limbs.
Muscles which works independently and are out of our control are called involunatary muscles.
eg : Muscles of the alimentary canal, muscles of the eyelids, etc.

Question 31.
Xylem vessels are thicker than phloem vessels. Why?
Answer:
The cell wall of the xylem vessels is comparatively thicker than that, of the phloem vessels. Lignin is used to make the cell wall strong. So the xylem vessels are thicker.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

You can Download Force Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

When an object is pushed or pulled, a force is being applied on it. The unit of force is newton. It is indicated by the letter N.

Force is that which changes or tends to change the shape, size, volume, state of rest or state of motion of a body.

There are various types of forces such as muscular force, magnetic force, electric force, gravitational force, frictional force etc.

Contact force and non contact force

The force applied by the contact between objects is contact force. The force applied on an object without contact on it is non contact force.
Ex:pushing a trolley- contact force, Magnet attracting a nail-non contact force

Kerala Syllabus 8th Standard Physics Notes Frictional force:

When a ball is rolling on the floor after some time it stops. This is because of the frictional force between the ball and floor. On the surface of the objects which are on contact with each other there are many mounts and pits get interlocked.

When we try to move a body by applying a force parallel to their surfaces an opposing force is experienced.
When a body moves or tends to move on the surface of another body, a force is experienced parallel to the surface which opposes the relative motion between them. This is friction.

Rolling friction and sliding friction

When a body rolls over the surface of another body, the force of friction which originates is rolling friction. When a body slides on the surface of another body the force of friction which originates is sliding friction.

Rolling friction is less than sliding friction. Striking a match stick on the side of a match box, wear and tear of machines, ability to hold objects, walking etc are advantages of the friction and treading of tyres of vehicles, wearing out of tyres, fuel loss are disadvantages of friction.

The materials like oil, grease which are used to reduce friction are the lubricants. Graphite is a solid lubricant.

Force Class 8 Kerala Syllabus Thrust and Pressure

The total normal force experienced by a ‘ surface is thrust. Thrust per unit area is pressure.
Pressure = \(\frac{Thrust}{area}\)

Unit of pressure is N/m2. This is known as pascal. The force acting on a larger area exerts a smaller pressure, while a force acting on a smaller area exerts a large pressure.

Hsslive Guru 8th Class Physics Kerala Syllabus Liquid Pressure

The pressure is experienced in liquids also. As the height of liquid column increases the pressure exerted by it increases.

The pressure exerted by a liquid column increases with increase in height. The thrust acting per unit area by a liquid is liquid pressure.

Liquids exert force on all sides of the container in which they are taken.

If the height of the liquid column is Ti’, density of liquid ‘d’ and acceleration due to gravity ‘g’, then liquid pressure is P = hdg

Class 8 Physics Kerala Syllabus Atmospheric pressure

Atmospheric air also can exert pressure. An envelope of air surrounds the earth. This is earth’s atmosphere. The density of atmospheric air near the surface of earth is greater and it decreases when we go up. The weight of air column over the unit area of earth’s surface is atmospheric pressure. One atmospheric pressure is the weight of mercury column of 0.76 m height and unit area (1m²). This is standard atmospheric pressure .

The unit of atmospheric pressure is* bar. The instrument used to measure atmospheric pressure is Barometre

Force Textbook Questions and Answers

Kerala Syllabus 8th Standard Physics Notes Pdf Question 1.
Classify the following situations into contact and noncontact forces.
a. Applying break in a bicycle.
b. A mango falling from a mango tree.
c. The earth revolving around the sun.
d. The speed of a hall rolling on ground is reduced.
Answer:
Contact force
a. Applying break in a bicycle.
d. The speed of a ball rolling on ground is reduced
Non contact force
b. A mango falling from a mango tree.
c. The earth revolving around the sun.

Class 8 Physics Notes Kerala Syllabus Question 2.
State reason
a. We are able to walk on the ground without slipping.
b. It is easy to cut vegetables using a sharp knife.
c. The number of tyres is more for goods vehicles.
d. The moving parts of machines experience wear and tear.
Answer:
a. Because of the friction between ground and feet
b. When the area decreases pressure increases. As the area of the edge of the knife is veiy less, it is easy to cut
c. When the area of surface increases pressure decreases, so when more wheels are used it won’t depressed in soil.
d. Because of the friction between the area of contacts.

Hss Live Guru 8th Physics Kerala Syllabus Question 3.
Match the following

ABC
Atmospheric pressurepascalcoconut oil
Lubricantattrationbarometer
Coconut falling downbarthermometer
Magnetfrictionnoncontact
gravitationrepulsion

Answer:

Atmospheric pressurebarbarometre
LubricantfrictionCoconut oil
Coconut falling downGravitational forceNon contact
Magnetattractionrepulsion

Hsslive Guru Physics 8th Standard Kerala Syllabus Question 4.
Bubbles rising from the bottom of the water filled in a bottle are depicted in the figure. Which is the correct figure? Justify your answer.
Kerala Syllabus 8th Standard Physics Notes

Answer:
Fig.c. There will be more pressure in the lower side of the liquid. The pressure decreases when it comes upward.

Hss Live Guru 8 Physics Kerala Syllabus Question 5.
A toy car of about 50 g placed on a polished table with threads attached to it carrying two pans passed through pulleys is depicted.
Force Class 8 Kerala Syllabus

(a) What do you observe if too g each is placed on both pans?
(b) What do you observe if too g is placed on pan A and 200 g in pan B
(c) Justify your answers.
Answer:
a. The car will be stationary because the force excreted from both sides are equal.
b. The pan of weight 200 will pull with more force and it moves that direction.
c. In the first phase same force is excerted. But in second phase force is exerted to one direction.

Force Additional Questions and Answers

Force And Pressure Class 8 Worksheets With Answers Pdf Kerala Syllabus Question 1.
Tabulate the following as contact force and non contact force.
1. pushing a trolley
2. falling coconut from coconut tree
3. magnet attract iron nail
4. Drowing water from the well
5. pushing the car
6. earth is moving around the sun
Answer:
contact force – 1,4,5
non contact force – 2,3,6

8th Standard Physics Notes Kerala Syllabus Question 2.
Ramu tried to push a round stone and rectangle stone of same weight. Which is easy to push.
Answer:
It is easy to push round stone because rolling friction is less than sliding friction.
3. Which are the following occation friction is beneficial and non beneficial
1. striking a match stick on the side of a match box
2. wear and tear of machines.
3. ability to hold objects
4. wearing out of tyres
5. walking 6. loss of fuel
Answer:
Beneficial : 1, 3, 5
non beneficial : 2, 4, 6

8th Std Physics Notes Kerala Syllabus Question 4.
Why does the design of ship and aeroplane are streamlining?
Answer:
To reduce friction while moving through air and water.

Kerala Syllabus 8th Standard Physics Question 5.
Write one solid lubricant and liquid lubricant
Answer:
Solid-graphite
Liquid-oil

Kerala Syllabus 8th Standard Science Notes Question 6.
Fill up suitably.
Force – newton
Pressure – ………..
Answer:
Pressure – pascal

8th Standard Physics Notes Pdf Kerala Syllabus Question 7.
Write the reason
1. Constructing a knife with sharp edge
2. Constructing the building with wide basement
Answer:
1. When the area of contact decreases the pressure increases
2. When area increases the chance to depress the base in soil is very less.

Class 8 Basic Science Solutions Kerala Syllabus Question 8.
Given figures, one can filled with water and other can filled with kerosine.
Hsslive Guru 8th Class Physics Kerala Syllabus
Which balloon is filled with water
Answer:
fig 2. When density of liquid increases the pressure increases, water is denser than kerosene.

Physics Class 8 Kerala Syllabus Question 9.
What are the factors effecting liquid pressure? Write the mathematical equation.
Answer:
The height of the liquid column is ‘h’, density of liquid ‘d’ and acceleration due to gravity ‘g’ are the factors. The equation is liquid pressure P = hdg

Basic Science Class 8 Kerala Syllabus Question 10.
As mountaineers climb higher altitude there is a possibility of nasal bleeding. Why?
Answer:
Going higher altitude the atmospheric pressure decreases. So blood veins are broken and blood is oozing out.

Question 11.
What is standard atmospheric pressure
Answer:
Atmospheric pressure at the sea level is equal to the pressure exerted by 0.76m of mercury column. This is standard atmospheric pressure.

Question 12.
When a hole is formed at the bottom of the water bottle, Anu closed the lid of the bottle tightly. The flow of water stops. Explain
Answer:
When the lid is closed tightly, the effect of atmospheric pressure is not experienced on the surface of the water in the bottle. Hence water cannot flow out from the bottle.

Question 13.
Mercury remains at a height of 76cm in a barometer. Raju made a hole at the top end of the tube. What will you observe? How can it be explained?
Answer:
Mercury will flow downwards when a hole is made at the top part of the tube. The atmospheric pressure exerted on the mercury in the vessels supports the mercury level in the tube. When a hole is made at the top, air enters the tube and the pressure is distributed throughout uniformly and hence the mercury level comes down.

Question 14.
A sharp nail pierces more easily through wood than a blunt nail. What is the reason?
Answer:
The surface area of the sharp tip of a nail is very small. Hence the pressure that it can exert on the wood is very great. That is why it pierces easily. For a blunt nail, the surface area of the tip is greater. Hence it can exert only a small amount of pressure.

Question 15.
What is meant by limiting friction.
Answer:
Limiting friction is the force of static friction acting just before the body starts moving.

Kerala Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World

Kerala State Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World

History Sslc Question 1.
‘American War of Independence was a movement of the English against England’. Substantiate the statement.
Answer:
The earlier colonies in America were established by a group of people who exiled to America from the religious persecution of the King of England in the 17th century on a ship called ‘Mayflower’. They were known as the Pilgrim Fathers. By the 18th century, England had established 13 colonies in the eastern coast of North America.

The British treated these colonies as centers for collecting raw materials for their industry and as markets for selling their products. The American War of Independence was the struggle of the American colonies against the wrong policies and taxes of the mother country. The forefathers of the American colonies were English. So American War of Independence is treated as a movement of the English against England.

Question 2.
How did the resentment of the middle class lead to the French Revolution?
Answer:
The French middle class included traders, writers, lawyers, officials, teachers, bankers and farmers. They enjoyed low social status. They were dissatisfied with the social inequality and the prevailing social order. They played a key role in the production of wealth and its ownership, but they had no role in the administration of the state.
They realised that it was impossible for them to achieve their demands under the existing structure. Therefore they decided to overthrow the government by leading the revolution.

Question 3.
Arrange the following events in the chronological order.
i. February Revolution
ii. Long March
iii. American Declaration of Independence
iv. Tennis Court Oath .
Answer:
i. American Declaration of Independence
ii. Tennis Court Oath .
iii. February Revolution
iv. Long March

Question 4.
How did the French Revolution influence the reforms of Napoleon?
Answer:
The concepts of French Revolution namely nationalism, liberty, equality and fraternity inspired the reforms of Napoleon. Nationalism strengthened in France under his rule. His reforms helped to bring about national unity in France. Farmers were made owners of land.

The middle class were allowed to operate with the nobles. Prepared a new and uniform code of law for France by codifying the existing laws. Exercised the state’s control over the clergy. Established the Bank of France to centralise finance. Formed sinking fund to avoid public debt. He also brought a change from feudal rule to republican rule.

Question 5.
How did the colonial rule influence the Latin American countries?
Answer:
The majority of the migrants from Europe to Latin America were Spanish and Portuguese. After colonizing the entire Latin America, the Spanish and the Portuguese propagated their language, religion and customs there. They built houses and churches in Spanish style. Several schools were started for imparting Spanish system of education. The Spanish farming methods and crops were introduced.

Racial discrimination was enforced to natives in all walks of life. Looted the resources and enslaved the people. The colonial powers took away gold, silver, etc. from the Latin American mines. Never allowed the people of the colonies to engage in trade with other countries other than Spain. Through such policies, the European countries implemented colonial domination in Latin American countries.

Question 6.
‘Russian Revolution helped the progress of the working class’. Do you agree with this statement? Why?
Answer:
The statement is true.
The leaders of working class led the revolution against the autocracy of the Tsarist emperors in Russia. In Russia, farmers and factory workers led a miserable life. Trade Unions were formed to find remedies for the plight of workers. Worker’s organizations called Soviets were formed all over Russia to conduct strikes.

The workers organised a protest March in 1917 and captured the city of Petrograd. The Bolsheviks propagated that only proletarians’ (workers) government could eradicate centuries old economic backwardness and inequality. The Bolsheviks attained power in Russia after the October Revolution of 1917.

Thus the first workers’ government assumed power in Russia under the leadership ofLenin. The new government distributed land among the workers, brought factories under public ownership and took many measures for the welfare of the workers.

Question 7.
How did China become a People’s Republic? Analyse.
Answer:
The Chinese Revolutions were against the foreign rule and monarchy. Dr. Sun Yat Sen led the revolution in 1911. It ended monarchy in China and made China a republic. Chiang Kai-Shek who succeeded Sun Yat-Sen asertained military autocracy in China. He gave opportunity for foreign powers including America to freely interfere in China and did not co-operate with the Communists.

Coal and iron industries, banking and foreign trade were all controlled by foreign countries. The Communists protested against the policy of Chiang Kai-Shek. They were brutally suppressed. In 1934, under the leadership of Mao Zedong, the Communist Party started a journey from Kiangsi in South China and ended at Yanan in North Western China. Throughout the journey they seized out agricultural land and villages from lords and distributed them among the workers.

The journey covered around 12000 kms. So it is known as the ‘Long March’. Chiang Kai-Shek had to seek political asylum in Taiwan, when the Red Army of Mao Zedong captured the centre of Kuomintang rule. China became the People’s Republic of China on 1 October 1949 under the leadership of Mao Zedong.

Sslc History Chapter 1 Pdf Question 8.
List the 13 British colonies in North America.
Answer:

  1. Georgia
  2. South Carolina
  3. North Carolina
  4. Virginia
  5. Maryland
  6. New Jersey
  7. Pennsylvania
  8. Connecticut
  9. Rhode Island
  10. New York
  11. Massachusetts
  12. New Hampshire
  13. Delaware

Question 9.
The mercantalist laws turned the American Colonial people against their mother land. Assess the statement.
Answer:
The British treated American colonies as centres for collecting raw materials for their industry and as market for selling their products. This policy implemented by the British merchants with the help of their motherland in the American colonies, is known as mercantilism. Several laws were implemented in the British colonies as part of this.

  • Restriction on cargo ships :
    Only British ships must be used to carry goods to and from.
  • Control over export:
    Products of colonies like sugar and wool could only be exported to England.
  • Stamp duty :
    British stamp must be affixed on all legal documents, newspapers, license, etc.
  • Import duty :
    Import tax must be paid for the import of tea, glass and paper.
    The above laws were harmful to the colonial people.

Question 10.
How did the laws implemented by the British affect the people of the colonies? Discuss based on the following hints.
i) Domestic trade in colonies
ii) Denial of rights
iii) Autocracy
iv) Inflation
Answer:
i) Domestic trade in colonies :
The controls and taxes in trade adversely affected domestic trade. The British treated the American colonies as centres for collecting raw materials for their industry and as market for selling their products. As part of Mereantalist Laws, products of colonies like sugar, wool, cotton and tobacco could only be exported to England, the goods to and from the colonies must be carried only in British ships and import tax must be paid for the import of tea, glass and paper.

ii)Denial of rights:
Imposed control over exports.

  • British troops to be maintained in the colonies.
  • British stamp must be affixed on all legal documents.

Exploitation :
Control over export, high tax, Stamp Act and the order to maintain British troops in the colonies paved the way for exploitation.

iii) Autocracy :
The Mereantalist laws denied the rights and freedom of the colonists. The imperialist policies of Britain led to the American War of Independence. Some of the autocratic policies of King George III of England led to the American revolution. The Townshend Act and the Stamp Act that Britain implemented in the colonies were examples of their autocracy.

iv) Inflation:
High tax and Stamp Act led to inflation.

Question 11.
The Revolution had been effected even before the war commenced. It was in the minds and hearts of people’. Evaluate the relevance of this remark made by John Adams, one of the leaders of the American War of Independence.
Answer:
The ideas of freedom and liberty that grew among the colonial people and the hatred towards the mother country were the main reasons behind the American War of Independence. The ideological conflict between the colonial people and the mother country influenced the American War of Independence. The attitude of freedom that grew among the colonial people forced them to oppose the laws of the mother country.

The ideologies of thinkers like John Locke and Thomas Paine also stimulated the people of the colonies to fight against the exploitative laws of the British. By raising the slogan ‘no taxation without representation’, the people of the colonies intensified their struggle against the British. From these, we can infer that the Revolution had been effected eyen before the war commenced, as stated by John Adams.
Kerala Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World 1

Class 10 History Chapter 1 Notes Kerala Syllabus Question 12.
Prepare a short note on the thinkers and their ideologies that inspired the American War of Independence.
Answer:
The ideologies of some thinkers inspired the people of the colonies to fight against the exploitive laws of the British. English thinkers like John Locke, Harrington and Milton stimulated the American War of Independence.

John Locke argued that ‘Every one has some fundamental rights. No government has the right to suspend them’. Thomas Paine said that ‘There is something absurd in supposing a continent (North America) be perpetually governed by a foreign power (England)’.

Thomas Paine through his Pamphlet ‘Common Sense’ declared that it was wise for the Americans to break the ties with Britain. Thomas Jefferson who was inspired by the ideologies of French thinkers asked the people of the colonies to organise the revolt and get independence.

Class 10 History Kerala Syllabus Question 13.
Prepare a flowchart illustrating the various events that led to the formation of the United States of America.
Answer:
Kerala Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World 2

Question 14.
Prepare a timeline on the various events that led to the American War of Independence.
Answer:
Kerala Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World 3

10th History Notes Kerala Syllabus Question 15.
Prepare a note on the ideologies of the American War of Independence that are reflected in the Constitution of India.
Answer:

  • Republican system
  • Written constitution
  • Federation
  • Fundamental Rights
  • Bill of Rights
  • Judicial Review

Question 16.
Analyse the chart given in Page 15 of Textbook and answer the following questions.
i. Which estate enjoyed higher social privilege and status and led a luxurious life in France?
ii. To which estate did the majority of people belong?
iii. Which estate exploited the common people and accumulated wealth?
iv. Which estate desired a change in the system? Why?
Answer:
i. Clergy (First Estate), Nobility (Second Estate)
ii. Third Estate
iii. Nobility and Clergy

iv. The members of the Third Estate desired a change in the system. The First and the Second Estates enjoyed all the rights and privileges. The Third Estate had no rights. They had to pay different types of feudal taxes to the clergy and the nobles. The members of the Third Estate who were dissatisfied with the social inequality in France led the revolution.

Question 17.
Examine the membership strength of each estate given in the table below. Relate it with the demand regarding the voting system.

States General

The First Estate285
The Second Estate308
The Third Estate621

Answer:
The French Parliament called ‘States General’ consisted of three estates. Traditionally, each Estate would vote as a group and had one vote. As a result, the nobility and the clergy could overrule the Third Estate. The first two Estates argued for Estate wise single voting system while the Third Estate demanded individual vote for each member of all the three estates. The total membership strength of the first two Estates was 593 and that of the Third Estate was 621.

The first two estates opposed the demand of individual vote for each member because the total membership strength of the Commons was more than that of theirs. Individual vote for each member meant that the Third Estate would get majority. Instead, Estate wise single voting system was allowed, the first and second Estate would get majority.

Question 18.
Analyse the events related to the French Revolution and evaluate how far it could attain the goals of liberty, equality and fraternity.
Answer:
The French Revolution

  • Threatened the autocratic rulers of Europe.
  • Ended the feudal system in Europe.
  • Helped the growth of middle class.
  • Led to the emergence of nationalism.
  • Contributed the concept of people’s sovereignty.
  • Proclaimed that nation is not merely a region, but the people.
  • Stimulated all the later revolutions in the world.

Question 19.
Analyse the aims and influence of French Revolution.
Answer:
French Revolution is the most important event in modern world history. It was a revolution that took place in France in 1789 against despotic rule of kings and feudal system. Its aim was to reconstruct the society based on the ideals of liberty, equality and fraternity.

It is also known as the ‘Mother of Revolutions’. The French Declaration of the Human Rights is an important episode in the history of the world. The French revolution influenced the whole of Europe.

Question 20.
Examine the reforms and policies of Napoleon and identify the ideologies of French Revolution reflected in them.
Answer:

The concepts of French Revolution

Policies and reforms of Napoleon

Rise of the middle classState’s control over clergy

Set up Bank of France

Formed Sinking fund

Codified the laws

Constructed roads

End of feudalismThe farmers were freed from the clutches of nobles and were made owners of land.
NationalismEstablished French national unity.

Tried to unify Europe under the leadership of France.

Progress in transport

Class 10 Social Science Chapter 1 Kerala Syllabus Question 21.
Discuss the discriminations faced by the Latin American people.
Or
How did the exploitation and discrimination faced by the Latin American people lead to revolution?
Answer:
After colonizing the entire Latin America, the Spanish and the Portuguese propagated their language, religion and customs there. They built houses and churches in Spanish style. Instead of Latin American system of education, Spanish schools were established for imparting Spanish system of education.

The Spanish farming methods and crops were adopted. The native people were enslaved to work in plantations. The Latin Americans were not allowed to engage in trade with countries other than Spain. Racial discrimination towards the people of the colonies was enforced in all walks of life.

Question 22.
Enlist the countries of Latin America liberated from the colonial rule of the European countries.
Answer:

Spanish colonies liberated

Portuguese colonies liberated

Equador

Argentina

Bolivia

Uruguay

Peru

Chilie

Colombia

Brazil

Question 23.
The Latin American Revolution was not only for political liberation, but also for the life and culture of the people. Substantiate the statement.
Answer:
The Spanish and the Portuguese who colonized South America destroyed the native culture there. They looted the wealth and resources of Latin American countries and even erased their traditional culture. Instead of Latin American languages, Spanish was made as the official language.

Schools were established for imparting Spanish system of education. They took away gold, silver, etc. from the Latin American mines. The things produced in Spain were not permitted to be made in the colonies. Enslaved the natives to work in plantations.

The Spanish farming methods and crops were adopted in the colonies. Racial discrimination towards the people of the colonies was enforced in all walks of life. So the Latin Americans organized the revolt against the Europeans in order to protect their life, culture and freedom.

Question 24.
Russian revolution was greatly motivated by the despotic rule of Tzar. Discuss.
Answer:

  • The Tzarist emperors who ruled Russia were autocrats. They denied all rights and privileges of the people.
  • The common people had no role in administration.
  • Power and wealth were concentrated in kings and nobles.
  • The landless fanners had to pay huge tax.
  • Low agricultural and industrial production.

Time line on Russian Revolution

  • 1894 : Coronation of Czar Nicholas 11.
  • 1898 : Formation of Social Democratic Labour Party.
  • 1903 : Split of Social Democratic Labour Party as Mensheviks and Bolsheviks.
  • 1905 : Bloody Sunday. Formation of Duma.
  • 1914 : Beginning of First World War. Ignoring the protest from Duma, Tsar Nicholas 11 decided to participate in the war.
  • 1917 March 12 : February Revolution. Nicholas 11  abdicated. A provisional government under Kerensky.
  • 1917 : October Revolution. Bolsheviks under Lenin assumed power.
  • 1924 : Formation of Union of Soviet Socialist Republic (USSR)
  • Majority of the Russian industries were controlled by foreingners.

Question 25.
Analyse the circumstances that led to form a provisional government in Russia.
Answer:
Ignoring the protest from Duma, the Legislative body, Nicholas II, the then Tsar decided to participate in the First World War that started in 1914. A lot of Russian soldiers were killed in this war. Workers’ organizations called ‘Soviets’ were formed all over Russia to conduct strikes when famine, poverty and unemployment became acute. Food shortage became severe by 1917.

Thousands of women marched along the streets of Petrograd on 8 March 1917 clamouring for bread. The workers organized protest march in Petrograd. The soldiers also joined with the workers and captured Petrograd. Nicholas II was thrown out of power. A provisional government was formed under Alexander Kerensky, the Menshevik leader. This is known as February Revolution.

Sslc History Chapter 1 Notes Question 26.
Discuss the role of the October Revolution in establishing the proletariat government in Russia.
Or
Write a note on October Revolution.
Answer:
The provisional government of Russia failed to solve the problems of the people. A group of Soviets did not approve the provisional government. Lenin, who had been in Switzerland, came to Russia and strongly opposed the provisional government.

The Bolsheviks under Lenin propagated that only a proletarians’ government could eradicate centuries old economic backwardness and inequality. Lenin declared that what the people wanted was not war and famine, but peace and food.

In October 1917, the Bolsheviks organised an armed rebellion against the provisional government. Kerensky fled from the country and Russia came under the control of the Bolsheviks. This event, through which the Bolsheviks attained power, is known as the October Revolution.

Following this, a proletarians’ government came to power in Russia under Lenin. The new government seized the land owned by the nobles and distributed it among farmers. Factories, banks, transportation facilities and foreign trade were brought under public sector.

Question 27.
Compare the strategies of colonial powers in China with the strategies that they adopted in other colonies.
Answer:
The colonial powers of Europe and America adopted policies in China different from the ones they had implemented in the colonies of other parts of the world. They were opium trade and open door policy. Chinese silk, tea and earthernware were in great demand all over Europe. China gained huge profit through the export of these items. There was nothing to import to China from Europe.

As a result, the European traders suffered a great loss. As a remedy for this, the British traders imported opium, an intoxicant to China. This affected the Chinese trade and its consumption made the Chinese people mentally imbalanced. It resulted in economic and mental subjugation of the Chinese. Taking advantage of the weakness of China, the imperial powers exploited her.

To acquire the privilege that the other European . countries enjoyed in China. John Hey, the then State Secretary of the USA proclaimed the ‘Open Door Policy’. As per this policy, America argued for equal rights and opportunities for all countries in Chinese market. China was divided into different regions to be controlled by various countries. The aim of this policy was to create an opportunity for the USA to interfere in China. China was divided as markets for Europeans. Chinese became slaves of Europeans.

The European countries established colonies in Asia and Africa and adopted a policy to destroy the wealth and culture of those regions. They treated the colonies as centres to produce the goods that they need and also as markets to sell their products.

Question 28.
Examine the role of the Chinese government led by Sun Yat Sen in ending the foreign domination.
Or
Chinese Revolution of 1911 led to the liberation of China from foreign powers and monarchy. Examine the statement.
Answer:
The Manchu dynasty that ruled China favoured foreign interference and domination. In 1911, a revolution took place in China under the leadership of Dr. Sun Yat Sen against the Manchu dynasty. This ended monarchy in China. After the revolution, Kuomintang Party established a republican government in Southern China under the leadership ofSunYatSen. He gave importance to ideologies like nationalism, democracy and socialism.

  • Nationalism : to expel the Manchu dynasty and the imperial powers.
  • Democracy : to establish democratic rule.
  • Socialism: to control capital and distribute land equally.

He decided to nullify the unjust treaties signed with the foreign countries and wanted to maintain equality with western countries.

Question 29.
Prepare a note on the role played by Mao Zedong to make China a peoples republic.
Answer:
China was under the despotic monarchy of Manchu dynasty. The revolution of 1911 under the leadership of Sun Yat Sen ended monarchy in China. The Kuomintang party under the leadership of Sun Yat Sen established a republican government in Southern China.

Following the Russian revolution, the Communist Party was formed in China also. The Party stood for the welfare of farmers and workers. In the beginning, the Kuomintang and the Communists co-operated with each other. But this co-operation was disturbed when Chiang Kai-Shek became the head of the Republic following the death of Sun Yat Sen.

The role of Mao Zedong:
Mao Zedong rose to the leadership of Chinese Communist Party. The Communists protested the policies of Chiang Kaishek. So they were brutally suppressed. In 1934. under the leadership of Mao Zedong, a journey started from Kiangsi in South China. The adventurous trip ended at Yanan in North Western China.

Throughout the journey, they seized out agricultural land and villages from lords and distributed them among farmers. The journey covered around 12000 kms. So it is know n as the ‘Long March’. Hence Mao Zedong and the Communist Party became the symbol of struggle of the Chinese against foreign power.

Chiang Kaishek had to seek political asylum in Taiwan when the Red Army of Mao Zedong captured the centre of Kuomintang rule. China became the People’s Republic of China on 1st October 1949 under the leadership of Mao Zedong.

Class 10 History Chapter 1 Questions And Answers Kerala Syllabus Question 30.
Prepare a Seminar paper on the liberation of China from foreign powers and monarchy and the formation of a republic.
Answer:
Seminar
Topic: Formation of People’s Republic of China Introduction:
In the twentieth century, China witnessed revolutions against the foreign rule and the monarchy. Ultimately it led to the formation of Chinese republic.

The colonial powers of Europe and America adopted policies in China different from the ones they had implemented in the colonies of other parts of the world. They were opium trade and open door policy.

Opium Trade:
China gained huge profit by the export of Chinese |ilk, tea and earthenware. The European traders Had nothing to export to China and so they suffered great loss. As a remedy, they exported opium, an intoxicant’to China. This affected the Chinese trade and its consumption made the Chinese people mentally imbalanced. It resulted in economic and mental subjugation of the Chinese.

Open Door Policy:
By the end of the 19th century, several European countries acquired many trade privileges in China. America could not acquire the privilege as it was not concerned about China at that time. To acquire the privilege, John Hey, the then State Secretary of the USA proclaimed the ‘Open Door Policy’.

As per this policy, America argued for equal rights and opportunities for all countries in Chinese market. The aim of this policy was to create an opportunity for the USA to enter China. As a result, China was divided into different regions to be controlled by the various countries.

The role of Dr. Sun Yat Sen:
The Mancliu dynasty in China favoured the foreign interference and domination. Some secret organizations in China revolted against this. This was known as Boxer Rebellion. Though this revolution failed, it stimulated the revolutions that came up later.

In 1911, another revolution took place under the leadership of Dr. Sun Yat Sen against the Manchu dynasty. This ended monarchy in China. After the revolution, the Kuomintang Party established a republican government in Southern China under the leadership of Sun Yat Sen. He decided to nullify the unjust treaties signed with the foreign countries and wanted to maintain equality with western countries.

The Kuomintang Republic adopted measures for the progress of agriculture and industry. China received assistance from Russia in various fields and the Chinese Communist Party was formed. In the beginning, the Kuomintang and the Communists co-operated with each other.

The Policies of Chiang Kaishek:
Following the death of Sun Yat Sen, Chiang Kaishek became the next ruler. The co-operation between, the Kuomintang and the Communists was disturbed during his rule. He was an anti-communist. Chiang ascertained military autocracy in China.

He gave opportunity for foreign powers including America to freely interfere in China and did not co-operate with the Communists. Coal and industries, banking and foreign trade were all controlled by foreign powers.

The role of Mao Zedong:
Mao Zedong rose to the leadership of Chinese Communist Party. The Communists protested the policies of Chiang Kaishek. So,they were brutally suppressed. In 1934, under the leadership of Mao Zedong, a journey started from Kiangsi in South China.

The adventurous trip ended at Yanan in North Western China. Throughout the journey, they seized out agricultural land and villages from lords and distributed them among farmers. The journey covered around 12000 kms. So it is known as the ‘Long March’.

Hence Mao Zedong and the Communist Party became the symbol of struggle of the Chinese against foreign power. Chiang Kaishek had to seek political asylum in Taiwan when the Red Army of Mao Zedong captured the centre of Kuomintang rule. China became the People’s Republic of China on 1stOctober 1949 under the leadership of Mao Zedong.

History Class 10 Chapter 1 Kerala Syllabus Question 31.
Compare the revolutions that took place in America, Europe and Asia and complete the table.
Kerala Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World 4
Answer:
Kerala Syllabus 10th Standard Social Science Solutions Chapter 1 Revolutions that Influenced the World 5

Kerala Syllabus 10th Standard Social Science Solutions

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life’s Mysteries in Little Chambers

You can Download life’s Mysteries in Little Chambers Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 1 life’s Mysteries in Little Chambers

life’s Mysteries in Little Chambers Questions and Answers

Microscope

Life's Mysteries In Little Chambers Kerala Syllabus 8th

The Microscope is a device used to magnify objects. The main parts of a microscope are Eye piece, Knobs, Objective Lens, Stage and clip, Diaphragm, Mirror etc.

The microscopes in which more than one lens is used are called compound microscopes: In a microscope there are two planes for the regulation of light. A plane mirror to reflect the sunlight and a concave mirror to reflect artificial light. The lens of the condenser that is fitted on the lower side of the stage focuses light on the material to be observed. Diaphragm regulates the intensity of light.

Indicators (Text Book Page No:9)

Life’s Mysteries In Little Chambers Kerala Syllabus 8th Question 1.
Why is a mirror fixed in a microscope?
Answer:
The Mirror is fixed to focus light on the objects to be observed.

Kerala Syllabus 8th Standard Biology Notes Question 2.
Why do we use a slide made of glass to place the material to be observed?
Answer:
A transparent surface is required to pass light of the material to be observed. So glass slide is used to place it.

Magnification Power of Microscope

Kerala Syllabus 8th Standard Biology Notes

Magnification Power is marked as tox and 45X in the objective lens and eyepiece. The magnification power of a microscope is the product of the numbers seen in the objective lens and eyepiece.
Preparation of Observation Material

Following are the steps in the preparation of a slide to observe plant cells.

8th Class Biology Notes Pdf Kerala Syllabus

  1. Put the observable material in a petridish containing water to prevent dehydration.
  2. Take thin cross sections of the material
  3. Put the sections to a watch glass con-taining water.
  4. Put the thinnest and complete sections to a watch glass containing stain.
  5. Add one or two drops of glycerin to the slide to avoid drying up of the material.
  6. Wash the stained section with clean water and place it in the glycerine on the slide.
  7. Put a cover glass over the material.
  8. Wipe out the excess glycerin using tissue paper.

Discovering of Cell

Robert Hooke discovered cells for the first time. The branch of science that deals with cells is called cell biology. The re markable findings that paved way for the development of cytology and the concerned scientists are indicated below.

  • Robert Brown – Discovered Nucleus (1831)
  • M. J. Schleiden – Discovered Plant Cells (1838)
  • Theodor Schwann- Discovered Animal Cells (1839)
  • Rudolf Virchow – Inferred that new cells arise only from existing cells (1858)

8th Class Biology Notes Pdf Kerala Syllabus Cell Theory

M.J. Schleiden and Theodor Schwann are the founders of cell theory. Mainly two concepts are put forward by cell Theory.

  • All living body is made up of cells.
  • Cells are the structural and functional units of organisms.

8th Class Biology 1st Chapter Kerala Syllabus Question 3.
What are the functions performed by cells?
Functions of Cells

8th Class Biology 1st Chapter Kerala Syllabus

  • Resist foreign particles,
  • Synthesis biomolecules.
  • Release energy from nutrients

Structure of cell is suitable to perform various biological functions. There are different parts inside the cell for this purpose.

  1. Nucleus – Centre of the cell.
  2. Ribosome – Helps in protein synthesis
  3. Chloroplast – Helps in photo synthesis
  4. Mitochondrion – Centre of energy production, storage and distribution.
  5. Vacuole – Stores water, salts, excretory materials etc.
  6. Golgi complex – Stores enzymes, hormones, mucous etc in small vesicles.

Basic Science Class 8 Pdf Notes Kerala Syllabus Question 4.
Complete the illustration given below related to different kinds of organelles.

Basic Science Class 8 Pdf Notes Kerala Syllabus

Hss Live Guru 8th Biology Kerala Syllabus Nucleus

Hss Live Guru 8th Biology Kerala Syllabus

Innumerable life processes take place within the cell. The biomolecules that control and coordinate these processes are synthesized under the control of the genes in the chromatin network of the nucleus.

Major parts of the nucleus are nucleoplasm, nuclear membrane, chromatin reticulum, nuclear pore, nucleolus etc.

Nucleoplasm is the fluid part of the nucleus. Nuclear membrane is the double-layered membrane that surrounds the nucleus. The minute pores in the nuclear membrane are called nuclear pores. The spherical nucleolus seen in the nucleus play a major role in the synthesis of ribosome. The chromatin reticulum seen as a network in the nucleoplasm encloses genes.

Prokaryotes and Eukaryotes

There is no definite nucleus in the cell of bacteria, cyanobacteria and mycoplasma etc. There are called prokaryotes. But in the cells of amoeba, plants, and animals a well defined nucleus is present. These organisms are called eukaryotes.

Biological Membranes

Cells are covered by membranes. They regulate the exchange of materials between cells and between cell organelles and cytoplasm. These membranes are called selectively permeable as they pass only the essential materials across them.

Indicators (Text Book Page No: 17)

Kerala Syllabus 8th Standard Physics Notes Question 5.
Plastids in the plant cells.
Answer:
Plastids are the cell organelles that are found only in plant cells. They are of 3 types – Chromoplast, Chloroplast and Leucoplast.

Class 8 Science Notes Kerala Syllabus Question 6.
Pigments in the chromoplasts.
Answer:
The pigments in Chromoplast are Xanthophyll (yellow), Carotene (yellowish Orange), Anthocyanin (Red, Purple) etc.

Kerala Syllabus 8th Standard Biology Notes Malayalam Medium Question 7.
Chromoplasts and sunlight.
Answer:
Chloroplast performs photosynthesis in the presence of sunlight. As the fruits ripen, the chloroplast in them changes to chromoplasts.

Class 8 Chemistry Chapter 1 Kerala Syllabus Question 8.
Change of colour and taste in fruits
Answer:
In the storage cells, leucoplasts are more. But as the fruit ripens the chlo- roplasts converted to chromoplasts. Starch changes to sugars. This is the reason behind the change in colour and taste when the fruit ripens.

life’s Mysteries in Little Chambers Let us complete the illustration (Text Book Page No: 18)

8th Class Biology Notes Pdf Malayalam Medium Question 1.
Compare the plant cell and the animal cell and complete the illustration given below.

Kerala Syllabus 8th Standard Physics Notes

Answer:

Class 8 Science Notes Kerala Syllabus

life’s Mysteries in Little Chambers Let us assess (Text Book Page No: 19)

Hsslive Guru 8th Class Biology Kerala Syllabus Question 1.
Complete the concept map related to the structure of animal cells.

Kerala Syllabus 8th Standard Biology Notes Malayalam Medium

Answer:

Class 8 Chemistry Chapter 1 Kerala Syllabus

Std 8 Biology Notes Malayalam Medium Question 2.
8th Class Biology Notes Pdf Malayalam Medium

(i) Identify the parts A,B, C,D,E marked in the plant cell.
Answer:
A – Chloroplast
B – Mitochondrion
C – Ribosome
D – Vacuole
E – Nucleus

(ii) Write the function of the parts denoted by the letters given below.
Answer:
A – Photosynthesis
B – Production, Storage and distribution of energy
C – Protein synthesis

(iii) In which part of the cell are genes found?
Answer:
Genes are seen in the chromatin reticulum in the nucleus.

life’s Mysteries in Little Chambers Additional Questions and Answers

Basic Science Class 8 Solutions Kerala Syllabus Question 1.
Find the odd one in each group. Also write the common characteristics of the others.
a. Bacteria, Cyanobacteria, Amoeba, Mycoplasma,
b. Golgi complex, Ribosome, Lysosome, Mitochondrion
c. Mitochondrion, Cell membrane, Ribosome, Golgi complex
d. Xanthophyll, Anthocyanin, Tonoplast, Carotene
Answer:
a. Amoeba; The others are prokary otes.
b. Lysosome; The others are present in both plant and animal cell.
c. Cell Membrane; Others are cell organelles

Kerala Syllabus 8th Standard Basic Science Textbook Solutions Question 2.
Find out the relation between the given word pairs and on that basis fill in the blanks.
a. M.J. Schleiden- ………… ; Theodor Schwann- Animal cell
b. Power house – Mitochondrion; Skeleton of cell- …………
c. Bacteria – ………… ; Amoeba – Eukaryote
d. Robert Hooke – Cell; ………… – Nucleus
e. Yellowish orange colour – Ca-rotene; Yellow colour – …………
f. Powerhouse of the cell – Mitochondrion; Transport of materials – …………
Answer:
a. Plant Cell
b. Endoplasmic reticulum
c. Prokaryote
d. Robert Brown
e. Xanthophyll
f. Endoplasmic reticulum

Basic Science Class 8 Chapter 1 Question Answer Question 3.
Copy the given figure, then identify and mark those parts, their features are given.|
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 16
a. Cytoskeleton
b. Controll various process inside a cell
c. Power house of cell
Answer:
a. Endoplasmic reticulum – understand, marking
b. Nucleus – understand, marking
c. Mitochondrion – understand, marking

Question 4.
It was Rudolf Virchow who dis-covered one of the prominent characteristics of cell. Analyse the statement.
Answer:
He found the process cell division. New cells are formed only from existing cell.

Question 5.
Which components of plant cell are not found in animal cell?
Answer:
Centrosome, Lysosome.

Question 6.
Some components of Animal cell are also found in the plant cell. List out them.
Answer:
Cell wall, large vacuole, plastids.

Question 7.
What change occur when a mango ripens? What are the causes?
Answer:
Colour changes.
Chloroplast is changed into chromoplast.
The sour taste is changed sweet taste. Starch is changed into sugar.

Question 8.
Choose the parts of microscope from the following terms.

Diaphragm, test tube, objective lens, eye piece, mirror, hammer, watch glass, cover glass, knobs, stage, clip

Answer:
Diaphragm, objective lens, eye piece, mirror, Knobs, Stage, Clip

Question 9.
Make a table showing the different parts of microscope and their function.
Answer:
Eye piece – helps to magnify objects; helps to observe objects
Knobs – to adjust the position of lens
Objective lens – helps to magnify objects
Stage & Clip – object is placed on the stage and the slide is fixed using clips
Diaphragm – helps to reduce the intensity of light Mirror – reflects sunlight

Question 10.
Why does mirror is fixed in the microscope?
Answer:
To allow sunlight on to the material observed.

Question 11.
Why a glass slide is used to place the material to be observed?
Answer:
A transparent surface is needed to fall sunlight reflected by the mirror on the material to be observed.

Question 12.
Why plane mirror and concave mirror are used in the microscope?
Answer:
Plane mirror reflects sunlight.
Concave mirror reflects artificial light.

Question 13.
Ananthu noted the markings 10x and 45x in the objective lens of the microscope. He also noted the markings on the eyepiece too. But he couldn’t understand this? Can you help him?
Answer:
These markings (10x, 45x) denote their magnification power.

Question 14.
How can we find out the magni-fication power of microscope?
Answer:
The product of the numbers seen in the objective lens and eyepiece de-notes its magnification power.

Question 15.
What is the difference between simple microscope and compound microscope?
Answer:
In simple microscope there is only one lens. But in compound micro-scope there are more than one lens.

Question 16.
Why the sections of plantain roots are put in the water taken in the petridish?
Answer:
To avoid dehydration of roots.

Question 17.
Anagha and Priya placed the sections of plantain root on glass slides. After sometime the materials on Anagha’s slide dried up. But Priya’s slide was suitable for observation. What may he the reason for this?
Answer:
To avoid drying up it is needed to add a few drops of glycerine on slide. Priya has done it.

Question 18.
Write procedure for observation of the section of plantain root.
Aim – to observe the cells in the roots of plantain.
Materials required -Plantain root, Petridish, Blade, Brush, Glass slide, Cover slip, Stain, Glycerine, Water, Microscope.
Procedure: –
Answer:

  • Take thin sections of the roots of plantain. Put it in the water in a petridish.
  • Transfer the sections to the water taken in a watch glass using a brush.
  • Put the thinnest and complete sections in a watch glass containing stain.
  • Add a few drops of glycerine on slide to avoid dehydration.
  • Place the stained section on the slide after putting it in fresh water.
  • Cover it using a cover glass to avoid air bubble.
  • Wipe out the excess glycerine using a tissue paper.

Question 19.
Match the Scientists and their contributions.
Rudolf Virchow – a. Found out that plant body is made up of cells.
M. J. Schleiden – b. Observed cells for the first time.
Robert Hooke – c. Found out that animal body is formed of cells.
Robert Brown – d. Formulated the inference that new cells form only from the existing ones.
Theodor Schwann – e.Invented nucleus.
Answer:
Rudolf Virchow – b
M. J. Schleiden – a
Robert Hooke – b
Robert Brown – e
Theodor Schwann – c

Question 20.
State Cell theory
Answer:
All living body is formed of cells. Cells are the structural and functional units of organisms.

Question 21.
Who are the founders of Cell Theory?
Answer:
MJ. Schleiden and Theodor Schwann.

Question 22.
“Our body is composed of millions of cells. Why these much of cells?” – Ananthu
‘They have various functions to perform’ – Bhadra
List out the functions of cells in the body.
Answer:

  1. Defend foreign particles.
  2. Releases energy from the cells.
  3. Synthesis biomolecules.

Question 23.
Classify the given cell organelles under suitable headings. Common terms can be included in both groups.
Ribosome, Nucleus, Cell membrane,Centrosome, Chloroplast, Vacuole, Golgi Complex, Mitochondrion, Lysosome
Answer:

Plant cellAnimal cell
Cell wallCentrosome
ChloroplastLysosome
VacuoleRibosome
RibosomeNucleus
NucleusGolgi complex
Golgi ComplexMitochondrion
MitochondrionCell membrane
Cell membrane

Question 24.
Identify the cell organelles that perform the functions mentioned below.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 8

  1. Encloses enzymes, hormones etc in small vesicles
  2. Gives strength and shape to cells.
  3. Power house of the cell.

Answer:

  1. b. Golgi complex
  2. c. Endoplasmic reticulum
  3. a. Mitochondrion

Question 25.
Complete the labelling of the given diagram of plant cell suitably.

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 18
Answer:
a. Nucleus
b. Centre of protein synthesis in the cell
c. Mitochondrion
d. Power house of the cell

Question 26.
Complete the illustration given below.
Vacuole, Mitochondrion, Endoplasmic Reticulum, Golgi complex
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 17

Answer:
Vacuole – Stores water, salts, and waste materials.

Mitochondrion – Power house of the cell, helps in the production and storage of energy.

Endoplasmic Reticulum – Pathway inside the cell. Provide rigidity and shape to cells.

Golgi Complex – Encloses enzymes, hormones & mucus in small vesicles.

Question 27.
While discussing about the cell Kunchu has a doubt” Are Protoplasm and Cytoplasm one and the same?” What is your response?
Answer:
Protoplasm constitutes all materials within the cell membrane. Cytoplasm is the part of protoplasm except the nucleus.

Question 28.
What is the difference between simple microscopes and electron microscopes?
Answer:
Simple Microscope

  • Magnifies an object 2000 times its actual size
  • Light source is used to see objects Electron Microscope
  • Magnifies objects million times
  • Electrons are utilized instead of light source.

Question 29.
Complete the table showing dif-ferent parts of nucleus and their functions.
Answer:
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 19

Question 30.
“Can organisms live without a definite nucleus”? I can’t believe it. Have you noticed Gopika’s doubt? How will you clarity this doubt?
Answer:
There are certain organisms without a definite nucleus. In bacteria, cyanobacteria and mycoplasma, there is no nucleus. They are called prokaryotes.

But in the cells of Amoeba, animals and plants there is a well defined nucleus covered with membrane. These are known as eukaryotes.

Question 31.
Complete the given flow chart.
Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 20
Answer:
a – Chloroplast
b – Leucoplast
c – Imparts colour to fruits and flowers.

Question 32.
Why do the cell membranes called selectively permeable membranes?
Answer:
Hence they allow only the essential materials to pass through.

Question 33.
How does the colour of fruits change when they ripen?
Answer:
As fruits ripens, the chloroplasts get converted to chromoplasts. Thus colour changes.

Question 34.
Describe the function of Centrosomes.
Answer:
Centrosomes is the cell organelle that is found only in animal cells. The centrioles that play a major role in cell division. It helps to the formation of spindle fibers.

Question 35.
Find out statements which are relate to animal cells from the following statements.
• seen cell wall.
• seen cell membrane.
• seen mitochondrion.
• seen vacuole and chloroplast.
Answer:
• seen cell membrane.
• seen mitochondrion.

Question 36.
How spindle fibres are formed in plant cells and fungus cells?
Answer:
In plant cells and fungus cells spindle fibres are formed through microtubule organising centre.

Question 37.
Why are biomembranes known as selectively permeable membrane?
Answer:
The exchange of materials between cells and between cell organelles and cytoplasm occurs through biomembranes. These membranes do not allow all but some materials to pass. Hence they are selectively permeable membrane.

Question 38.
The discovery of electron microscope enables the detailed study on the cells easy. How?
Answer:
The electron microscope can magnify an object into millions of time of the original size. So it helps to understand more about the micro structures of the cell.

Question 39.
What are unicellular organisms? Write examples for this?
Answer:
Some organisms have only a single cell in their body. Such organisms are called unicellular organisms,
eg: Amoeba, Paramoecium and Euglena.

Question 40.
What are multicellular organisms?
Answer:
There are a number of cells in the body of these creatures. They are therefore know as multicejlular organisms.

Question 41.
What is cell wall?
Answer:
Cell wall is the covering seen outer to the plasma membrane in plants. It is composed of cellulose. It has numerous pores. Movement of molecules is possible through these pores.

Question 42.
How do animal cells differ from the plant cells in structure?
Answer:
Plant cell and animal cell are often structurally same. But the plant cell is surrounded by a rigid protective cell wall formed of cellulose. Cell wall is the major factor by which we can identify the animal cells from plant cells. In plant cell large vacuole and chloroplasts are seen.

Question 43.
Draw the diagrammatic repres-entation of animal cells and plant cells and illustrate the parts?
Answer:
Animal cell

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 7

Plant cell

Kerala Syllabus 8th Standard Basic Science Solutions Chapter 1 life's Mysteries in Little Chambers 12

Kerala Syllabus 10th Standard Social Science Important Questions Chapter 1 Revolutions that Influenced the World

Kerala State Syllabus 10th Standard Social Science Important Questions Chapter 1 Revolutions that Influenced the World

Answer the following. Score 1 each.  

A Choose the correct answer.

Sslc History Chapter 1 Important Questions Question 1.
The sailor who discovered North America
Vasco da Gama
Megallan
Christopher Columbus
Answer:
Christopher Columbus

Question 2.
To which revolution is Boston Tea Party related?
French Revolution
American War of Independence
Russian Revolution
Answer:
American War of Independence

Question 3.
Under whose leadership was the Constitution of USA framed?
James Madison
James Ottis
Thomas Paine
Answer:
James Madison

Question 4.
The Commander-in-Chief of the colonial army against the British army.
Thomas Jefferson
George Washington
James Madison
Answer:
George Washington

Question 5.
The ruler who said: 7 am the State’.
Lotus XVI
Mary Antoinette
Louis XIV
Answer:
Louis XIV

Question 6.
The thinker who said: ‘Man is born free, but everywhere he is in chains’.
Rousseau
Voltaire
Montesquieu
Answer:
Rousseau

Question 7.
The emperor known as ‘Justinian of modern France’.
Louis XVI
Napoleon
Louis XIV
Answer:
Napoleon

Question 8.
The leader known as the ‘Protector’ in the Latin American countries.
Simon Bolivar
Jose De San Martin
Francisco Miranda
Answer:
Jose De San Martin

Sslc Social Science Important Questions Question 9.
The famous Russian writer who wrote the novel ‘The Mother’
Leo Tolstoy
Maxim Gorky
Frederick Engels
Answer:
Maxim Gorky

Question 10.
The leader of October Revolution
Vladimir Lenin
Alexander Kerensky
Nicholas II
Answer:
Vladimir Lenin

Question 11.
To which revolution is Boxer Rebellion related?
Russian Revolution
French Revolution
Chinese Revolution
Answer:
Chinese Revolution

Question 12.
Under whose leadership did China become the People’s Republic of China?
Mao Zedong
Chiang Kaishek
Sun Yat Sen
Answer:
Mao Zedong

Question 13.
The first President of USA
George Washington
James Madison
Thomas Jefferson
Answer:
George Washington

B. Answer in a sentence each.

Question 1.
Who were known as ‘Pilgrim Fathers’?
Answer:
A group of people who exiled to America from the religious persecution of the King of England in the 17th century on a ship called Mayflower and established colonies there were known as Pilgrim Fathers.

Question 2.
Who prepared the American Declaration of Independence?
Answer:

  • Thomas Jefferson
  • Benjamin Franklin

Question 3.
Which were the three Estates in French society?
Answer:

  1. First Estate : Clergy
  2. Second Estate; Nobles
  3. Third Estate : Middle class, farmers and craftsmen

Question 4.
Name the famous thinkers who influenced the French Revolution.
Answer:

  • Voltaire
  • Rousseau
  • Montesquieu

Question 5.
Name the Russian writers who depicted the plight of workers and farmers in their works.
Answer:

  • Maxim Gorky
  • Leo Tolstoy
  • Ivan Turgenev
  • Anton Chekov

Social Science Class 10 Kerala Syllabus Chapter 1 Question 6.
Who were the famous leaders of Bolshevik party?
Answer:

  • Lenin
  • Trotsky

Question 7.
Which were the basic ideals upheld by Dr. Sun Yat Sen?
Answer:

  • Nationalism : to expel the Manchu dynasty and the imperial powers
  • Democracy : to establish democratic rule
  • Socialism: to control capital and distribute land equally

Question 8.
What was Boxer Rebellion? Why is it considered as anti-colonial struggle in China?
Answer:
Some secret organizations in China revolted against the foreign interference and domination in 1900. The emblem of these organizations was the Boxer’s fist. So this was known as the Boxer Rebellion.
It was against foreign dominance and the Manchu dynasty.

Question 9.
Which was the famous slogan of American War of Independence? Who framed it?
Answer:

  • No taxation without representation
  • James Ottis

Question 10.
Name the book of Rousseau which influenced the French Revolution greatly. What was its content?
Answer:

  • Social Contract
  • The government is the outcome of the contract between the ruler and the ruled.

Question 11.
Write the demands put forward by the First Continental Congress.
Answer:

  • The revocation of the regulations enforced on industry and trade.
  • Not to impose taxes without the approval of the people.

Question 12.
Why did the European countries under the – leadership of England organize themselves ” against Napoleon?
Answer:
The European countries feared that the reforms of Napoleon would spread all over Europe. It was the ideals of French Revolution and not Napoleon that they were afraid of.

Answer the following. Score 2 each.

Question 1.
Complete the following table.
Kerala Syllabus 10th Standard Social Science Important Questions Chapter 1 Revolutions that Influenced the World 1
Answer:
Kerala Syllabus 10th Standard Social Science Important Questions Chapter 1 Revolutions that Influenced the World 2

Kerala Syllabus 10th Standard History Chapter 1 Question 2.
Renaissance made significant changes in human thought, views and life. What are they?
Answer:

  • Humanism
  • Scientific temper
  • Spirit of criticism
  • Rational thinking

Question 3.
What did the people demand through the slogan ‘No taxation without representation’ framed by James Ottis?
Answer:

  • Representation in government
  • Membership in British Parliament
  • Self rule
  • Withdrawal of high taxes

Question 4.
Prepare a note on the ideologies of thinkers that stimulated the colonial people to fight against the exploitive laws of the British.
Answer:

Thinkers

Ideologies

John LockeEveryone has some fundamental rights. No government has the right to suspend them.
Thomas PaineThere is something absurd, in supposing a continent (North-America) be perpetually governed by a foreign power (England)

Question 5.
What all things are included in the American Declaration of Independence regarding human rights and freedom?
Answer:

  • All are equal.
  • Everyone has certain rights.
  • Everyone has the right to life, liberty and the pursuit of happiness
  • To secure these rights, governments should be formed with the consent of the governed.

Question 6.
What were the factors that aggravated the financial crisis in France at the time of the revolution?
Answer:

  • The luxurious life and squander of the Bourbon kings, clergy and lords.
  • The frequent wars waged.
  • Frequent spells of drought and crop failure.
  • Financial and military assistance given to American colonies in the American War of Independence.

Question 7.
What is meant by ‘Tennis Court Oath’? Evaluate its significance in the history of France.
Answer:
The members of the Third Estate assembled in a tennis court. They declared themselves as the National Assembly of France. They also swore not to leave until they had framed a constitution for France. This event is known as the ‘Tennis Court Oath’.
The Third Estate became the National Assembly and prepared the constitution.

Question 8.
Name the leaders of Latin American Revolution.
Answer:

  • Francisco Miranda
  • Simon Bolivar
  • Jose De San Martin

Question 9.
What were the demands put forward by the Russian workers?
Answer:

  • Abolish the private property system
  • Everyone has to work
  • All powers be given to workers and people.
  • All means of production should be owned by people.

Question 10.
What is known as the ‘Bloody Sunday’?
Answer:
The workers organized a huge march in Petrograd on 9th January 1905 demanding political rights and economic reforms. The march was fired at by the soldiers and hundreds of demonstrators were massacred. This event is known as the ‘Bloody Sunday’.

Class 10 Social Science Chapter 1 Kerala Syllabus Question 11.
What is meant by ‘Open Door Policy’?
Answer:

  • To acquire trade privileges in China, US State Secretary John Hey made a declaration. This is known as open door policy.
  • As per this policy, America argued for equal rights and opportunities for all countries in Chinese market. The aim of this policy was to create an opportunity for the USA to interfere in China.

Answer the following. Score 3 each.

Question 1.
What is Mercantalism?
Answer:
The British treated the colonies in North America as centres for collecting raw materials for their industry and as market for selling.their products. This policy implemented by the British merchants with the help of their motherland in the American colonies is known as Mercantalism.

Question 2.
What is Boston Tea Party?
Answer:
The high tax levied by the British government on tea fired up strong protest in America. On 16 December 1773, a group of people disguised as the Red Indians, boarded the ships at night in the Boston Harbour and threw 342 chests of tea into the sea. This protest is known as the Boston Tea Party.

Question 3.
Match the statements of the French rulers correctly.

Ruler

Statement

Louis XIVAfter me the deluge
Louis XVIf they can’t eat bread, let them eat cake.
Mary AntoinetteI am the state

Answer:

Ruler

Statement

Louis XIVI am the state
Louis XVAfter me the deluge
Mary AntoinetteIf they can’t eat bread, let them eat cake

Question 4.
Prepare a note on the reign of terror in France.
Or
The period in France from July 1793 to July 1794 is known as reign of terror. Why?
Answer:
In July 1793, a Committee of Public Safety was constituted to control the internal affairs of France under the leadership of Robespierre. Leaders like Mirabeau and Danton were members in it.

Using guillotine, they did mass executions of suspected enemies of the Revolution. A lot of nobles and clergy including Louis XVI and his wife Mary Antoinette were killed. At last Robespierre was also executed. The reign lasted until July 1794 and is known as the Reign of Terror.

Question 5.
Why were the European countries afraid of Napoleon?
Answer:
Napoleon defeated the European alliance which was formed under the leadership of Britain against France. He seized the power in France in 1799. Though an autocrat, he instituted several reforms in France. In France, nationalism strengthened during the time of Napoleon. Napoleon invaded the European countries one by one. The European countries feared that the reforms of Napoleon would spread all over Europe. It was the ideals of French Revolution and not Napoleon that they were afraid of.

Role of Simon Bolivar in Latin American Revolution:
Simon Bolivar took up the leadership of Latin American Revolution from Miranda. He dedicated his life for the freedom of Venezuela, He defeated Spain in many battles after taking up the leadership of Columbian army.

He organized revolutions in Venezuela, Equador, Columbia and Peru and liberated them from Spain. He is reverently called the ‘Liberator’
in these countries. He is also known as the j George Washington of South America.

Question 6.
What were the circumstances that led the Russian workers to make different demands?
Answer:

  • Farmers and factory workers in Russia led a tragic life under the autocracy of the Tsarist emperors who ruled Russia.
  • The low agricultural production affected the farmer’s income. Moreover the landless farmers had to pay huge tax.
  • Though Russia was rich in mineral resources, the industrial production was meagre. It was the foreigners who controlled majority of the industries.

Class 10 History Chapter 1 Kerala Syllabus Question 7.
What were the demands put forward by the Bolsheviks under Lenin against the provisional government? Why did they not approve the provisional government?
Answer:

  • Russia should withdraw from the First World War.
  • Power should be handed over to the Soviets.
  • Seize the lands owned by the lords and distribute them among farmers.
  • Make factories public property.
  • The provisional government rejected all these demands.

Question 8.
The Cabinet under Lenin took some revolutionary measures. What were they?
Answer:

  • Withdrew from the First World War.
  • Seized the land owned by the lords and distributed them among the farmers.
  • Factories, banks, transport facilities and foreign trade were brought under public ownership.

Question 9.
Arrange the items given in the boxes suitably.

Bastille prisonThe Preamble of the French Constitution
States GeneralThe beginning of French Revolution
Tennis Court OathThe symbol of despotism
EstatesSocial classes in France
MontesquieuThinker
Declaration of Rights of ManFrench Parliament

Answer:

  • Bastille prison – The symbol of despotism
  • States General – French Parliament
  • Tennis Court Oath – The beginning of French Revolution
  • Estates – Social classes in France
  • Thinker – Montesquieu
  • Declaration of Rights of Man – The Preamble of the French Constitution

Question 10.
Explain the role of thinkers in France in making the people aware of the inequalities and exploitation.
Or
Explain how the thinkers and their ideologies paved the wav for French Revolution.
Answer:
The French thinkers who influenced the French Revolution were Rousseau, Voltaire and Montesquieu.

Rousseau

  • His statement ‘man is born free but everywhere he is in chains’ spelled the importance of freedom.
  • Declared that the people are the sovereign.
  • Through his book ‘Social Contract’, stated that the government is the outcome of the contract between the ruler and the ruled.

Voltaire

  • Ridiculed the exploitation of the clergy.
  • Promoted rational thinking, ideals of equality and humanism.

Montesquieu

  • Encouraged democracy and republic.
  • Suggested division of powers of the government into legislature, executive and judiciary.

History Chapter 1 Class 10 Kerala Syllabus Question 11.
What is October Revolution? Explain its significance in the history of Russia.
Answer:
In October 1917, the Bolsheviks organized an armed rebellion against the provisional government. Kerensky fled from the country and Russia came under the control of the Bolsheviks. This event, through which the Bolsheviks attained power in Russia, is known as October Revolution.

Significance

  • Russia withdrew from first world war.
  • Land was distributed to farmers.
  • Factories and banks under public ownership.
  • Emergence of Soviet Union.

Question 12.
Analyse the circumstance which led to the transition of the Third Estate of the French Parliament called ‘Commons’ into National Assembly.
Answer:

  • The Commons had no role in the administration.
  • They had to pay tax to government, nobles and clergy.
  • The status of one vote for each estate was against the interests of the Commons.
  • The Commons wanted one vote for each person instead of one vote for each estate.
  • The other two estates opposed this demand.
  • Under this circumstance, the Commons declared themselves as the National Assembly of France and assembled in the nearby tennis court.

Question 13.
What was the social background that forced the Third Estate to struggle against the First and Second Estates of the French Society?
Answer:

  • The Third Estate had no role in administration.
  • Low social status.
  • Paid taxes to government, nobles and clergy.
  • First and Second Estates were exempted from all taxes.
  • The King, nobles and clergy occupied majority of the cultivated land.
  • The nobility lived by exploiting the peasants.

Answer the following. Score 4 each.

Question 1.
Which were the Mercantalist Laws implemented by Britain in the colonies of North America? How did they become a cause for the American War of Independence?
Or
How did Britain control the trade in American colonies through Mercantalist Laws?
Answer:

  • The goods to and from the colonies must be carried only in British ships or ships built in the British colonies.
  • Import tax must be paid for the import of tea, glass and paper.
  • Products of the colonies like sugar, wool, cotton, tobacco, etc. could only be exported to England.
  • British stamp must be affixed on all the legal documents, newspapers, pamphlets and license.
  • The colonial people opposed these exploitations which resulted in American War of Independence.

Question 2.
Write the results of American War of Independence.
Or
The American War of Independence greatly influenced the later history of the world in different ways. Explain how?
Answer:

  • Gave direction and motivation to the later freedom struggles and revolutions all over the world.
  • Put forward the concept of republican form of government.
  • Prepared the first written constitution.
  • Contributed to the concept of federal system that ensured freedom and authority of states in the union.

Question 3.
What were the reforms instituted by Napoleon in France?
Or
What were the changes brought about in the social, economic and political fields of France during the time of Napoleon?
Answer:

  • Farmers were made the owners of land.
  • Prepared a new code of law by codifying the existing laws.
  • Established Bank of France to centralise finance.
  • Formed ‘sinking fund’ with the aim to avoid public debt.
  • Exercised state’s control over the clergy.
  • Constructed several roads for transportation.

Question 4.
Write the features of February Revolution.
Answer:

  • Workers and soldiers joined together against the Tsar.
  • Opposition to Tsarist rule grew strength.
  • Tsar was thrown out of power.
  • A provisional government was formed.

Sslc History Chapter 1 Question 5.
How far were the middle class of France responsible for the French Revolution?
Or
The inequalities and exploitation suffered by the Third Estate was the basic reason for the French Revolution. Substantiate.
Answer:
The French middle class included traders, writers, lawyers, officials, teachers, bankers, farmers and craftsmen.

  • They had no role in the administration.
  • Paid land tax namely, ‘Taille’ to the government.
  • Had low social status.
  • Paid taxes to clergy and nobles.

The middle class were dissatisfied with the prevailing social order. They realised that it was impossible for them to achieve their demands under the existing order. Therefore they decided to overthrow the government by leading the revolution.

Question 6.
Evaluate the results of the Russian Revolution.
Answer:

  • Seized the land of lords and distributed among the peasants.
  • The Russian Revolution transformed Russia from centuries old backwardness and helped her to attain economic, scientific and technological progress.
  • Formation of Union of Soviet Socialist Republic (USSR) by consolidating different Soviet Republics and a new constitution.
  • New constitution came into force in 1924.
  • Spread the socialist ideas all over the world.
  • Gave importance to public sector.
  • Introduced centralized planning.

Question 7.
Evaluate the importance of the American War of Independence in the struggle for freedom by people all over the world.
Answer:
American War of Independence was a historic struggle launched by the English settlers of the colonies of North America against the mother country. It was a mass movement of the people against denial of fundamental rights, denial of freedom, high taxes and denial of representation in the government.

The historic American Declaration of Independence contains many provisions for the enforcement of rights and freedom. Its main provisions were All are equal, Everyone has certain rights, People have the right to change the governments that deny them their rights’.

This declaration became a model for the protection of human right activities. It also put forward the concepts of republican form of government and federal state to the world. Its yet another important noble idea was the concept of a written constitution to rule the nation.

Question 8.
Evaluate the background of the French
Revolution.
Answer:

  • Autocratic rule of the kings.
  • Luxury and extravagance of kings, nobles and clergy.
  • Social inequality – the division of the French society into three estates.
  • Financial crisis due to luxurious life and squander of kings, nobles and clergy, frequent wars by kings, frequent spells of drought and crop failure.
  • Emergence of middle class due to inequality, denial of rights and freedom and exorbitant taxes.
  • Influence of thinkers

Question 9.
Discuss the contributions of French Revolution to subsequent history of the world.
Or
Analyse the influence of the French Revolution on the socio political system that existed in Europe.
Answer:

  • End of feudal system in Europe.
  • Proclaimed that nation is not merely a region, but the people.
  • Contributed the concept of people’s sovereignty.
  • Led to the emergence of nationalism.
  • Helped the growth of the piiddle class.
  • Threatened the autocratic rulers in Europe.
  • Stimulated all the later revolutions in the world.
  • Spread the ideas of liberty, equality and fraternity.

Question 10.
Arrange the table suitably.

A

B

NapoleonBolshevik
Sun Yat SenCommon Sense
LeninWaterloo
Thomas PaineKuomintang

Answer:

A

B

NapoleonWaterloo
Sun Yat SenKuomintang
LeninBolshevik
Thomas PaineCommon Sense

Question 11.
Compare the influence of Russian Revolution with that of the French Revolution.
Answer:

Russian Revolution

French Revolution

End of despotism of kingsEnd of despotism of kings
End of feudal systemEnd of feudal system
Put an end to class divisionPut an end to social division
Ownership by workersSovereignty of people
Emergence of SovietsEmergence of middle class
Mother of socialist movementsMother of revolutions
Spread of socialist ideologySpread of republican ideology and human rights

Social Science Class 10 Chapter 1 Question Answer Question 12.
Compare the February Revolution and the October Revolution.
Answer:

February Revolution

October Revolution

Led by MensheviksLed by Bolsheviks
Russia did not withdraw from the First World WarRussia withdrew from the First World War
Put an end to Tsarist regimePut an end to Menshevik rule
Lords continued as owners of landLand was distributed among farmers
Did not give importance to public ownershipFactories, banks, etc. were brought under public ownership

Question 13.
“When France sneezes, the rest of Europe catches cold : Metternich”
This was the remark made by Austrian Chancellor Metternich about the impact of French Revolution. Based on this remark, analyse how French Revolution influenced France and Europe.
Answer:
French Revolution is considered as the mother of all revolutions. The ideals of French Revolution namely liberty, equality and fraternity influenced and gave strength to democratic believers of not only Europe, but the whole world. Following the French Revolution, Napoleon became the ruler of France and he introduced many reforms stimulated by the ideals and aims of French Revolution.

The European countries under the leadership of England tried to restore the old political structure and social order. The triumph of the slogans of liberty, equality and fraternity frightened them. Besides England, European countries like Russia, Prussia’ and Austria had to ensure that similar revolutions as that of France should not take place in their countries.

France became the centre of attraction to Europe following the revolution. New ideas and ideologies always came from France and it used to influence the rest of Europe.

Question 14.
Prepare a note on the background of Latin American Revolution.
Answer:
The Spanish and Portuguese

  • Looted the resources and wealth of the people.
  • Enslaved the natives to work.
  • Used the people and resources of the colonies for the comfort of the chiefs of colonies.
  • Destroyed the indigenous Latin American culture.
  • Implemented European culture and life style in the colonies.
  • Racial discrimination towards the colonial people.
  • Imposed trade regulations.

Question 15.
Write the features of the reforms introduced by Dr. Sun Yat Sen in China.
Answer:

  • Nationalism : To end monarchy and imperialist rule
  • Establish democratic rule
  • Socialism
  • Opposition to foreign domination
  • Measures for the progress of agriculture and industry
  • Co-operation with the Communists

Question 16.
Evaluate the changes that occurred in the administrative structure of France following the French Revolution.
Answer:

  • End of Feudalism.
  • Uniform system of administration in France.
  • Equal taxation over all sections of people.
  • Abolished the special privileges enjoyed by higher Estates.
  • Church was brought under the control of the state.

Question 17.
Identify the correlation between the two items in section A and similarly complete section B.
i. A : Thomas Paine : American War of Independence
B : Rousseau : ……………….

ii. A : Jose De Sanmartin : Latin American Revolution
B : Mao Zedong : …………………

iii. A . October Revolution : Russian Revolution
B : Long March : …………….

iv. A : French Revolution : 1789
B : Russian Revolution – : ……………….
Answer:
i. A : Thomas Paine : American War of Independence
B : Rousseau : French Revolution

ii. A : Jose De Sanmartin : Latin American Revolution
B : Mao Zedong : Chinese Revolution

iii.A : October Revolution : Russian Revolution
B : Long March : Chinese Revolution

iv.A : French Revolution : 1789
B : Russian Revolution : 1917

Answer the following. Score 5, 6 each.

Question 1.
Prepare a description showing the causes that led to the American War of Independence.
Answer:
By the 18th century, England had established 13 colonies in the eastern coast of North America. The British treated these colonies as centres for collecting raw materials for their industry and as market for selling their products. The policies of the mother country England led to the freedom struggle in the colonies.

Mercantalist Laws:

  • The policy adopted by the British merchants with the help of their motherland in the American colonies is known as Mercantilism.
  • The goods to and from the colonies must be carried only in British ships or ships built in the colonies.
  • British stamps must be affixed on all the legal documents, newspapers, pamphlets, license, etc.
  • Import tax must be paid for the import of tea, glass, paper, etc.
  • Products of the colonies like sugar, wool, cotton, tobacco, etc. could only be exported to England.
  • Colonies must provide food and quarters for the British troops which were maintained in the colonies.

Influence of thinkers:
The ideologies of thinkers like John Locke and Thomas Paine stimulated the people of the colonies to fight against the exploitative laws of the British. John Locke said : ‘Everyone has some fundamental rights. No government has the right to suspend them.’ Thomas Paine said : ‘There is something absurd, in supposing a continent (North America) be perpetual ly governed by a foreign power (England).

No taxation without representation:
This was a slogan framed by James Ottis that thundered along different parts of North America in the 18th century. The slogan showed the desire for freedom and the hatred towards the mother country nurtured by the colonial people.

Boston Tea Party:
The high tax levied by the English government on tea fired up strong protest in America. On 16 December 1773, a group of people disguised as the Red Indians, boarded the ships at night in the Boston Harbour and threw 342 chests of tea into the sea. This protest is known as the Boston Tea Party.

Continental Congress:
The delegates of all the colonies except Georgia met at Philadelphia in 1774 to protest against the policies and rules imposed by England. It is known as the First Continental Congress. Subsequently, people of the colon ies submitted a petition to the king of England.

They demanded the revocation of the regulations enforced on industry and commerce and not to impose tax without the approval of the people. But the king sent a military force to suppress the people. This led to the war between England and the colonies. The Second Continental Congress held at Philadelphia in 1775 elected George Washington as the Commander-in-chief of the Continental Army.

The American Continental Congress issued the famous Declaration of Independence on 4 July 1776: The war between England and the colonies in North America that began with the Declaration of Freedom, ended in 1781.

By the Treaty of Paris in 1783, England ratified the freedom of 13 colonies. George Washington became the first President of USA. The United States of America came into being with the first written constitution in the world.

History Class 10 Chapter 1 Kerala Syllabus Question 2.
Assess the reasons that led to the French Revolution.
Or
Explain the causes of the French Revolution on the basis of the following hints.
i) Despotic rulers
ii) Influence of thinkers
iii) Social inequality
iv) Financial crisis
Answer:
The French Revolution of 1789 was the movement against centuries old despotic kingship and feudal order in France. The aim of the Revolution was to rebuild France on the basis of the ideals of Liberty, Equality and Fraternity. The following are the causes for the Revolution.

i) Despotic rule of kings:
It was the despotic misrule of the Bourbon kings who ruled France in the 17th and 18thcenturies that led mainly, to the revolution. These kings believed in the divine right of kingship. Louis XIV who ruled France in the 17th centuiy said: ‘I am the state’.

He also said ‘God has given power to the king over his subjects and only God has the authority to question him’. Louis XV of France had the attitude ‘After me, the deluge’. The kings who concentrated all powers led a life of squander and extravagance. The majority of the people of France led a life of misery. A minority including the rulers led a life of luxury and extravagance.

iii) Social inequality:
The social inequality that prevailed in France was another reason for the Revolution. In French society, the people did not have equal rights. The French society was divided into three strata known as Estates.

The clergy formed the First Estate. They held vast land. They collected the tax called ‘Tithe’ from farmers. The clergy were exempted from all taxes. They also controlled higher positions in administration. The nobility formed the Second Estate. They engaged in military service. Collected various taxes from farmers. Made farmers work without wages. Exempted from taxes. Led luxurious life and held vast lands.

The Third Estate consisted of the middle class including traders, writers, lawyers, officials, teachers, bankers, farmers and craftsmen. They had no role in the administration. Paid land tax namely ‘Taille’ to the government. Paid taxes to clergy and nobles and had low social status.

ii) Influence of thinkers :
Various thinkers in France played an important role in making the people aware of the inequalities and exploitation.

Rousseau

  • Spelled out the importance of freedom with the statement ‘Man is bom free but everywhere he is in chains’.
  • Declared that the people are sovereign.

Voltaire

  • Ridiculed the exploitation of clergy.
  • Promoted rational thinking, ideals of equality and humanism.

Montesquieu

  • Encouraged democracy and the republic.
  • Suggested division of powers of the government into legislature, executive and judiciary.

iv) Financial crisis:
The following brought France to the brink of bankruptcy:

  • The luxurious life and squander of Bourbon kings, nobles and clergy.
  • Frequent wars
  • Frequent spells of drought and crop failure.
  • Financial and military assistance given to American colonies in theAmerican War of Independence.

Summoning of States General:
To overcome the financial crisis, Louis XVI, the king decided to propose new taxes. So to levy new taxes upon the commoners, Louis XVI summoned the French Parliament called States General.

Similar to the French society, the States General also had three estates. The first two Estates argued for Estate wise single voting system. The Third Estate demanded individual vote for each member of all the three estates.

Tennis Court Oath:
While the arguments went
on, the members of the Third Estate declared themselves as the National Assembly of France. They assembled in the tennis court nearby and swore not to leave until they had framed a new constitution for France. This event is known as the ‘Tennis Court Oath’. This was the beginning of the French Revolution.

Question 3.
Analyse how the February Revoltuion and October Revolution led to the formation of Soviet Union.
Or
Explain the causes of the Russian Revolution on the basis of the following indicators.
• Autocratic rulers
• Influence of marxists ideology
• Pathetic life of people
• Role of Bolshevik party
Or
Explain the circumstances that led Bolsheviks into power in Russia.
Answer:
Russian Revolution of 1917 has greatly influenced the world. Its causes are discussed below.

Autocracy of Tsars:
Russia was ruled by emperors known as Tsars. They were despots. Farmers and factory workers in Russia led a tragic life under the autocracy of the Tsarist emperors.

Economic causes:
The low agricultural production affected the farmer’s income. Moreover, the landless farmers had to pay huge tax. Though Russia was rich in natural resources, their industrial production was meagre.

Influence of thinkers and writers:
Writers like Maxim Gorky and Leo Tolstoy depicted the plight of the workers and farmers in their works. The Marxist ideologies formulated by Karl Marx and Frederick Engels stirred the workers. They called for establishing the supremacy of the workers instead of that by the capitalists.

Social Democratic Party:
Trade unions were formed to find remedies for the plight of the workers. Based on the Marxist ideologies, the Social Democratic Workers Party was formed. Later this party was split into the Mensheviks and the Bolsheviks. The main leaders of the Bolsheviks were Lenin and Trotsky. Alexander Kerensky led the Mensheviks.

Rousseau – Japanese war:
The crisis reached its worst when Russia was defeated in the Russia- Japan war in 1905. The workers organized a huge march at Petrograd on 9 January 1905 demanding political rights and economic reforms. The march was fired at by the soldiers and hundreds of demonstrators were massacred. This event is known as the Bloody Sunday. Workers’ organizations called ‘Soviets’ were formed all over Russia to conduct strikes.

February Revolution:
When strikes gained strength, the emperor Tsar Nicholas II formed a legislative assembly called Duma’. When the First World War started in 1914, the Tsar decided to participate in it, ignoring the protests from Duma. A lot of Russian soldiers were killed in this war.

Food shortage became severe by 1917. In February 1917, workers and soldiers joined together and captured Petrograd. Tsar Nicholas II was thrown out of power. A provisional government was formed under Alexander Kerensky, the Menshevik leader. This event is known as Februaiy Revolution.

Lenin and Bolshevik Party:
A group of Soviets did not approve the provisional government. Lenin, the leader of the Bolsheviks argued that the entire power should be transferred to the Soviets if they were to realize the aims of the revolution. The Bolsheviks and the Soviets supported him. They propagated that only a proletarians’ government could eradicate centuries old economic backwardness and inequality. They put forward demands like:

  • Withdraw Russia from the First World War.
  • Seize the lands owned by the lords and distribute them to farmers.
  • Make factories public property.

October Revolution:
In October 1917, the Bolsheviks started an armed rebellion against the provisional government. Kerensky fled from the country. Russia came under the control of the Bolsheviks. This event in which the Bolsheviks attained power is known as the October Revolution. A proletarians’ government came under the control of Lenin in Russia.

Class 10 History Chapter 1 Question 4.
Evaluate the role of Dr. Sun Yat Sen and Mao Zedong in the Chinese Revolution.
Or
Describe the establishment of the People ’s Republic of China by analyzing different phases of the Chinese Revolution.
Answer:
The Chinese Revolutions were organised under the leadership of Dr. Sun Yat Sen in 1911 and in the 1930’s and 1940’s under the leadership of Mao Zedong. The reasons are discussed below.

In the 20th century, China witnessed revolutions against foreign rule and monarchy. The colonial powers of Europe and America adopted policies in China, different from the ones they had implemented in the colonies of other parts of the world. They were opium trade and open door policy. As a result, many parts of China came under tile control of imperial powers. Chinese trade and industries were controlled by foreigners.

Role of Sun Yat Sen:
The Manchu dynasty that ruled China favoured the foreign interference and domination. In 1911, a revolution took place under the leadership of Sun Yat Sen against the Manchu dynasty. This ended monarchy in China. After the revolution, Kuomintang Party established a republican government in Southern China under the leadership of Sun Yat Sen.

He gave importance to ideologies like nationalism, democracy and socialism. He advocated to nullify the unjust treaties signed with the foreign countries and wanted to maintain equality with the western countries. The Kuomintang republic adopted measures for the progress of agriculture and industry.

The Chinese Communist Party was formed during this time. In the beginning, the Kuomintang and the Communists co-operated with each other. But this co-operation was disturbed when Chiang Kaishek became the head of the republic following the death of Sun Yat Sen.

Role of Mao Zedong:
Mao Zedong was the leader of Chinese Communist Party. Chiang Kaishek ascertained military autocracy in China. He gave opportunity for foreign powers including America to freely interfere in China and did not co-operate with the Communists. Coal and industries, banking and foreign trade were all controlled by foreign countries. The Communists protested against the policy of Chiang. They were brutally suppressed.

At that time, Mao Zedong rose to the leadership of the Communist Party. In 1934, under the leadership of Mao Zedong, an adventurous journey started from Kiangsi in South China to Yanan in North Western China. Throughout the journey, they seized out agricultural lands and villages from lords and distributed them among the farmers.

The journey covered around 12000 kms. So it is known as the ‘Long March’. Mao Zedong and the Communist Party became the symbol of the struggle of the Chinese against foreign power. The Red Army under the leadership of Mao Zedong captured Nanking, the capital of Kuomintang government in April 1949. So Chiang-Kaishek had to seek political asylum in Taiwan. China became the People’s Republic of China on 1st October 1949 under the leadership of Mao Zedong.

Question 5.
Explain the causes of the Latin American Revolution and the consequent liberation from ‘ European domination.
Answer:
Latin America consists of South and Middle America and the West Indies Islands. Spain and Portugal established colonies throughout these regions. So the people of Latin America rose in revolt against the imperialist powers. The reasons for the revolt are discussed below.

Policies of Imperial Powers
The Europeans looted the wealth and resources of Latin American countries and even erased their traditional culture. After colonizing the entire Latin America, the Spanish and the Portuguese propagated their language, religion and customs there. They built houses and churches in Spanish style.

Several schools were established for imparting Spanish system of education. The Spanish farming methods and crops were adopted. Racial discrimination towards the people of the colonies was enforced in all walks of life.

The Europeans took away gold, silver, etc. from the Latin American colonies. They never allowed the people of the colonies to engage in trade with countries other than Spain. The things produced in Spain were not permitted to be produced in the colonies. People had to work in hazardous environment in mines without any safety measures. The Europeans enslaved the natives to work in plantations.

Influence of other revolutions:
The success of American War of Independence and the French Revolution inspired the Latin American people to fight against the European domination.

Leaders of the revolution:
Francisco Miranda, Simon Bolivar and Jose De San Martin were the leaders of Latin American Revolution. Under their leadership, Latin American colonies were liberated. By 1825 almost all the Latin American colonies became free.

Question 6.
Choose the most suitable items related to Column A from Columns B. and C and arrange the table.

A

B

C

American War of IndependenceSimon BolivarSocial Contract
French RevolutionLeninLong March
Latin American RevolutionGeorge WashingtonSecond Continental Congress
Chinese RevolutionRousseauOctober Revolution
Russian RevolutionMao ZedongSpain

Answer:

A

B

C

American War of IndependenceGeorge WashingtonSecond Continental Congress
French RevolutionRousseauSocial Contract
Latin American RevolutionSimon BolivarSpain
Chinese RevolutionMao ZedongLong March
Russian RevolutionLeninOctober Revolution

Std 10 History Chapter 1 Question 7.
Match the columns suitably.

A

B

Maxim GorkyThe Ten Days that Shook the World
Pablo NerudaThe Mother
Thomas PaineBattleship Potemkin
John ReedHeights of Macchu Picchu
Sergei EisensteinCommon Sense

Answer:

A

B

Maxim GorkyThe Mother
Pablo NerudaHeights of Macchu Picchu
Thomas PaineCommon Sense
John ReedThe Ten Days that Shook the World
Sergei EisensteinBattleship Potemkin

Question 8.
Prepare a Timeline including the following events.
Peoples’ Republic of China – 1815
Boston Tea Party – 1900
February Revolution – 1949
Battle of Waterloo – 1773
Boxer Rebellion – 1917
Kerala Syllabus 10th Standard Social Science Important Questions Chapter 1 Revolutions that Influenced the World 3

Question 9.
Arrange the following events in chronological order.
i) Tennis Court Oath
ii) First Continental Congress
iii) October Revolution
iv) Formation of Chinese Republic
v) Declaration of American Independence
Answer:
i) First Continental Congress (1774)
ii) Declaration of American Independence (1776)
iii) Tennis Court Oath (1789)
iv) Formation of Chinese Republic (1911)
v) October Revolution (1917)

Additional Information / Extra Reading

Bastille Fort:
Bastille was a medieval fortress on the east side of Paris, built by the end of the 14th century. The first stone of fortress was laid on April 22, 1370 during the reign of Charles V of France. When completed, it had 8 towers 30 metres high, linked by walls of equal height and surrounded by a moat more than 24 metres wide. Cardinal Richelieu was the first to use Bastille as a state prison.

Later Bastille was considered a symbol of the despotism of the ruling Bourbon dynasty. It was a rumour that Bastille had great storage of arms and ammunitions that forced the Parisian mob to attack it. On the morning of July 14,1789 when only seven prisoners were in the jail, a crowd stormed Bastille prison. Since Bastille was considered as the symbol of royal power, 14th July, the day of its destruction is celebrated as National Day in France.

Jacobins and Girondins:
Jacobins and Girondins were two prominent political parties in the French National Assembly. Jacobins were extreme revolutionaries and Girondins were moderates. The Jacobins wanted to make France a Republic by overthrowing monarchy. The Jacobin Society formed in 1789 had leaders like Marat, Danton and Roberspierre.

The leaders of Girondins were from Girond, the south west part of France. They aimed at making France a republic by abolishing monarchy. The main leaders of Girondins were Brissot, Condorcet and Petion.

Tipu Sultan and French Revolution:
The French Revolution had echoes in India when the ruler of Mysore Tipu Sultan declared his solidarity with the Revolution and planted the Tree of Liberty in his Capital Srirangapattanam. He also became a member of French Club Jacobin.

Bourgeoisie:
The word bourgeoisie originated from the French word bourgeois which means dwellers in cities. The feudal society was divided into two: lords and serfs. A new class of traders emerged between the two. This middle class was called bourgeoisie. They amassed great wealth through trade and industry. So they were the founders of capitalism.

Ten Days that Shook the World:
Ten Days that Shook the World is a book written by American journalist and socialist John Reed about the October Revolution in Russia in 1917.
Winter was coming on .. the terrible Russian winter …. on the freezing front miserable armies continued to starve and die, without enthusiasm. The railways were breaking down, food lessening, factories closing. The desperate masses cried out that the bourgeoisie was sabotaging the life of the people ………. The internal affairs of the country becoming miserable…. Lenin asked the people to raise the slogan ‘peace and food for Soviets’ and wanted to usurp power. Following this, the Bolsheviks captured Russian State Bank, Telephone Exchange and the other government institutions ………

Kerala Syllabus 10th Standard Social Science Important Questions

Plus Two Chemistry Notes Chapter 3 Electrochemistry

Students can Download Chapter 3 Electrochemistry Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Notes Chapter 3 Electrochemistry

Electrochemistry-
branch of chemistry which deals with the inter-relationship between electrical energy and chemical changes.

Electrolysis – The chemical reaction occuring due to the passage of electric current (i.e., electrical energy is converted into chemical energy).

Electrochemical reaction –
The chemical reaction in which electric current is produced (i.e., chemical energy is converted into electrical energy). Example: Galvanic cell

Electrochemical Cell: – (Galvanic Cell/Voltaic Cell) :
It converts chemical energy into electrical energy during redox reaction, e.g. Daniell Cell
The cell reaction is
Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s)
It has a potential equal to 1.1 V.
3 Electrochemistry
If an external opposite potential is applied in the Daniell ce|l, the following features are noted:
a) When Eext < 1.1 V,
(i) electrons flow from Zn rod to Cu rod and hence current flows from Cu rod to Zn rod.
(ii) Zn dissolves at anode and Cu deposits at cathode.

b) When Eext= 1.1 V,
(i) No flow of electrons or current,
(ii) No chemical reaction.

c) When Eext > 1.1 V
(i) Electrons flow from Cu to Zn and current flows from Zn to Cu.
(ii) Zn is deposited at the Zn electrode and Cu dissolves at Cu electrode.

Plus Two Chemistry Notes Chapter 3 Electrochemistry

Galvanic Cells :
In this device, the Gibbs energy of the spontaneous redox reaction is converted into electrical work.

The cell reaction in Daniell cell is a combination of the following two half reactions:

  1. Zn(s) → Zn2+(aq) + 2 \(\overline { e } \) (oxidation half reaction/ anode reaction)
  2. Cu2+(aq) + 2 \(\overline { e } \) → Cu(s) (reduction half-reaction/ cathode reaction)

These reactions occur in two different vessels of the Daniell cell. The oxidation half reaction takes place at Zn electrode and reduction half reaction takes place at Cu electrode. The two vessels are called half cells or redox couple. Zn electrode is called oxidation half cell and Cu electrode is called reduction half cell. The two half-cells are connected externally by a metallic wire through a voltmeter and switch. The electrolyte of the two half-cells are connected internally through a salt bridge.

Salt Bridge :
It is a U-shaped glass tube filled with agar-agar filled with inert electrolytes like KCl, KNO3, NH4NO3.

Functions of Salt Bridge :

  1. It maintains the electrical neutrality of the solution by intermigration of ions into two half-cells.
  2. It reduces the liquid-junction potential.
  3. It permits electrical contact between the electrode solutions but prevents them from mixing.

Electrode potential –
potential difference developed between the electrode and the electrolyte. According to IUPAC convention, the reduction potential alone is called electrode potential and is represented as \(E_{M^{n+} / M}\)

Standard Electrode Potential :
The electrode potential understandard conditions, (i.e., at 298 K, 1 atm pressure and 1M concentrated solution) is called standard electrode potential. It is represented as EΘ.

Representation of a Galvanic Cell :
A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge.

For example, the Galvanic cell can be represented as,
Zn (s)|Zn2+(aq)||Cu2+(aq)|Cu(s)

Cell Potential or EMF of a Cell :
The potential difference between the two electrodes of a galvanic cell is called cell potential (EMF) and is measured in volts.
EMF = Ecell = Ecathode – Eanode = ERjght – ELeft
Consider a cell, Cu(s) | Cu22+ (aq) || Ag+ (aq) | Ag(s)
Ecell = Ecathode – Eanode = EAg+/Ag – ECu2+/Cu

How to calculate emf of a cell when concentration varies from standard conditions.

Measurement of Electrode Potential using Standard Hydrogen Electrode (SHE)/Normal Hydrogen Electrode :
SHE or NHE consists of a platinum electrode coated with platinum black. The electrode is dipped in an exactly 1 M HCl solution and pure H2 gas at 1 bar is bubbled through it at 298 K. The electrode potential is arbitrarily fixed as zero at all temperatures.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 2

Representation of SHE/NHE :
When SHE acts as anode:
Pt(s), Hsub>2(g, 1 bar) / H+(aq, 1 M)
When SHE acts as cathode:
H+(aq, 1 M)/H2(g, 1 bar), Pt(s)

Plus Two Chemistry Notes Chapter 3 Electrochemistry

Electrochemical Series/Activity series :
The arrangement of various elements in the increasing or decreasing order of their standard electrode potentials.

Applications of Electrochemical Series:
1. To calculate the emf of an electrochemical cell – The electrode with higher electrode potential is taken as cathode and the other as anode.
\(E_{\mathrm{cell}}^{\Theta}=E_{\mathrm{cathode}}^{\Theta}-E_{\mathrm{anodo}}^{\Theta}\)

2. To compare the reactivity of elements – Any metal having lower reduction potential (electode potential) can displace the metal having higher reduction potential from the solutions of their salt, e.g. Zn can displace Cu from solution.
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

3. To predict the feasibility of cell reactions -If EMF is positive, the cell reaction is feasible and if it is negative the cell reaction is not feasible.

4. To predict whether H2 gas will be evolved by reaction of metal with acids – All the metals which have lower reduction potentials compared to that of H2 electrode can liberate H2 gas from acids.

5. To predict the products of electrolysis.

Nernst Equation :
It gives a relationship between electrode potential and ionic concentration of the electrolyte. For the electrode reaction,
Mn+ (aq) + n \(\overline { e } \) → M(s)
the electrode potential at any concentration measured with respect to SHE can be represented by,
Plus Two Chemistry Notes Chapter 3 Electrochemistry 3
R = gas constant (8.314 J K-1 mol-1), T=temperaturein kelvin, n = number of electrons taking part in the electrode reaction, F = Faraday constant (96487 C mol-1)

By converting the natural logarithm to the base 10 and subsitituting the values of R(8.314 J K-1 mol-1),T (298 K) and F (96487 C mol-1) we get,
Plus Two Chemistry Notes Chapter 3 Electrochemistry 4

Nernst Equation for a Galvanic Cell :
In Daniell cell, the electrode potential for any concentration of Cu2+ and Zn2+ ions can be written as,
Plus Two Chemistry Notes Chapter 3 Electrochemistry 5
Converting to natural logarithm to the base 10 and substituting the values of R, F and T=298 K, it. reduces to
Plus Two Chemistry Notes Chapter 3 Electrochemistry 6
Consider a general electrochemical reaction,
Plus Two Chemistry Notes Chapter 3 Electrochemistry 7

Equilibrium Constant and Nernst Equation:
Plus Two Chemistry Notes Chapter 3 Electrochemistry 8
where Kc is the equilibrium constant.

Electrochemical Cell and Gibbs Energy of the Reaction (∆rG):
Plus Two Chemistry Notes Chapter 3 Electrochemistry 9

Conductance of Electrolytic Solutions: .Conductors:
A substance which allows the passage of electricity through it. Conductor are classified as,

Metallic or Electronic Conductors:
In these the conductance is due to the movement of electrons and it depends on:

  1. The nature and structure of the metal
  2. Number of valence electrons per atom
  3. Temperature (it decreases with increase in temperature)
    e.g. Ag, Cu, Al etc.

ii. Electrolytic Conductors
Electrolytes – The substances which conduct electricity either in molten state or in solution, e.g. NaCl, NaOH, HCl, H2SO4 etc. The conductance is due to the movement of ions. This is also known as ionic conductance and it depends on:

  1. Nature of the electrolyte
  2. Size of the ions and their solvation
  3. Nature of the solvent and its viscosity
  4. Concentration of the electrolyte
  5. Temperature (it increases with increase in temperature)

Ohm’s law – It states that the current passing through a conductor (I) is directly proportional to the potential difference (V) applied.
i.e., I ∝ V or I = \(\frac{V}{R}\)
where R – resistance of the conductor- unit ohm. In SI base units it is equal to kg m²/s³ A²

Plus Two Chemistry Notes Chapter 3 Electrochemistry

The electrical resistance of any substance/object is directly proportional to its length T, and inversly proportional to its area of cross section ‘A’.
R ∝ \(\frac{\ell}{\mathrm{A}}\) or R = ρ\(\frac{\ell}{\mathrm{A}}\) where,

ρ – (Greek, rho) – resistivity/specific resistance – SI unit ohm metre (Ω m) or ohm cm (Ω cm).

Conductance (G):
inverse or reciprocal of resistance (R).
\(G=\frac{1}{R}=\frac{A}{\rho \ell}=\kappa \frac{A}{\ell}\)
where K = \(\frac{1}{\rho}\) called conductivity or specific conductance (K – Greek, kappa)

SI unit of conductance – S (siemens) or ohm-1.
SI unit of conductivity – S m-1
1 S cm-1 = 100 S m-1

Molar Conductance of a Solution (Λm):
It is the conductance of the solution containing one mole of the electrolyte when placed between two parallel electrodes 1 cm apart. It is the product of specific conductance (K) and volume (V) in cm³ of the solution containing one mole of the electrolyte.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 10
where M is molarity of the solution.
Unit of Λm is ohm’1 cm2 mol’1 Or S cm² mol-1
Λm = \(\frac{K}{C}\) [C-Concentration of the solution.]

Measurement of the Conductivity of Ionic Solutions :
The measurement of an unknown resistance can be done by Wheatstone bridge. To measure resitance of the electrolyte it is taken in a conductivity cell. The resistance of the conductivity cell is given by the equation.
\(R=\rho \frac{\ell}{A}=\frac{1}{\kappa A}\)
Plus Two Chemistry Notes Chapter 3 Electrochemistry 11
The quantity \(\frac{\ell}{\mathrm{A}}\) is called cell constant and isdenoted A by G*. It depends on the distance (/) between the electrodes and their area of cross-section (A).

Variation of Conductivity and Molar Conductivity with Concentration :
Conductivity (K) always decreases with decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity (Λm) increases with decrease in concentration. This is because the total volume, V of the solution containing one mole of electrolyte also increases.

The variation of molar conductance is different for strong and weak electrolytes,

1. Variation of Λm with Concentration for Strong Electrolytes:
The molar conductance increases slowly with decrease in concentration (or increase in dilution) as shown below:
Plus Two Chemistry Notes Chapter 3 Electrochemistry 12
There is a tendency for Λm to approach a certain limiting value when concentration approaches zero i. e., dilution is infinite. The molar conductance of an electrolyte when the concentration approaches zero is called molar conductance at infinite dilution, Λm or Λ°m. The molar conductance of strong electrolytes obeys the relationship.
Λm = Λ°m -AC1/2 where C = Molar concentration, A = constant for a particular type of electrolyte.
This equation is known as Debye-Huckel-Onsagar equation.

2. Variation of Λm with Concentration for Weak Electrolytes :
For weak electrolytes the change in Λm with dilution is due to increase in the degree of dissociation and consequently increase in the number of ions in total volume of solution that contains 1 mol of electrolyte. Here, Λm increases steeply on dilution, especially near lower concentrations.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 13
Thus, the variation of Λm with √c is very large so that we cannot obtain molar conductance at infinite dilution Λ°m by the extrapolation of the graph.

Kohlrausch’s Law:
The law states that, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar ionic conductivities of the cations and anions at infinite dilution.
Λ°m = γ+ λ°+ + γ λ°
λ°+ and λ° are the molar conductivities of cations and anions respectively at infinite dilution, Y+ and V. are number of cations and anions from a formula unit of the electrolyte.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 14

Applications of Kohlaransch’s Law
1) To calculate Λ°m of weak electrolytes

2) To calculate degree of dissociation of weak electrolytes
\(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)

3) To determine the dissociation constant of weak electrolytes
Plus Two Chemistry Notes Chapter 3 Electrochemistry 15

Electrolytic Cell and Electrolysis:
In an electrolytic cell, external source of voltage is used to bring about a chemical reaction. Electrolysis is the phenomenon of chemical decomposition of the electrolyte caused by the passage of electricity through its molten or dissolved state from an external source.

Plus Two Chemistry Notes Chapter 3 Electrochemistry

Quantitative Aspects of Electrolysis
Faraday’s Laws of Electrolysis First Law:
The amount of any substance liberated or deposited at an electrode is directly proportional to the quantity of electricity passing through the
electrolyte.
w α Q where ‘Q’ is the quantity of electric charge in coulombs.
w = ZQ .
w = Zlt
(∵ Q = It) where T is the current in amperes , ‘t’ is the time in seconds and ‘Z’ is a constant called electrochemical equivalent.

Second Law:
The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 16
The quantity of electricity required to liberate/deposit 1 gram equivalent of any substance is called Faraday constant ‘F’.
1 F = 96487 C/mol ≈ 96500 C/mol

Products of Electrolysis:
It depend on the nature of the material being electrolysed and the type of electrodes being used.

Electrolysis of Sodium Chloride:
When electricity is passed through molten NaCl, Na is deposited at the cathode and Cl2 is liberated at the anode.
Na+(aq) + \(\overline { e } \) → Na(s) (Reduction at cathode)
Cl(aq) → ½ Cl2(g) + \(\overline { e } \) (Oxidation at anode)

When concentrated aqueous solution of NaCl is electrolysed, Cl2 is liberated at anode, but at cathode H2 is liberated instead of Na deposition due to the high reduction potential of hydrogen.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 17
The resultant solution is alkaline due to the formation of NaOH.

Electrolysis of CuSO4 :
When aqueous CuSO4 solution is electrolysed using Pt electrodes, Cu is deposited at the cathode and O2 is liberated at the anode.
Cu2+(aq) + 2 \(\overline { e } \) → Cu(s) (at cathode)
H2O(l) → 2H+(aq) + 1/2 O2(g) + 2 \(\overline { e } \) (at anode)

If Cu electrode is used, Cu is deposited at cathode and an equivalent amount of Cu dissolves in solution from the anode (because oxidation potential of Cu is higherthan that of water).
Cu2+(aq) + 2 \(\overline { e } \) → Cu(s) (at cathode)
Cu(s) → Cu2+(aq) + 2\(\overline { e } \) (atanode)

Commercial Cells (Batteries)
The electrochemical cells can be used to generate electricity. They are two types:
i) Primary Cells:
Cells in which the electrode reactions cannot be reversed by external energy. These cells cannot be recharged, e.g. Dry cell, Mercury cell.

ii) Secondary Cells :
Cells which can be recharged by passing current through them in the opposite direction so that they can be used again.
e.g. Lead storage battery, Nickel-Cadmium cell.

Primary Cells
a) Dry Cell:
Anode – Zn container
Cathode – Carbon (graphite) rod surrounded by powdered MnO2 and carbon.
Electrolyte – moist paste of NH4Cl and ZnCl2
The electrode reactions are :
Anode : Zn → Zn2+ + 2 \(\overline { e } \)
Cathode: MnO2 + NH4+ + \(\overline { e } \) → MnO(OH) + NH3
Dry cell has a potential of nearly 1.5 V.

Plus Two Chemistry Notes Chapter 3 Electrochemistry

b) Mercury Cell:
Anode – Zn amalgam (Zn/Hg)
Cathode – paste of HgO and carbon
Eelectrolyte – paste of KOH and ZnO. The electrode reactions are,
Anode : Zn/Hg + 2OH → ZnO(s) + H2O + 2 \(\overline { e } \)
Cathode : HgO + H2O + 2 \(\overline { e } \) → Hg(l) + 2 OH
Overall reaction : Zn/Hg + HgO(s) → ZnO(s)+ Hg(l)
The cell potential = 1.35 V

2. Secondary Cells
a) Lead Storage Battery :
Anode – lead plates
Cathode – grids of lead plates packed with lead dioxide (PbO2)
Electrolyte – 38% (by weight) soution of H2SO4.
The cell reactions when the battery is in use are,
Anode: Pb(s) + SO42-(aq) → PbSO4 + 2 \(\overline { e } \)
Cathode: PbO2(s) + SO42-(aq) + 4H+(aq) + 2 \(\overline { e } \) → PbSO4(s) + 2H2O(I)
The overall cell reaction is,
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

The emf of the cell depends on the concentration of H2SO4. On recharging the battery the reaction is reversed and PbSO4(s) on anode is converted to Pb and PbSO4(s) at cathode is converted into PbO2.

b) Nickel-Cadmium Cell:
Anode- Cd
Cathode – metal grid containing nickel (IV) oxide. Electrolyte – KOH solution. The overall cell reaction during discharge is,
Cd(s) +2 Ni(OH)3(s) → CdO(s) + 2Ni(OH)2(s) + H2O(l)

3) Fuel Cells :
These are Galvanic cells designed to convert the energy of combustion of fuels directly into electrical energy.

H2 – O2 fuel cell – In this, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous NaOH solution, which acts as the electrolyte.
Plus Two Chemistry Notes Chapter 3 Electrochemistry 18
The electrode reactions are,
Anode : 2H2(g) + 4OH(aq) → 4H2O(l) + 4\(\overline { e } \)
Cathode : O2(g) + 2H2O(l) + 4\(\overline { e } \) → 4OH(aq)
Overall reaction : 2H2(g) + O2(g) → 2H2O(l)

Advantages of Fuel Cells –
pollution free, more efficient than conventional methods, Runs continuously as long as the reactants are supplied, electrodes are not affected.

Plus Two Chemistry Notes Chapter 3 Electrochemistry

Other examples:
CH4 – O2 fuel cell, CH3OH – O2 fuel cell

Corrosion :
Any process of destruction and consequent loss of a solid metallic material by reaction with moisture and other gases present in the atmosphere. More reactive metals are corroded more easily. Corrosion is enhanced by the presence of impurities, air & moisture, electrolytes and defects in metals.
Examples: Rusting of iron, tarnishing of Ag.

Mechanism:
In corrosion a metal is oxidised by loss of electrons to O2 and form oxides. It is essentially an electro chemical phenomenon. At a particular spot of an object made of iron, oxidation take place and that spot behaves as anode.
2 Fe(s) → 2 Fe2+ + 4\(\overline { e } \)E° = -0.44 V

Electrons released at anodic spot move through metal and go to another spot on the metal and reduce 02 in presence of H+. This spot behaves as cathode.
O2(g) + 4 H+(aq) + 4\(\overline { e } \) → 2 H2O(l) E° = 1.23 V

The overall reaction is,
2 Fe(s) + O2(g)+ 4H+(aq) → 2 Fe2+ + 2H2O(I) E° = 1,67V

The ferrous ions are further oxidised by atmospheric 02 to ferric ions and form hydrated ferric oxide (rust) Fe2O3.xH2O

Prevention of Corrosion
1) Barrier Protection:
Coating the surface with paints, grease, metals like Ni, Cr, Cu etc.

2) Sacrificial Protection:
Coating the surface of iron with a layer of more active metals like Zn, Mg, Al etc. The process of coating a thin film of Zn on iron is known as galvanisation.

3) Anti-rust Solutions:
Applying alkaline phosphate/ alkaline chromate on iron objects which provide a protectve insoluble film. Also, the alkaline nature of the solutions decreases the availability of H+ ions and thus decreases the rate of corrosion.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Students can Download Chapter 3 Classification of Elements and Periodicity in Properties Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties

Plus One Chemistry Classification of Elements and Periodicity in Properties One Mark Questions and Answers

Plus One Chemistry Chapter 3 Questions And Answers Question 1.
Which of the following is not a Dobereiner triad?
a) Cl, Br, I
b) Ca, Sr, Ba
c) Li, Na, K
d) Fe, Co, Ni
Answer:
d) Fe, Co, Ni

Plus One Chemistry Chapter 3 Question 2.
The elements of s-block and p-block are collectively called ___________
Answer:
Representative elements

Classification Of Elements And Periodicity In Properties Questions And Answers Question 3.
The cause of periodicity of properties is
a) Increasing atomic radius
b) Increasing atomic weights
c) Number of electrons in the valence shell
d) The recurrence of similar outer electronic configuration
Answer:
d) The recurrence of similar outer electronic configuration

An online valence electrons calculator finds the abbreviated or condensed electron configuration of an element with these instructions.

Classification Of Elements And Periodicity In Properties Important Questions Question 4.
Halogen with highest ionization enthalpy is ___________ .
Answer:
Fluorine

Classification Of Elements And Periodicity In Properties Question 5.
Which of the following represents the most electropositive element?
a) [He]2s1
b) [He]2s2
c) [Xe]6s1
d) [Xe]6s2
Answer:
c) [Xe]6s1

Periodic Classification Of Elements Class 10 Extra Questions With Answers Question 6.
Second electron gain enthalpy is
Answer:
always positive

Classification Of Elements And Periodicity In Properties Class 11 Pdf Question 7.
Correct order of polarising power is
a) Cs+ < K+ < Mg2+ < Al3+
b) Al3+ < Mg2 + K+ < Cs+
c) Mg2+ < Al3+ < K+ < Cs+
d) K+ < Cs+ < Mg2+ < Al3+
Answer:
a) Cs+ < K+ < Mg2+ < Al3+

Chemistry Chapter 3 Class 11 Important Questions Question 8.
The IUPAC name of the element with atomic number is 109 is ___________
Answer:
Une

Important Questions Of Periodic Classification Of Elements Class 11 Question 9.
The size of iso electronic species F~, Ne and Na+ is affected by
a) Nuclear charge
b) Principal quantum number.
c) Electron – electron interaction in outer orbitals.
d) None of the factors because their size is the same.
Answer:
Nuclear charge as nuclear charge is high the size is small.

Chapter 3 Chemistry Class 11 Question 10.
In transition elements the differentiating electron occupies (n-1)d sublevel in preference to ______________
Answer:
np level

Plus One Chemistry Classification of Elements and Periodicity in Properties Two Mark Questions and Answers

Important Questions For Class 11 Chemistry Chapter 3 Question 1.
The arguments made by two students are as given:
Student 1: ‘Hydrogen belongs to Group 1.’
Student 2: ‘Hydrogen belongs to Group 17.’
1. Who is right?
2. What is your opinion?
Answer:
1. Nobody is right.

2. Hydrogen has a one s-electron and hence can be placed in group 1 (alkali metals). It can also • gain an electron to achieve a noble gas arrangement and hence it can behave similar to a group 17 (halogen family) element. Because it a special case, hydrogen is placed separately at the top of the Periodic Table.

Class 11 Chemistry Chapter 3 Important Questions With Answers Question 2.
Match the following:

Sodiumf-block
Oxygens-block
Uraniumd-block
Silverp-block

Answer:
Sodium – s-block
Oxygen – p-block
Uranium – f-block
Silver – d-block

Periodic Classification Of Elements Important Questions Question 3.

  1. Which one has greater size, Na or K?
  2. Justify your answer.

Answer:
1. K

2. K comes below Na in the Periodic Table. The atomic size increases down the group due to the fact that the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus.

Questions On Periodic Classification Of Elements Class 11 Question 4.
The general characteristics of a particular block of elements is as given:
They are highly electropositive, soft metals. They are good reducing agents. They lose the outermost electron(s) readily to form 1+ ion of 2+ ion.

  1. Which block has this general characteristics?
  2. Write down two general characteristics of p-block.

Answer:
1. s-block

2. The p-block contains metals, non-metals and metalloids. They form ionic as well as covalent compounds.

Class 11 Chemistry Chapter 3 Important Questions Question 5.
The variation in electron gain enthalpies of elements is less systematic than for ionization enthalpies. Out of oxygen and sulphur which has greater negative value for electron gain enthalpy? Justify.
Answer:
Sulphur has greater negative value (-200 kJ mol1) for electron gain enthalpy compared to that of oxygen (-141 kJ mol1). This is because, due to smaller size of oxygen the added electron experiences much repulsion from the electrons present in the shell. Due to large size, the electron-electron repulsion is much less in sulphur.

Question 6.
Some elements are given. Li, Cs, Be, C, N, O F, I, Ne, Xe.

  1. Arrange the above elements in the increasing order of ionization enthalpy.
  2. Arrange the given elements in the decreasing order of negative electron gain enthalpy.

Answer:

  1. Cs < I < Li < Xe < Be < C < N < O < F < Ne
  2. Cl > F > O > N > C > Be > Li > I >C s > Xe > Ne

Plus One Chemistry Classification of Elements and Periodicity in Properties Three Mark Questions and Answers

Question 1.
Analyse the given figure and answer the questions that follow.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 1

  1. What is meant by atomic radius?
  2. Explain covalent radius.
  3. Write down another two types of terms expressed as atomic size.

Answer:

  1. Atomic radius is defined as the distance from the centre of the nucleus of the atom to the outermost shell of electrons.
  2. Covalent radius is defined as one half of the distance between the centre of nuclei of two similar atoms bonded by a single covalent bond.
  3. Vander Waals’ radius, Metallic radius

Question 2.
Consider the statement: The element with 1s2
configuration belongs to the p-block.’

  1. Identify the element.
  2. Do you agree with this statement?
  3. Justify.

Answer:

  1. Helium
  2. Yes
  3. Strictly, helium belongs to the s-block but its positioning in the p-block along with other group 18 elements is justified because it has a completely filled valence shell (1s²) and as a result, exhibits characteristic of other noble gases.

Question 3.
The properties of elements are a periodic function of
their atomic weights.

  1. Who proposed this law?
  2. Can you see anything wrong in this law? If yes, justify your answer.
  3. State modem periodic law.

Answer:

  1. Mendeleev
  2. Yes, atomic number is the more fundamental property of an element than atomic mass.
  3. The physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 4.
1. Define ionisation enthalpy.
2. IE1 <IE2 <IE3
What is meant by this? Justify.
Answer:
1. Ionisation enthalpy is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

2. It shows the increasing order the magnitude of successive ionization enthalpies. As the positive charge of the ion increases, it becomes more difficult to remove the valence electron due to increase in effective nuclear charge.

Question 5.
Say whether the following are true or false:

  1. On moving across a period ionization enthalpy decreases.
  2. Mg is biggerthan Cl.
  3. Ionization enthalpy of Li is less than that of K.

Answer:

  1. False
  2. True
  3. False

Question 6.
Analyze the graph given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 2
1. Identify the graph.
2. Account forthe following observations:
i) ‘Ne’ has the maximum value of ∆iH.
ii) In the graph from Be to B, ∆iH decrease.
Answer:
1. Graph showing the variation of ionisation enthalpy (∆iH) with atomic number (Z) of the elements of second period.

2. i) ‘Ne’ is an inert gas and it has closed electron shell with stable octet electronic configuration. Hence it has the maximum ionisation enthalpy in the second period.
ii) Be has completely filled electronic configuration and a more stable. So ionization enthalpy is high

Question 7.
Study the graph and answer the questions that follow:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 3
1. On moving down a group what happens to electron gain enthalpy?
2. Why chlorine shows more negative electron gain enthalpy than fluorine? ,

Answer:
1. On moving down a group,electron gain enthalpy becomes less negative because the size of the atom increases and the added electron would be farther from the nucleus.

2. Due to small size of F, the added electron goes to the smaller 2p quantum level and suffers significant replusion from the other electrons present in this level. But due to big size of Cl, the added electron goes to the n = 3p quantum level and occupies a larger region of space and the electron-electron repulsion is much less.

Question 8.
Electron gain enthalpy is an important periodic property.
1. What is meant by electron gain enthalpy?
2. What are the factors affecting electron gain enthalpy?
3. How electron gain enthalpy varies on moving across a period? Justify.

Answer:
1. Electron gain enthalpy(∆egH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion.

2. Effective nuclear charge, atomic size, electronic configuration

3. Electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently, it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the nucleus. Thus more energy is released.

Question 9.
1. The electron gain enthalpies of Be and Mg are positive. What is your opinion? Justify.
2. Electron gain enthalpies of nobles gases have large positive values. Why?
Answer:
1. The statement is correct. Electron gain enthalpies of Be (+240 kJ/mol) and Mg (+230 kJ/mol) are positive because they have stable electronic configurations with fully filled s-orbitals (Be – 2s², Mg – 3s²). Hence, the gain of electron is highly endothermic.

2. Noble gases have stable octet electronic configuration of ns2 np6 (except He -1 s²) and they have practically no tendency to accept additional electron. They have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. Hence, energy has to be supplied to add an extra electron.

Question 10.
During a group discussion, a student argued that ionization enthalpy depends only upon electronic configuration.
1. Do you agree? Comment.
2. Define shielding effect/screening effect.
3. Is there any relation between ionization enthalpy and shielding effect/screening effect? Explain.
Answer:
1. No. In addition to electornic configuration, ionization enthalpy depends upon other factors like atomic size, nuclear charge and shielding effect/screening effect.

2. In multi-electron atoms, the nuclear charge experienced by the valence electron will be less than the actual charge on the nucleus because it is shielded by inner core of electrons. This is called shielding effect or screening effect.

3. Yes. Ionization enthalpy decreases with increase in shielding effect/screening effect of inner electrons. This is because as a result of shielding of the valence electron by the intervening core of electrons it experiences a net positive charge which is less than the actual charge of the nucleus. In general, shielding is effective when the orbitals in the inner shells are completely filled.

Question 11.
a) Which of the following has higher first ionization enthalpy, N or O? Justify.
b) Which one is bigger, For F ? Why?
Answer:
1. N. This is because in N, three 2p-electrons reside in different atomic orbitals in accordance with Hund’s rule (\(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)) whereas in O, two the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion (\(2p_{ x }^{ 2 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)). Consequently, it is easier to remove the fourth 2p-electron from O than it is, to remove one of the three 2p-electrons from N.

2. F (136 pm) is bigger than F (72 pm). An anion is bigger than its parent atom. Addition of one electron in F results in increased repulsion among the electrons and a decrease in effective nuclear charge. Thus, attraction between nucleus and the electrons decreases and hence size increases.

Question 12.
Electronegativity differs from electron gain enthalpy.

  1. Do you agree?
  2. What do you mean by electronegativity?

Answer:

  1. Yes.
  2. Electronegativity is defined as the tendency of an atom in a chemical compound to attract the shared pair of electrons to itself.

Question 13.
Ionization enthalpy is an important periodic property.
1. What is the unit in which it is expressed?
2. What are the factors influencing ionization enthalpy?
3. How ionisation enthalpy varies in the periodic table?
Answer:
1. kJ mol-1.

2. Atomic/ionic radius, nuclear charge, shielding effect/screening effect, penetration effect and electronic configuration.

3. Ionization enthalpy generally increases with increase in atomic number across a period due to regular increase in nuclear charge and decrease in atomic size. Thus, the attractive force between the nucleus and the electron cloud increases. Consequently, the electrons are more and more tightly bound to the nucleus.

Ionisation enthalpy generally decreases from top to bottom a group: This is because the effect of increased nuclear charge is cancelled by the increase in atomic size and the shielding effect. Consequently, the nucelar hold on the valence electron decreases gradually and ionization enthalpy decreases.

Question 14.
The second period elements show anomalous ‘ behaviour.
1. Give reason.
2. What are the anomalous properties of second period elements?
Answer:
1. The first element is each group belong to the second period. The difference in behaviour of these elements from the other elements of the same group can be attributed to the following factors:

  • Small atomic size
  • Large charge/radius ratio
  • High electronegativity
  • Absence of d-orbtials in the valence shell

2. The important anomalous properties of second period elements are: diagonal relationship, maximum covalence of four and ability to form pπ —pπ multiple bonds.

Question 15.
Atomic size, valency, ionization enthalpy, electron gain enthalpy and electronegativity are the important periodic properties of elements.
1. What do you mean by periodicity?
2. Periodic properties are directly related to ___________
Answer:
1. The periodical repetition of elements with similar properties after certain regular intervals when the elements are arranged in the order of increasing atomic numbers is called periodicity,

2. Electronic configuration.

Question 16.
1. Which is the element among alkali metals having lowest ionization enthalpy?
2. What is meant by valence of an element? How it varies in the periodic table?
3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 3 Classification of Elements and Periodicity in Properties 4

Identify the elements A, B, C and D, if the graph represents halogens.
Answer:
1. Fr

2. Valence of an element is the combining capacity of that element. In the case of representative elements the number of valence electrons increases from 1 to 8 on moving across a period, the valence to the element with respect to H and Cl increases from 1 to 4 and then decreases from 4 to zero. On moving down a group, the number of valence electrons remains same and, therefore, all the elements in a group exhibit same valence.

3. A = F, B = Cl, C = Br, D = I

Question 17.
Li and Mg belonging to first and second group in periodic table respectively resemble each other in many respects.
1. Name the relationship.
2. B can only form [BF4] ion while Al can form [AIF6]3-, though both B and Al belong to group 13. Justify.
Answer:
1. Diagonal relationship.

2. This is because B, being a second period element has a maximum covalence of 4. It cannot expand its covalence beyond 4 due to absence of d-orbitals. But Al, being a third period element has vacant d-orbitals in its valence shell and hence can expand its covalence beyond 4.

Question 18.
During a group discussion a student argues that both oxidation state and valence are the same.
1. Do you agree?
2. Justify taking the case of [AICI(H2O)5]2+.
Answer:
1. No. Valence refers to the combining capacity of an element whereas oxidation state is the charge assigned to an element in a compound based on the assumption that the shared electron in a covalent bond belongs entirly to the more electronegative element.

2. In [AICI(H2O)5]2+the valence of Al is 6 while its oxidation state is +3.

Question 19.
Among the elements of the third period, identify the element

  1. With highest first ionization enthalpy.
  2. That is the most reactive metal.
  3. With the largest atomic radius.

Answer:

  1. Ar
  2. Na
  3. Na

Question 20.
A cation is smaller than the corresponding neutral atom while an anion is larger. Justify.
Answer:
A cation is smaller than its parent atom because it has fewer number of electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased . repulsion among the electrons and a decrease in effective nuclear charge.

Question 21.
1. How the metallic character varies in the periodic table?
2. Categorize the following oxides into acidic, basic, neutral and amphoteric:
Al2O3, Na2O, CO2, Cl2O7, MgO, CO, As2O3, N2O
Answer:
1. The metallic character decreases from left to right across the period due to increase in ionization enthalpy along a period which makes loss of electrons difficult. From top to bottom a group metallic character increases due to decrease in ionization enthalpy. Thus, metallic character decreases diagonally from left bottom to right top of the periodic table.

2. Acidic oxides: CO2, Cl2O7
Basic oxides: Na2O, MgO
Neutral oxides: CO, N2O
Amphoteric oxides: Al2O3, As2O3

Question 22.
A group of ions are given below:
Na+, Al3+, O2-, Ca2+, Mg2+, F, N3-, Br
1. Find the pair which is not isoelectronic.
2. Arrange the above ions in the increasing order of size.
Answer:
1. Ca2+ and Br
2. Al3+ < Mg2+ < Na+ < F < O2- < N3- < Ca2+ < Br

Plus One Chemistry Classification of Elements and Periodicity in Properties Four Mark Questions and Answers

Question 1.
Statement 1: ‘Atomic mass is the fundamental property of an element.’
Statement 2: ‘Atomic number is a more fundamental property of an element than its atomic mass.’
1. Which statement is correct? Justify your answer.
2. Name the scientist who proposed this statement? What observation led him to this conclusion?
Answer:
1. Statement 2. Atomic number indicates the number of electrons present in an element. Most of the chemical properties of an element depend on its electronic configuration,

2. Henry Moseley. He observed regularities in the characteristic X-ray spectra of the elements. A plot of √υ (where u is the frequency of the X-rays emitted) against atomic number (Z) gave a straight line and not the plot of √υ vs atomic mass.

Question 2.
Atoms possessing stable configuration have less tendency to loss electrons and consequently will have high value of ionization enthalpy.
1. Justify this statement by taking the case of half-filled and completely filled electronic configurations.
2. The noble gases have highest ionization enthalpies in each respective periods. Why?
Answer:
1. Atoms with half-filled and completely filled electronic configurations have extra stability due to symmetric distribution of electrons and maximum exchange energy. Hence, more energy is required for the removal of their electrons. Elements like N (1s² 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)), P (1s² 2s² 2p6 3s² \(3p_{ x }^{ 1 }3{ p }_{ y }^{ 1 }3{ p }_{ z }^{ 1 }\)) etc. possessing half-filled shells have high ionization enthalpies.
Elements like Be ((1s² 2s²), Mg(1s² 2s² 2p6 3s²) etc. having completely filled shells show high values of ionization enthalpy.

2. Noble gases have closed electron shells and very stable octet electronic configurations (except He). Hence, maximum amount of energy is required to remove their valence electron.

Question 3.
Mendeleev arranged the elements in the order of increasing atomic weights.
a) Write down the merits of Mendeleev’s periodic table.
b) What are the demerits of Mendeleev’s periodic table?
Answer:
a) Merits of Mendeleev’s periodic table:

  • Study of elements – Elements are classified into groups with similar properties, thus facilitating the study of properties of elements.
  • Prediction of new elements – Mendeleev left certain vacant places in his table which provided a clue for the discovery of new elements, e.g. Eka-AI (for Ga), Eka-Si (for Ge).
  • Determination of correct atomic weights – With the help of this table, doubtful atomic weights of some elements were corrected.

b) Demerits of Mendeleev’s periodic table:

  • Position of hydrogen was not certain.
  • Anomalous pairs of elements – Certain elements of higher atomic weight preceed those with lower atomic weight, e.g. I (at.wt. 127) was placed after Te (at.wt. 128).
  • Lanthanides and Actinides are not given proper places in this periodic table.
  • No proper position for isotopes.

Question 4.
1. Name any three numerical scales of electronegativity.
2. How electronegativity varies in the periodic table? Justify.
Answer:
1. Pauling scale, Mullimen-Jaffe scale, Allred- Rochow scale.
2. Electronegativity generally increases from left to right a period and decreases from top to bottom in a group. This is because, from left to right across a period atomic size decreases and attraction between the valence electrons and the nucleus increases. From top to bottom in a group atomic size increases and attraction between the valence electrons and the nucleus decreases.

Question 5.
1. First ionization enthalpy of Na is lower than that of Mg. But its second ionization enthalpy is higher than that of Mg. Explain.
2. Which one is smaller, Na or Na+? Give reason.
Answer:
1. By the removal of one electron from Na it gets
the stable octet configuration of Ne. But when the first electron is removed from Mg it gets the unstable configuration of Na. It requires more energy due to small size and greater nuclear charge of Mg. In the case of Na the second electron is to be removed from a stable octet configuration which requires more energy than the removal of second electron from Mg.

2. Na+ (95 pm) is smaller than Na (186 pm). Acation is smaller than its parent atom. Na+ has fewer number of electrons (10 electrons) compared to Na (11 electrons). But the nuclear charge remains the same in both. Thus, effective nuclear charge per electron is greater in Na+. Thus, the attraction between nucleus and the remaining electrons increases and size decreases.

Question 6.
Removal of electron becomes easier on moving down the group.
1. Comment the above statement based on ionization enthalpy.
2. How electronic configuration influences the ionization enthalpy value?
Answer:
1. On moving from top to bottom in a given group, size of the atom increases and ionisation enthalpy decreases. Hence, it becomes easier to remove the valence electron.

2. Atoms with octet configuration, half-filled and completely filled configurations have extra stability and hence have higher values of ionization enthalpy.

Question 7.
The energy released during the addition of an electron to an isolated neutral atom is called electron gain enthalpy.
1. Explain how electron gain enthalpy differ from electronegativity.
2. The second ionisation enthalpy of an element is always greater than the first ionisation enthalpy. Give reason.
Answer:
1. Electron gain enthalpy(AegH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion. It is a quantitative property of an isolated gaseous atom, which can be measured. Whereas, electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. It is not a measureable quantity.

2. This is because to remove second electron from a positively charged ion more amount of energy is required due to increase in effective nuclear charge.

Question 8.
The physical and chemical properties of elements are periodic functions of their atomic numbers.
1. The atomic number of an element ‘X’ is 19. Write the group number, period and block to which X’ belong in the periodic table.
2. Name the element with
i) highest electronegativity and
ii) highest electron gain enthalpy
Answer:
1. The element is K.
19K= 1s² 2s² 2p6 3s² 3p6 4s1
Group number = 1
Period number = 4
Block = s-block

2. i) Fluorine
ii) Chlorine

Plus One Chemistry Classification of Elements and Periodicity in Properties NCERT Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table? (2)
Answer:
The basic theme of organisation of elements in the periodic table is to facilitate the study of the properties of all the elements and their compounds. On the basis of similarities in chemical properties, the various elements have been divided into different groups. This has made the study of elements simple because their properties are now studied in the form of groups rather than individually.

Question 2.
What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? (2)
Answer:
Mendeleev Periodic Law states that the properties of the elements are a periodic function of their atomic weights whereas Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of arrangements of elements from atomic weight to atomic number.

Question 3.
Consider the following species. N3-, O2-, F, N2+, Mg2+ and Al3+ (2)
1. What is common in them?
2. Arrange them in order of increasing ionic radii.
Answer:
1. Each one of these ions contains 10 electrons and
hence these are isoelectronic ions,

2. The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. Among the isoelectronic ions: N3-, O2-, F, Na+, Mg2+ and Al3+, nuclear charge increase in the order:
N3-< O2- < F < Na+ < Mg2+ < Al3+
Therefore, the ionic radii decrease in the order:
N3- > O2- > F > Na+ > Mg2+ > Al3+

Question 4.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons (1)
Answer:
Nuclear mass does not affect the valence shell. Thus, option (c) is the correct answer.

Question 5.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: (1)
a) F > Cl > O > N
b) F > O > Cl > N
c) Cl > F > O > N
d)0>F>N>CI
Answer:
a) F > Cl > O > N
Across a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is F > Cl > O > N and the choice (a) is correct.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Students can Download Chapter 7 Equilibrium Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Plus One Chemistry Equilibrium One Mark Questions and Answers

Question 1.
Equilibrium in a system having more than one phase is called _________
Answer:
heterogeneous

Question 2.
Addition of a catalyst to a chemical system at equilibrium would result in
a) Increase in the rate of forward reaction
b) Increase in the rate of reverse reaction
c) A new reaction path
d) Increase in the amount of heat evolved in the reaction
Answer:
c) A new reaction path

Question 3.
With increase in temperature, equilibrium constant of a reaction
a) Always increases
b) Always decreases
c) May increase or decrease depending upon the number of moles of reactants and products
d) May increase or decrease depending upon whether reaction is exothermic or endothermic
Answer:
d) May increase or decrease depending upon whether reaction is exothermic or endothermic

Question 4.
Water is a conjugate base of ____________ .
Answer:
H3O+

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Question 5.
Which of the following substances on dissolving in water will give a basic solution?
a) Na2CO3
b) Al2(SO4)3
c) NH4Cl
d) KNO3
Answer:
a) Na2CO3

Ged exam books to download, equilibrium concentration calculator, Finding the Least Common Denominator, combining like expressions.

Question 6.
Choose the correct answer for the reaction,
N2(g) + 3H2(g) \(\rightleftharpoons \) 2NH3(g); ∆rH = -91.8 kJ mol-1 The concentration of H2(g) at equilibrium can be increased by
1) Lowering the temperature
2) Increase the volume of the system.
3) Adding N2 at constant volume.
4) Adding H2 at constant volume.
Answer:
2) and 4) are correct.

Question 7.
Conjugate base of a strong acid is a
Answer:
Weak base

Question 8.
The expression forostwald dilution law is
Answer:
Ka = Cα²

Question 9.
The hydroxyl ion concentration in a solution having pH = 4 will be
Answer:
10-14

Question 10.
A mono protic acid in 1M solution is 0.01 % ionized the dissociation constant of this acid is
Answer:
10-8

Question 11.
A certain buffer solution contains equal concentration of x and Hx. The Kafor Hx is 10-6 pH of buffer is
Answer:
6

Question 12.
In the equilibrium reaction
CaCO3(s) → CaO(s) + CO2(g) the equilibrium constant is given by —–
Answer:
PCO2

Question 13.
Congugate base of a strong acid is a _________ .
a) Strong base
b) Strong acid
c) Weak acid
d) Weak base
e) Salt
Answer:
d) Weak base

Question 14.
The species acting both as bronsted acid and base is _________ .
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 1
Answer:
b) HSO4

Question 15.
PH of .01 M KOH solution will be _________ .
Answer:
12

Plus One Chemistry Equilibrium Two Mark Questions and Answers

Question 1.
“High pressure and low temperature favours the formation of ammonia in Haber’s process.” Analyse the statement and illustrate the conditions using Le-Chatliers principle?
Answer:
The given statement is correct.
N2(g) + 3H2(g) \(\rightleftharpoons \) 2NH3(g); ∆rH =-91.8 kJ mol-1 Since the number of moles decreases in the forward reaction a high pressure of ~ 200 atm is applied. Since the forward reaction is exothermic the optimum temperature of ~ 700 K is employed for maximum yield of ammonia.

Question 2.
“Chemical equilibrium is dynamic in nature”. Analyse the statement and justify your answer.
Answer:
At equilibrium the reaction does not stop. Both forward and backward reactions are taking place at equal rates. Thus, at equilibrium two exactly opposite changes occur at the same rate. Hence, chemical equilibrium is dynamic in nature.

Question 3.
Pressure has no influence in the following equilibrium: N2(g) + O2(g) \(\rightleftharpoons \) 2NO(g)

  1. Do you agree with this?
  2. What is the reason for this?

Answer:

  1. Yes.
  2. Here the total number of moles of the reactants is equal to that of the products. Hence pressure is having no influence in this equilibrium.

Question 4.
During a class room discussion a student is of the view that the value of equilibrium constant can be influenced by catalyst.

  1. Do you agree with the statement?
  2. Justify the role of catalyst in an equilibrium reaction?

Answer:

  1. No.
  2. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression. It only helps to attain the equilibrium state in a faster rate.

Question 5.
What is the equilibrium constant (K) in the following cases?

  1. Reaction is reversed.
  2. Reaction is divide by 2.
  3. Reaction is multiplied by 2.
  4. Reaction is splitted into two.

Answer:

  1. 1/K
  2. √K
  3. K2
  4. K1K2

Question 6.
1. What is homogeneous equilibrium?
2. Suggest an example for this.
Answer:
1. The equilibrium in which the reactants and products are in the same phase,

2. N2(g) + 3H2(g) \(\rightleftharpoons \) 2NH3(g)
In this equilibrium, the reactants and products are in the gaseous phase.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Question 7.
The equilibrium constants for two reactions are given. In which case the yield of product will be the maximum?
For first reaction: K1 = 3.2 × 10-6
For second reaction: K2 = 7.4 × 10-6
Answer:
Higher the value of K, greater will be the yield of product. So maximum yield will be in the second case.

Question 8.
Write an expression for equilibrium constant, Kc forthe ‘ reaction, 4NH3(g) + 5O2(g) \(\rightleftharpoons \) 4NO(g) + 6H2O(g)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 2

Question 9.
1. State Henry’s law.
2. Suggest an example fora gas in liquid equilibrium.
Answer:
1. The mass of a gas dissolved in a given mass of a solvent at any temperature is directly proportional to the pressure of the gas above the solvent.

2. Equilibrium between the CO2 molecules in the. gaseous state and the CO2 molecules dissolved in water under pressure,
CO2(g) \(\rightleftharpoons \) CO2(in solution)

Question 10.
1. What is heterogeneous equilibrium?
2. Suggest an example forthis.
Answer:
1. Equilibrium in a system having more than one phase is called heterogeneous equilibrium

2. Equilibrium between solid Ca(OH)2 and its saturated solution:
Ca(OH)2(s) + (aq) \(\rightleftharpoons \) Ca2+(aq) + 2OH(aq)

Question 11.
For the equilibrium 2SO3(g) → 2SO2(g) + O2(g), Kc at 47 °C 3.25 × 10-9 mol per litre. What will be the value of Kp at this temperature (R = 8.314 J K-1mol-1).
Answer:
R = 8.314 J K-1 mol-1 ∆n = 3 – 2 = 1
T = 47 °C = 273 + 47 = 320 K
Kc = 3.25 × 10-9
Kp = Kc (RT)∆n
= 3.25 × 10-9 (8.314 × 320)1
= 3.25 × 10-9 × 8.314 × 320 = 8.65 × 10-12

Question 12.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ions in it.
Answer:
pH= -log[H+] = 3.76
log[H+] = – 3.76
[H+] = antilog of (- 3.76) = 1.738 × 10-4 mol L-1

Question 13.
The equilibrium constant can be expressed in terms of partial pressure as well as concentration.
1. Give the relation between Kp and Kc.
2. What is the relation between Kp and Kc for the reaction, N2(g) + O2(g) \(\rightleftharpoons \) 2NO(g)?
Answer:
1. Kp = Kc(RT)∆n, where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants).

2. Here, ∆n = 2 – 2 = 0
Kp = Kc(RT)∆n , Kp = Kc (RT)°
∴ Kp = Kc

Question 14.
1. Explain Arrhenius concept of acids and bases with suitable examples.
2. How proton exists in aqueous solution? Give reason.
Answer:
1. According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions, H+(aq) and bases are substances that produce hydroxyl ions, OH(aq). Forexample, HCl is an Arrhenius acid and NaOH is an Arrhenius base.

2. In aqueous solution the proton bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H3O+(aq). This is because a bare proton, H+ is very reactive and cannot exist freely in aqueous solutions.

Question 15.
1. What is an acidic buffer?
2. Suggest an example for an acidic buffer.
Answer:
1. An acidic buffer is a buffer solution having pH less than 7. It is prepared by mixing a weak acid and its salt formed with a strong base.

2. Mixture of acetic acid (CH3COOH) and sodium acetate (CH3COONa) is an example for an acidic buffer. Its pH is around 4.75.

Question 16.
1. What is a basic buffer?
2. Suggest an example for basic buffer.
Answer:
1. A basic buffer is a buffer solution having pH greater than 7. It is prepared by mixing a weak base and its salt formed with a strong acid.

2. Mixture of ammonium hydroxide (NH4OH) and ammonium chloride (NH4CI) is an example for a basic buffer. Its pH is around 9.25.

Plus One Chemistry Equilibrium Three Mark Questions and Answers

Question 1.
The concentration of reactant and products for the reaction, H2(g) + l2(g) \(\rightleftharpoons \) 2Hl(g) are recorded as follows:

Reactant or ProductMolar Concentration
H20.080
l20.060
HI0.490

a) Write down the expressions for equilibrium constant of the above reaction.
b) Calculate the equilibrium constant at the temperature 298 K if [Hl] = 0.49 M, [H2]=0.08 M and [l2]=0.06 M.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 3

Question 2.
1. When equilibrium is reached in a chemical reaction?
2. What is the influence of molar concentration in a reaction at equilibrium?
3. Write the expression for equilibrium constant for the decomposition of NH4CI by the reaction,
NH4Cl \(\rightleftharpoons \) NH3+HCl
Answer:
1. When the rate of forward reaction is equal to rate of backward reaction, the chemical reaction is said to be in equilibrium.

2. Rate of chemical reaction is directly proportional to the product of molar concentration of the reactants.

3. \(\mathrm{K}=\frac{\left[\mathrm{NH}_{3}\right] \mathrm{HCl}}{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}\)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Question 3.
1. What is meant by Kp?
2. How Kp is related to Kc?
Answer:
1. Kp is the equilibrium constant in terms of the partial pressures of the reactants and products (Pressure should be expressed in bar as standard state is 1 bar). It is used for reactions involving gases.

2. Kp = Kc (RT)∆n
where R = universal gas constant, T = absolute temperature and ∆n = number of moles of gaseous product(s) – number of moles of gasesous reactant(s).

Question 4.
a) What do you mean by equilibrium constant?
b) Write any two characteristics of equilibrium constant.
c) Write an expression for equilibrium constant of the reaction, 2SO2(g) + O2(g) \(\rightleftharpoons \) SO3(g).
Answer:
a) Equilibrium constant at a given temperature is the ratio of product of molar concentrations of the products to that of the reactants, each concentration term being raised to the respective stoichiometric coefficients in the balanced chemical equation.

b) 1. The value of equilibrium constant is independent of the initial concentrations of the reactants and products.
2. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

Question 5.
2NO2(g) \(\rightleftharpoons \) N2O4(g); ∆H = -52.7 kJ mol-1
1. What change will happen if we increase the temperature?
2. What is the effect of increase in pressure in the above equilibrium?
3. What happens when N2O4 is removed from the reaction medium?
Answer:
1. Since the forward reaction is exothermic, on increasing temperature the rate of backward reaction (endothermic reaction) increases.

2. Since the number of moles decreases in the forward reaction, on increasing pressure, the rate of forward reaction increases.

3. Rate of forward reaction increases.

Question 6.
Consider this reaction:
CO(g) + 2H2(g) \(\rightleftharpoons \) CH3OH(g); ∆rH = -92 kJ mol-1
Explain the influence of the following on the basis of
Le Chatelier’s principle.
1. Decrease in pressure.
2. Increase in temperature.
3. Increase in the partial pressure of hydrogen.
Answer:
1. On decreasing pressure the reaction shifts in the direction in which there is increase in the number of moles. Thus, the rate of backward reaction increases on decreasing pressure.

2. On increasing temperature, the rate of endothermic reaction increases. Here, backward reaction is endothermic. Hence, on increasing temperature the rate of backward reaction increases.

3. Hydrogen, being a reactant increase in its partial pressure increases the rate of forward reaction.

Question 7.
The equilibrium showing dissociation of phosgene gas is given below:
COCl2(g) \(\rightleftharpoons \) CO(g) + Cl2(g)
When a mixture of these three gases at equilibrium is compressed at constant temperature, what happens to
1. The amount of CO in mixture?
2. The partial pressure of COCl2?
3. The equilibrium constant for the reaction?
Answer:
1. Amount of CO decreases, because the system favours the reaction in which number of moles decreases with increase of pressure i.e., backward reaction.

2. Increases.
3. Equilibrium constant remains the same since temperature is constant.

Question 8.
The equilibrium constant of the reaction H2(g) + l2(g) \(\rightleftharpoons \) 2Hl(g) is 57 at 700 K. Now, give the equilibrium constants for the following reactions at the same temperature:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 4
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 5

Question 9.
1. What are buffer solutions?
2. Which of the following are buffer solutions?
NaCl + HCl
NH4Cl + NH4OH
HCOOH + HCOOK
3. What is the effect of pressure on the following equilibria?
i) Ice \(\rightleftharpoons \) Water
ii) N2(g) + O2(g) \(\rightleftharpoons \) 2N0(g)
Answer:
1. These are solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali.

2. NH4Cl + NH4OH

3. i) When pressure is increased the melting point of ice decreases and hence the rate of forward reaction will increase.
ii) Pressure has no effect in this equilibrium because there is no change in the number of moles of the gaseous reactants and products.

Question 10.
The aqueous solution of the compounds NaCl, NH4Cl and CH3COONa show different pH.

  1. Identify the acidic, basic and neutral solution among them.
  2. The concentration of hydrogen ion in a soft drink is 4 × 10-4. What is its pH?

Answer:

  1. Acidic-aqueous solution of NH4Cl Neutral – aqueous solution of NaCl Basic – aqueous solution of CH3COONa
  2. pH = – log[H+] = – log[4 × 10-4] = 3.398

Question 11.
1. What is pH? What is its significance?
2. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3. What is its pH?
Answer:
1. pH is a logarithmic scale used to express the hydronium ion concenration in molarity more conveniently. The pH of a solution is defined as negative logarithm to the base 10 of the activity of hydrogen ion. pH = — log \({ { a }_{ { H }^{ + } } }\) = —log[H+]

2. [H+] = 3.8 × 10-3
pH = -log[H+]
= -log [3.8 × 10-3] = -(-2.42) = 2.42

Question 12.
1. State the Le-Chatelier’s principle.
2. Apply the above principle in the following equilibrium and predict the effect of pressure.
CO(g) + 3H2(g) \(\rightleftharpoons \) CH4(g) + H2O(g)
Answer:
1. The Le Chateliers principle states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.

2. On increasing pressure the rate of forward reaction increases. This is because number of moles decreases in the forward reaction. In other words, the value of Qc decreases on increasing pressure. As Qc < Kc, the reaction proceeds in the forward direction.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Question 13.
1. Explain Lewis concept of acids and bases.
2. Why does BF3 act as a Lewis acid?
Answer:
1. A Lewis acid can be defined as a species which accepts electron pair and a Lewis base is a species which donates an electron pair.

2. In BF3, the boron atom is electron deficient and it accepts a lone pair of electron. So it acts as a Lewis acid.

Question 14.
1. How the value of AG influence the direction of an equilibrium process?
2. The equilibrium constant for a reaction is 8. What will be the value of ∆G at 27 °C?
Answer:
1. If ∆G is negative, then the reaction is spontaneous and proceeds in the forward direction.
If ∆G is positive, then reaction is considered non- spontaneous. instead, as reverse reaction would have a negative ∆G, the products of the forward reaction shall be converted to the reactants.
If ∆G is 0, reaction has achieved equilibrium. At this point, there is no longer any free energy to drive the reaction.
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 6

Question 15.
1. What is common ion effect?
2. Suggest an example for this effect.
Answer:
1. Common ion effect may be defined as the suppression of the dissociation of a weak electrolyte by the addition of some strong electrolyte containing a common ion.

2. The dissociation equilibrium of NH4OH is shifted towards left in presence of NH4Cl having the common ion, NH4+.

Question 16.
1. Predict whether an aqueous solution of (NH4)2SO4 is acidic, basic or neutral?
2. Justify your answer.
Answer:
1. An aqueous solution of (NH4)2SC4 is acidic in nature.

2. (NH4)2SO4 is formed from weak base, NH4OH, and strong acid, H2SO4. In water, it dissociates completely
(NH4)2SO4(aq) → 2NH4+ (aq) + SO42- (aq)
NH4+ ions undergoes hydrolysis to form NH4OH and H+ ions.
NH4+(aq) + H2O(l) \(\rightleftharpoons \) NH4OH(aq) + H+(aq)
NH4OH is a weak base and therefore remains almost unionised in solution. This results in increased H+ ion concentration in solution making the solution acidic.

Question 17.
1. What are sparingly soluble salts? Suggest an example.
2. Define solubility product constant, Ksp.
3. Obtain the relation between solubility product constant (Ksp) and solubility (S), of a solid salt of general formula Mxp+ Xyq-.
Answer:
1. Sparingly soluble salts are those salts with solubility less than 0.01 M.
e.g. BaSO4

2. The solubility product of a sparingly soluble salt at a given temperature is defined as the product of the concentrations of its ions in the saturated solution, with each concentration term raised to the power equal to the number of times the ion occurs in the equation representing the dissociation of the electrolyte.

3. The equilibrium in the saturated solution of the salt can be represented as,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 17

Question 18.
The Le Chatelier’s principle is applicable to physical and chemical equilibria.
1. What are the factors which can influence the equilibrium state of a system?
2. Explain the factors affecting the chemical equilibrium on the basis of Le Chatelier’s principle taking Haber’s process for the manufacture of ammonia as an example.
Answer:
1. The following factors can influence the equilibrium state of a system:

  • Change in concentration of the reactants or products.
  • Change in temperature.
  • Change in pressure.
  • Addition of inert gas.
  • Presence of catalyst.

2. N2(g) + 3H2(g) \(\rightleftharpoons \) 2NH3(g); ∆rH = -91.8 kJ mol-1
When concentration of N2 or H2 is increased, a good yield of NH3 can be achieved. The rate of forward reaction can also be increased by removing NH3 from the reaction mixture.

When pressure is increased, the system will try to decrease pressure and for this system will proceed in that direction where there is minimum number of moles i.e., forward reaction. Thus, a good yield of NH3 can be achieved by increasing pressure.

Since the formation of NH3 is an exothermic reaction, a good yield of NH3 can be achieved by decreasing the temperature. But if the temperature is decreased to very low value the reactant molecules do not have sufficient energy to interact. Hence, an optimum temperature of 500°C is used.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Question 19.
1. Soda water is prepared by dissolving CO2 in water under high pressure. What is the principle involved in this process?
2. At 1000 K, equilibrium constant Kc for the reaction 2SO3(g) \(\rightleftharpoons \) 2SO2(g) + O2(g) is 0.027. What is the value of Kp at this temperature?
Answer:
1. Henry’s law
2. Kp =Kc(RT)∆n
∆n = 3.2 = 1
Kp = 0.027 × (0.0831 × 1000)1 = 2.2437

Question 20.
1. For the reaction PCL \(\rightleftharpoons \) PCl3 +Clc
i) Write the expression of Kc.
ii) What happens if pressure is increased?
2. Write the conjugate acid and base of the following species:
i) H20 ii) HCO;
3. Name the phenomenon involved in the preparation of soap by adding NaCI.
Answer:
1. i) \(\kappa_{c}=\frac{\left[\mathrm{PC}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
ii) If we increase the pressure the system will try to decrease the pressure. For this system will proceed in the direction where there is minimum number of moles, i.e., rate of backward reaction increases by decreasing the pressure.

2. i) Conjugate acid of H2O is H3O+
Conjugate base of H2O is OH”
ii) Conjugate acid of HCO,” is H2CO3 Conjugate base of HCO3 is CO32-

3. Common ion effect.

Question 21.
a) The pH of black coffee is 5.0. Calculate the hydrogen ion concentration.
b) The Ksp of barium sulphate is 1.5 × 10-9. Calculate the solubility of barium sulphate in pure water.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 7

Question 22.
1. What is conjugate acid-base pair?
2. Illustrate with an example.
Answer:
1. The acid-base pairthat differs only by one proton is called conjugate acid-base pair. Such acid-base pairs are formed by loss or gain of a proton.

2. Consider the ionization of hydrochloric acid in water.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 8
HCl(aq) acts as an acid by donating a proton to H2O molecule which acts as a base because it accepts the proton. The species H2O+ is produced when water accepts a proton from HCl. Therefore, Cl is the conjugate base of HCl and HCl is the conjugate acid of Cl. Similarly, H2O is the conjugate base of H2O+ and H3O++ is the conjugate acid of the base H2O.

Question 23.
1. What are the applications of equilibrium constant?
2. What is meant by reaction quotient, Qc?
3. Predict the direction of net reaction in the following cases:
i) Qc < Kc
ii) Qc > Kc
iii) Qc = Kc
Answer:
1. The applications of equilibrium constant are:
• To predict the extent of a reaction on the basis of its magnitude.
• To predict the direction of the reaction.
• To calculate equilibrium concentrations.

2. Reaction quotient, Qc at a given temperature is defined as the ratio of the product of concentrations of the reaction products to that of the reactants, each concentration term being raised to their individual stoichiometric coefficients in the balanced chemical equation, where the concentrations are not necessarily equilibrium values.

3. i) When Qc > Kc, the reaction will proceed in the
direction of reactants (reverse reaction), i.e., net reaction goes from right to left.
ii) When Qc < Kc, the reaction will proceed in the direction of products (forward reaction), i.e., net reaction goes from left to right.
iii) When Qc = Kc, the reaction mixture will be at equilibrium, i.e., no net reaction occurs.

Question 24.
Solubility product helps to predict the precipitation of salts from solution.
1. Find the relation between solubility (S) and solubility product (Ksp) of calcium fluoride and zirconium phosphate.
2. The solubility product of two sparingly soluble salts XY2 and AB are 4 × 10-15 and 1.2 × 10-16 respectively. Which salt is more soluble? Explain.
Answer:
1. The equilibrium in the saturated solution of calcium fluoride can be represented as,
CaF2(s) \(\rightleftharpoons \) Ca2+(aq) + 2F(aq)
Ksp = [Ca2+][F]2 = S.(2S)2 = 4S3
The equilibrium in the saturated solution of zirconium phosphate can be represented as,
Zr3(PO4)4(s) \(\rightleftharpoons \) 3Zr4+(aq) + 4PO43-(aq)
Ksp = [Zr4+]3[PO43-]4 = (3S)3.(4S)4 = 6912S7

2. XY2 is more soluble than AB.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 9

Ionic Strength Calculator … Ionic strength of a solution indicates the concentration of ionic charge in the solution.

Question 25.
a) How common ion effect can influence the solubility of ionic salts?
b) What is the application of common ion effect in gravimetric estimation?
Answer:
1. In a salt solution, if we increase the concentration of any one of the ions, according to Le Chatelier’s principle, it should combine with the ion of its opposite charge and some of the salt will be precipitated till Ksp = Qsp. Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till Ksp = Qsp.

2. The common ion effect is used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation.

Plus One Chemistry Equilibrium Four Mark Questions and Answers

Question 1.
Match the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 10
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 11

Question 2.
In Contact process, SO3 is prepared by the oxidation of SO2 as per the following reaction:
2SO2(g) + O2(g) \(\rightleftharpoons \) 2SO3(g); ∆H = -189.4
a) What happens to the rate of forward reaction when i) temperature is increased?
ii) pressure is decreased?
iii) a catalyst V2O5 is added?
b) Calculate the pH of 0.01 M H2SO4 solution. Also, calculate the hydroxyl ion concentration in the above solution.
Answer:
1. D When temperature is increased, the rate of forward reaction decreases since it is exothermic.
ii) When pressure is decreased the rate of forward reaction decreases since it is associated with decrease in number of moles.
iii) When a catalyst V2O5 is added the rate of both forward and backward reactions are increased by the same extent and equilibrium is reached earlier.
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 12

The method presented in our buffer pH calculator allows you to compute the pH of both arterial and venous blood.

Question 3.
Calculate the [H+] in the following biological fluids whose pH are given in brackets.
i) Human muscle fluid (6.83)
ii) Human stomach fluid (1.22)
iii) Human blood (7.38)
iv) Human saliva (6.4)
Answer:
i) pH =-log[H+] = 6.83
log[H+] = – 6.83
[H+] = antilog (- 6.83) = 1.48 × 10-7 mol L-1
ii) [H+] = antilog (- 1.22) = 6.03 × 10-2 mol L-1
iii) [H+] = antilog (- 7.38) = 4.17 × 10-8 mol L-1
iv) [H+] = antilog (- 6.4) = 3.98 × 10-7 mol L-1

Question 4.
The pH value of a solution determines whether it is acidic, basic or neutral in nature.
1. The concentration of hydrogen ion in the sample of a soft drink is 3.8 × 10-3 mol/L. Calculate its pH. Also predict whether the above solution is acidic, basic or neutral.
2. The dissociation constants of formic acid (HCOOH) and acetic acid (CH3COOH) are 1.8 × 10-4and 1.8 × 10-4 respectively. Which is relatively more acidic? Justify your answer.
Answer:
1. pH = – log[H+] = – log[3.8 × 10-3] = 2.42
Since pH is less than 7, it is an acidic solution,

2. HCOOH is more acidic.
Ka value is directly proportional to the acid strength, i.e., greater the Ka value, stronger is the acid.

Question 5.
a) Write the expression for Henderson – Hasselbalch equation for i) An acidic buffer & ii) A basic buffer.
b) Calculate the pH of a solution which is 0.1 M in
CH3COOH and 0.5 M in CH3COONa. Ka for CH3COOH is 1.8 × 10-6.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 13

Plus One Chemistry Equilibrium NCERT Questions and Answers

Question 1
A liquid is in equilibrium with its vapour in a sealed containerat a fixed temperature. The volume of the container is suddenly increased. (3)
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
a) Vapour pressure decreases due to increase in volume.
b) Rate of evaporation remains same and rate of condensation decreases.
c) Finally the same vapour pressure is restored and the rate of evaporation becomes equal to the rate

Question 2.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3 M. What is its pH? (2)
Answer:
pH = -log[H+]
= -log (3.8 × 10-3) = 2.42

Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Question 3.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. (2)
Answer:
H= -log [H+]
or log [H+] = -3.76 = -4.24
[H+] = antilog (-3.76) = 1.74 × 10-4M

Question 4.
The ionization constants of HF, HCOOH and HCN at 298 K are 6.8 × 10-4, 1.8 × 10-4 and 4.8 × 10-9 respectively. Calculate the ionization constants of the corresponding conjugate base. (3)
Answer:
The relation between ionization constant of an acid and that of its conjungate base is Ka x Kb = Kw
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 14

Question 5
The ionization constant of nitrous acid is 4.5 x 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. (2)
Answer:
Sodium nitrite is a salt of strong base and weak acid. Its degree of hydrolysis, h is given by the relation.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 15

Question 6
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine. (2)
Answer:
Pyridinium hydrochloride is a salt of a weak base (pyridine) and a strong acid (HCl). The pH of an aqueous solution of this salt is given by the relation:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium 16

Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Students can Download Chapter 1 Some Basic Concepts of Chemistry Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry

Plus One Chemistry Some Basic Concepts of Chemistry One Mark Questions and Answers

Plus One Chemistry Chapter Wise Questions And Answers Question 1.
Which of the following is a mixture?
a) Graphite
b) Sodium chloride
c) Distilled water
d) Steel
Answer:
d) Steel

Plus One Chemistry Chapter 1 Questions And Answers Question 2.
1 µ g = __________ g
[10-3, 10-6, 10-9, 10-12] ‘
Answer:
10-6

Plus One Chemistry First Chapter Questions And Answers Question 3.
The number of significant figures in 0.00503060 is __________ .
Answer:
6

Plus One Chemistry Chapter Wise Questions And Answers Pdf Question 4.
The balancing of chemical equations is based on which of the following law?
a) Law of multiple proportions
b) Law of conservation of mass
c) Law of definite proportions
d) Gay-Lussac law
Answer:
b) Law of conservation of mass

By using the percent/actual yield calculator, you can get the accurate amount of percent, actual, and theoretical yield produced in a chemical reaction.

Plus One Chemistry Questions And Answers Question 5.
Which among the following is the heaviest?
a) 1 mole of oxygen
b) 1 molecule of sulfur trioxide
c) 100 u of uranium
d) 44 g carbon dioxide
Answer:
d) 44 g carbon dioxide

Plus One Chemistry Previous Year Question Papers And Answers Chapter Wise Question 6.
Calculate the number of atoms in 48 g of He?
Answer:
Gram atomic mass of He = 4 g.
Thus, numberofatomsin4g (1 mol) He = 6.02 × 1023
So number of atoms in 48 g of He = \(\frac{48}{4}\) × 6.02 × 1023
=12 × 6.02 × 1023
= 7.224 × 1024

Hsslive Chemistry Previous Questions And Answers Chapter Wise Plus One Question 7.
One mole of CO2 contains how many gram atoms?
Answer:
3 gram atoms.

Plus One Chemistry Previous Year Questions And Answers Chapter Wise Question 8.
The ratio of gram atoms of Au and Cu in 22ct gold is __________
Answer:
7 : 2

Plus One Chemistry Chapter Wise Questions And Answers Pdf Download Question 9.
A compound contains 69.5% oxygen and 30.5% nitrogen and its molecular weight is 92. The compound will be
Answer:
N2O4

Plus One Chemistry Chapter 1 Question 10.
The total number of electrons present in 1 mole of water is
Answer:
6 × 1024.

Hsslive Plus One Chemistry Chapter Wise Questions And Answers Question 11.
40g NaOH is present in 100 ml of a solution. Its molarity is __________
Answer:
10 M

Plus One Chemistry Some Basic Concepts of Chemistry Two Mark Questions and Answers

Plus One Chemistry Some Basic Concepts Of Chemistry Questions Question 1.
Classify the following substances into homogeneous and heterogeneous mixtures.

  • Milk
  • Iron
  • Air
  • Gasoline
  • Kerosene
  • Muddy water

Answer:

HomogeneousHeterogeneous
Milk, Iron
Gasoline
Air
Kerosene
Muddy Water

Plus One Chemistry Previous Year Questions Chapter Wise Question 2.
Calculate the volume occupied by 4.4 g of CO2 at STP?
Answer:
1 mole CO2 = 44 g
4.4 CO2 = 0.1 mole CO2
Volume occupied by 1 mol CO2 at STP = 22.4 L
∴ Volume occupied 0.1 mol CO2 at STP = 0.1 × 22.4 L
= 2.24 L

Plus One Chemistry Chapter 1 Previous Questions And Answers Question 3.
During a group discussion a student argued that “the water of sea and river should have different chemical composition”.

  1. What is your opinion?
  2. Which law would you suggest to support your answer?
  3. State the law.

Answer:

  1. I can’t join with him.
    The water of sea and water of river must have the same chemical composition.
  2. Law of definite proportions.
  3. A given compound always contains exactly the same proportion of elements by weight.

Plus One Chemistry Question And Answer Question 4.
“When science developed some theories are also modified”.
Write the modified atomic theory.
Answer:

  1. Atom is no longer considered as indivisible, it has been found that atom is made up of sub atomic particles called protons, neutrons and electrons.
  2. Atoms of the same element may not be similar in all respects.
  3. Atoms of different elements may be similar in one or more respects.
  4. The ratio in which atomic unit may be fixed and integral but may not be simple.
  5. The mass of atom can be changed into energy.

Plus One Chemistry Chapter Wise Previous Questions And Answers Question 5.
Carbon combines with oxygen to form CO and CO2.

  1. What is the law behind this?
  2. State the law.

Answer:

  1. Law of multiple proportions.
  2. If two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Hss Live Plus One Chemistry Chapter Wise Questions And Answers Question 6.
Calculate the volume occupied by 6.02×1025 molecules of oxygen at STP.
Answer:
Volume occupied by 1 mole of oxygen gas at STP = 22.4 l
i.e., Volume occupied by 6.02×1023 molecules of oxygen gas at STP = 22.4 l
Hence the volume occupied by 6.02 × 1025 molecules
of oxygen gas at STP = \(\frac{22.4 \times 6.02 \times 10^{25}}{6.02 \times 10^{23}}\) = 2240 l.

Plus One Chemistry Chapter Wise Questions And Answers Hsslive Question 7.
Calculate the molality of a solution of NaOH containing 20g of NaOH in 400 g solvent.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 1

Question 8.
Calculate the mole fraction of NaOH in a solution containing 20 g of NaOH per 360 g of water.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 2

Question 9.
12 g of carbon reacts with 32 g of oxygen to form 44g of carbon dioxide.

  1. Which law of chemical combination is applicable here?
  2. State the law.

Answer:

  1. Law of conservation of mass.
  2. Matter can neither be created nor destroyed. Or, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.

Question 10.
When hydrogen and oxygen combine to form water, the ratio between volume of reactants and products is 2:1:2.

  1. Which law of chemical combination is applicable here?
  2. State the law.

Answer:

  1. Gay Lussac’s law of gaseous volumes.
  2. When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Question 11.
Carbon form two oxides, the first contains 42.9% C and the second contains 27.3% carbon. Show that these are in agreement with the law of multiple proportions.
Answer:
In the first compound:
C = 42.9%
O = 100-42.9 = 57.1%
So, the ratio between the masses of C and O = 42.9:57.1 = 1:1.33
In the second compound:
C = 27.3%
O= 100-27.3= 72.7%
So, the ratio between the masses of C and O = 27.3 : 72.7= 1:2.66
Hence, the ratio of masses of oxygen which combines with a fixed mass of carbon is 1.33:2.66 or 1:2, a simple whole number ratio. This illustrates the law of multiple proportions.

Question 12.
Match the following:

AB
1 amu1.008 x 1.66 x1024
Mass of 1 H atom6.02 x 1023
Molar volume of O2 at STP11.2L
Volume of 14g of N2 at STP1.66 x 1024
Avogadro number22.4L

Answer:

AB
1 amu1.66 x 1024
Mass of 1 H atom1.008 x 1.6 x1024
Molar volume of O2 at STP22.4L
Volume of 14g of N2 at STP11.2L
Avogadro number6.02 x 1023

Question 13.
Calculate the molality of a solution containing 10 g ofNaOH in 200 cm3 of solution. Density of solution is 1.4 g/mL. (Molar mass of NaOH = 40)
Answer:
Mass of the solution = 200 × 1.04 = 208 g
Mass of NaOH (WB) = 10g Molar mass of NaOH (MB) = 40 g mol-1
Mass of water (WA) = (208 -10) g = 198 g = 0.198 kg
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 3

Question 14.
Calculate the mass percentage of oxygen in CaCO3.
Answer:
Molecular mass of CaCO3 = 100 g mol-1
Mass of oxygen in 100 g CaCO3=3 × 16 g = 48 g
Percentage of oxygen in CaCO3 =\(\frac{48}{100}\)×100 = 48%

Question 15.
KCIO3 on heating decomposes to KCI and O2. Calculate the mass and volume of O2 produced by heating 50 g of KCIO3.
Answer:
The reaction is represented as,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 4
According to the equation, 96 g of oxygen is obtained from 245 g of KCIO3.
Hence mass of oxygen obtained from 50 g KCIO3 is \(\frac{96 \times 50}{245}=19.6 \mathrm{g}\)
According to the equation 245 g of KCIO3 gives 3 moles of O2 at STP which is 3 × 22.4 L = 67.2 L
Volume of oxygen liberated by 50g of KCIO3
= \(\frac{67.2 \times 50}{245}=13.71 \mathrm{L}\)

Question 16.
Calculate the number of molecules present in

  1. 11g of CO2.
  2. 56 mL Of CO2 at STP.

Answer:
1. \(\frac{11}{44}\) = 0.25 mole
1 mole of CO2 contains 6.022 × 1023 molecules.
∴ 0.25 mole CO2 contains 6.022 × 1023 × 0.25
= 1.51 × 1023 molecules.

2. 56 mL = 0.056 L
\(\frac{0.056}{22.4}\) =0.0025 mole
= 6.022 × 1023 × 0.0025=1.5 × 1021 molecules

Question 17.
Calculate the number of moles of 02 required to produce 240 g of MgO by burning Mg metal. [Atomic mass: Mg=24, 0=16]
Answer:
2 Mg + O2 → 2MgO
No. of moles of MgO = \(\frac{240}{40}\)=6
No. of moles of 02 required = 6/2 = 3

Question 18.
Arrange the following in the increasing order of their mass.
(a) 1 g of Ca
(b) 12 amu of C
(c) 6.022 × 1023 mol-ecules of CO2
(d) 11.2 L of N2 at STP
Answer:
a) Mass of 1 g Ca = 1 g
b) Mass of 12 amu C = \(\frac{12}{6.022 \times 10^{23}}\) = 2 × 10-23
c) Mass of 6.022 × 1023 molecules of CO2 = 44 g
d) Mass of 11.2 L of N2 at NTP = \(\frac{28 \times 11.2}{22.4}\) = 14 g
(b) < (a) < (d) < (c)

Question 19.
Complete the table:

42g N21.5 mole N233600mLN2 (STP)
16g 0: – – – – mole O211.2 L of O2 (STP)
….g CO21 mole CO2– – – – L of C O2 (STP)
28g CO1 mole CO– – – – mLCO (STP)

Answer:

42g N21.5 Mole N233600mLN2 (STP)
16g O20.5 Mole O211.2 L of O2 (STP)
44 g CO21 mole CO222.4 L of CO2 (STP)
28g CO1 mole CO22400mLCO (STP)

Question 20.
1. Irrespective of the source, pure sample of H20 always contains 88.89% by mass of oxygen and 11.11% by mass of hydrogen.
a) Which law is illustrated here?
b) State the law.
2. Complete the table by filling in the blanks:

48 g O21.5 mol O2……mL O2 (at STP)
…… g Na2 gram atom Na2NA Na atoms
…….g CO22.5 mol CO256 L (at STP)
8.5 g NH3……mol NH311.2 L (at STP)

Answer:
1. a) Law of definite proportions.
b) The same chemical compound always contains the same elements combined in the same fixed proprotion by mass.
2.

48 g O21.5 mol O233600 mL O2 (at STP)
46g Na2 gram atom Na2Na Na atoms
110 g CO22.5 mol CO256 L (at STP)
8.5 g NH30.5 mol NH311.2L(atSTP)

Question 21.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?.
Answer:
Mass of HCI in 0.25 mL 0.75 M HCl \(=\frac{36.5 \times 0.75 \times 0.25}{1000}=0.6844 \mathrm{g}\)
As per reaction 100 g CaCO3 reacts with 2 × 36.5 = 73 g of HCl
∴ Mass of CaCO3 reacting with 0.6844 g HCl \(=\frac{100 \times 0.6844}{73}=0.9375 \mathrm{g}\)

Plus One Chemistry Some Basic Concepts of Chemistry Three Mark Questions and Answers

Question 1.
During a Seminar, a student remarked that “Dalton’s atomic theory has some faulty assumptions”.
a) Do you agree with him?
b) What is the present status of Dalton’s atomic theory?
c) Write any two wrong postulates of Dalton’s atomic theory.
Answer:
a) I agree with him. Out of 6 Dalton’s postulates, 5 postulates are faulty and only one is correct.
b) Dalton’s atomic theory has undergone many modifications.
c)

  • All substances are made up of small indivisible particles called atoms.
  • Atoms of the same elements are identical in mass and other properties.

Question 2.
One gram atom of an element contains 6.023 × 1023 atoms.

  1. Find the number of atoms in 8 g oxygen.
  2. Which is heavier, 1 oxygen atom or 10 hydrogen atoms?
  3. Define mole and Avogadro number.

Answer:
1. 16 g oxygen contains 6.022 × 1023 atoms
∴ 8 g oxygen contains \(\frac{6.022 \times 10^{23}}{2}\) = 3.011 × 1023

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 5

3. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
Avogadro number – It is the number of discrete particles present in 1 mole of any substsnce. (Avogadro number, NA = 6.022 × 1023)

Question 3.

  1. Classify the following as homogeneous and heterogeneous mixtures.
    Air, Smoke, Gunpowder, NaCI solution, Petrol, Bronze, Mixture of sugar and sand.
  2. State and explain law of multiple proportions with example.

Answer:
1. Homogeneous-Air, NaCI solution, Bronze, Gun powder, Petrol.
Heterogeneous – Mixture of sugar and sand, Smoke.

2. When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other element bear a simple ratio. ,
eg. Carbon reacts with oxygen to form two compounds viz. CO and CO2. In CO mass ratio is 12:16. In CO2 mass ratio is 12:32. Then mass ratio between oxygen in the 2 compounds is 16:32 or 1:2 which is a simple whole number ratio. Hence, the law is verified.

Question 4.
1. One mole of an ideal gas occupies 22.4 L at STP
a) Calculate the mass of 11.2 L of oxygen gas at STP.
b) Calculate the number of atoms present in the above sample.
2. 21 g of nitrogen gas is mixed with 5 g of hydrogen gas to yield ammonia according to the equation.
N2 + 3H2 → 2NH3
Calculate the maximum amount of ammonia that can be formed.
Answer:
1. a) Mass of 22.4 L oxygen at STP = 32 g
∴ Mass of 11.2 L oxygen at STP = 16 g
b) No. of atoms present in 16 g of O2
\(\frac{6.02 \times 10^{23}}{2}\) ×2 = 6.02 × 1023 atoms

2. N2 + 3H2 → 2NH3
1 mole N2 + 3 mole H2 → 2moles NH3
1 mole N2 requires 3 mol H2
i.e., 28g N2 requires 6 g H2
Hence, 21 g N2 requires \(\frac{6 \times 21}{28}\) = 4.5 g H2
21 g N2 reacts completely and 0.5g H2 remains unreacted.
Hence, N2 is the limiting reagent.
28g N2 gives 2 × 17 g NH3
∴ 21 g N2 gives \(\frac{2 \times 17 \times 21}{28}\) = 25.5 g NH3

Question 5.
When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other bear a simple ratio.
i) Name the above law.
ii) Explain the above law by taking oxides of carbon.
Answer:
i) Law of multiple proportions.
ii) Carbon reacts with oxygen to form two compounds viz. CO and CO2.
In CO, mass ratio is 12:16
In CO2,mass ratio is 12:32
Ratio of the masses of oxygen combining with a fixed mass of carbon in the two compounds is 16:32 or 1:2, which is a simple whole number ratio.

Question 6.
A compound contains 80% carbon and 20% hydrogen. If the molecular mass is 30 calculate empirical formula and molecular formula.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 6

Question 7.
A compound contains 4.07% of hydrogen, 24.27% of carbon and 71.65% of chlorine. The molar mass is 98.96. What is the empirical and molecular formula?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 7

Question 8.
Nitrogen forms various oxides.
1. Identify the law of chemical combination illustrated here. Also state the law.
2. Determine the formula of each oxide from the given data and illustrate the law.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 8

Answer:
1. Law of multiple proportions.
When two elements combine to form more than one compound the different mass of one of the elements which combine with the fixed mass of the other element bear a simple ratio.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 9
In NO and NO2, the masses of oxygen combining with a fixed mass (14 g) of nitrogen are in the ratio, 16:32 = 1:2. Similarly, in N2O and N2O3, the masses of oxygen combining with a fixed mass (28 g) of nitrogen are in the ratio, 16:48 = 1:3. These are simple whole number ratios. Hence, the law of multiple poportions is verified.

Plus One Chemistry Some Basic Concepts of Chemistry Four Mark Questions and Answers

Question 1.
Which of the following weighs more?
a) 1 mole of glucose
b) 4 moles of oxygen
c) 6 moles of N
d) 5 moles of sodium
Answer:
a) 1 mole glucose = (72 + 12 + 96) g = 180 g
b) 4 moles of oxygen = 4 × 32g = 128g
c) 6 moles of nitrogen = 6 × 14 g = 84 g
d) 5 moles of Na = 5 × 23 g = 115 g.
Thus, 1 mole glucose weighs more.

Question 2.
3 g of H2 is mixed with 29 g of O2 to yield water.
1. Which is the limiting reagent?
2. Calculate the maximum amount of water that can be formed.
3. Calculate the amount of the reactants which remains unreacted.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 10
According to the equation, 4 g H2 requires 32 g
So 3 g H2 requires \(\frac{377 \times 33}{44}\) = 24 g O2.
Here 3 g H2 is mixed with 29 g of O2. All H2 will react. Hence H2 is the limiting reagent.

2. According to the equation, 4 g H2 gives 36 g H2O. Hence 3 g H2 will give 36 × 3/4 = 27 g H2O.

3. Amount of O2 unreacted = (29 – 24)g = 5 g

Question 3.
a) Calculate the mass of oxygen required for the complete burning of 2 g of carbon.
b) Calculate the molar mass of (i) CO2 (ii) CH4
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 11

Question 4.
One gram mole of a substance contains 6.022 x 1023 molecules.
1. 24 g of carbon is treated with 72 g of oxygen to form CO2. Identify the limiting reagent.
2. Find the number of molecules of CO2 formed in this situation.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 12

2 mol of C requires 2 mol of O2.
2 mol C completely reacts with 2 mol of O2 and 0.25 mol O2 and 0.25 mol O2 remains unreacted. Hence, C is the limiting reagent.

2. No. of moles of CO2 formed = 2
∴ No. of molecules of CO2 formed
= 2 × 6.022 × 1023 = 1.2044 × 1024

Question 5.
One gram mole of a substance contains 6.022×1023 molecules.
i) Find out the number of molecules in 2.8 g of nitrogen.
ii) Which is the heavier-one SO2 molecule or one CO2 molecule?
Answer:
i) No. of molecules in1 mole of N2 = 6.022 × 1023
i.e., No. of molecules in 28 g of N2 = 6.022 × 1023
∴ No. of molecules in 2.8 g N2
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 13

Question 6.
a) How can you illustrate the law of multiple proportions by using oxides of metals containing 78.7% and 64.5% of the metal?
b) Match the following:
1/12th the mass of C12 atom – 1 mole
1 g of hydrogen atom – amu
22.4 L O2 at NTP – gram mole
180 g of glucose – gram atom
6.022 × 1023 particles – molar volume
Answer:
a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively.
First Oxide :
Mass of oxygen = 100 – 78.7 = 21.3 g
No. of parts by mass of oxygen combining with one part by mass of metal =\(\frac{78.7}{21.3}=3.7 \mathrm{g}\)

Second oxide:
Mass of oxygen = 100 – 64.5 = 35.5 g
No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}=1.9 \mathrm{g}\)
The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio.

b) 1/12th the mass C12 atom – amu
1 g of hydrogen atom – gram atom
22.4 L O2 at NTP – molar volume
180 g of glucose – gram mole
6.022 × 1023 particles – 1 mole

Question 7.
Calculate
1. The number of molecules present in 1 g of water.
2. The volume of 0.2 mole of sulphur dioxide at STP.
Answer:
1. Number of moles in 1 g water = \(\frac{1}{8}\)
∴ No. of molecules in 1 g water
\(=\frac{1 \times 6.022 \times 10^{23}}{18}=3.35 \times 10^{22}\)

2. Volume of 0.2 mol S02 at STP = 0.2 × 22.4 litre
= 4.48 litre

Question 8.
“One mole of all substances contain the same number of specified particles.”
a) Justify the statement.
b) Howto connect mole, gram mole, and gram atom?
c) What is the relation between number of moles and volume?
d) Calculate the number of moles of a gas in 11.2 L at • STP.
Answer:
a) This statement is true i.e., one mole of all sub-stances contain the same number of specified particles. According to Avogadro’s law.
1 mole of any substance contains 6.022 × 1023 specified particles.

b)
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 14
1 gram mole is the molecular mass expressed in gram. It is the mass of 1 mole molecules in gram. Thus, 1 gram mole contains 1 mole molecules.
1 gram atom is the atomic mass expressed in gram. It is the mass of 1 mole atoms in gram. Thus, 1 gram atom contains 1 mole atoms.

c) Number of moles is directly proportional to volume (according to Avogadro law).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 15

Plus One Chemistry Some Basic Concepts of Chemistry NCERT Questions and Answers

Question 1.
Calculate the molecular mass of the following : (3)
1. H2O
2. CO2
3. CH4
Answer:
1. Molecular mass of H2O = 2(1.008 u) +16.00 u
= 18.016u

2. Molecular mass of CO2 = 12.01 u + 2(16.00 u)
= 44.01 u

3. Molecular mass of CH4 = 12.01 u + 4(1.008 u)
= 16.042 u

Question 2.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 16

Question 3
Calculate the amount of carbon dioxide that could be produced when (3)
1. 1 mole of carbon is burnt in air.
2. 1 mole of carbon is burnt in 16 g of dioxygen.
3. 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
1. The balanced equation for the combustion of carbon dioxide in dioxygen in air is
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 17

In air combustion is complete.
Hence, Amount of CO2 produced when 1 mole of carbon is burnt in air = 44 g

2. As only 16 g dioxygen is available it is the limiting reagent.
Hence, amount of CO2 produced = \(\frac{44}{32}\)×16 = 22 g

3. Here again, dioxygen is the limiting reactant. Therefore, amount of CO2 produced from 16 g dioxygen = \(\frac{44}{32}\)×16 = 22g

Question 4.
Chlorine is prepared in the laboratory by treating manganase dioxide (MnO2) with aqueous hydrochloric acid according to the reaction,
4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
(Atomic mass of Mn = 54.94 u) (2)
Answer:
1 mol of MnO2, i.e., 54.94 + 32 = 86.94 g MnO2 react with 4 moles of HCl, i.e., 4 × 36.5 g = 146 g of HCl.
∴ Mass of HCl reacting with 5.0 g of MnO2
\(=\frac{146}{86.94} \times 5 \mathrm{g}=8.4 \mathrm{g}\)

Question 5.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 1 Some Basic Concepts of Chemistry 18

Plus Two Chemistry Notes Chapter 2 Solutions

Students can Download Chapter 2 Solutions Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Notes Chapter 2 Solutions

Solutions:
homogeneous mixtures of two or more pure substances, having uniform composition and properties throughout. The substances forming a solution are called components.

Solvent and Solute:
The component that is present in the largest quantity is known as solvent.

One or more components present in the solution other than solvent are called solutes. e.g. In sugar solution, water is the solvent and sugar is the solute.

Binary solution:
A solution containing only two components.

Aqueous solutions:
solutions in which the solvent is water.

Types of Solutions
Plus Two Chemistry Notes Chapter 2 Solutions 1

Percent Concentration. One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute.

Expressing Concentration of Solutions :
The concentration of a solution is defined as the amount of solute present in the given quantity of the solution.

1. Mass percentage (w/w) :
The mass % of a component in a given solution is the mass of the component (solute) per 100 g of solution.
Plus Two Chemistry Notes Chapter 2 Solutions 2
e.g. 10% glucose solution means 10 g of glucose dissolved in 90 g of water resulting in a 100 g solution.

2. Volume percentage (v/v) :
The volume % of a component in a given solution is the volume of the component per 100 volume of solution.
Plus Two Chemistry Notes Chapter 2 Solutions 3
Example:
10% ethanol solution means 10 mL of ethanol dissolved in 90 mL of water.

Plus Two Chemistry Notes Chapter 2 Solutions

3. Mass by volume percentage (w/v):
It is the mass of solute dissolved in 100 mL of the solution. Used in medicine and pharmacy.

4. Parts per million (ppm):
It is the parts of a solute (component) per million parts of the solution. When a solute is present in very minute amounts, parts per million (ppm) is used.
Plus Two Chemistry Notes Chapter 2 Solutions 4

Use of Mole Fraction Equation … This formula is easy to use if you know the number of moles of all the solutes and solvents.

5. Mole fraction (X):
ratio of number of moles of one component to the total number of moles of all the components present in the solution.
Plus Two Chemistry Notes Chapter 2 Solutions 5

For a binary solution, nA be the number of moles of A and nB be the number of moles of B.
Plus Two Chemistry Notes Chapter 2 Solutions 6

The sum of mole fractions of all the components present in the solution is always equal to 1.
i.e., χA + χB = 1
Fora solution containing ‘i’ number of components,
χ1 + χ2 +……………… + χi = 1
Mole fraction is independent of temperature.

6. Molarity (M):
number of moles of solute dissolved in one litre of the solution.
Plus Two Chemistry Notes Chapter 2 Solutions 7

7. Molality (m):
number of moles of solute per kilogram of the solvent.
Plus Two Chemistry Notes Chapter 2 Solutions 8

Solubility:
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a particular temperature.

Factors affecting solubility
Nature of the solute, nature of the solvent, temperature ‘ and pressure

Solubility of Solids in Liquids :
Like dissolves like:
Polar solutes are soluble in polar solvents and non-polar solutes are soluble in non-polar solvents.

Unsaturated solution:
Solution in which more solute can be dissolved at the same temperature.

Plus Two Chemistry Notes Chapter 2 Solutions

Saturated solution:
Solution in which no more solute can be dissolved at the same temperature and pressure.

Effect of temperature :
Solubility increases with temperature if the reaction is endothermic. Solubility decreases with temperature if the reation is exothermic.

Effect of pressure :
Pressure does not have any significant effect on solubility of solids in liquids because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

Solubility of a Gas in a Liquid :
It is greately affected by pressure and temperature.

Effect of pressure
Henry’s law :
The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
The most commonly used form of Henry’s law states that the partial pressure of the gas in the vapour phase (p) is proportional to the molefraction of the gas (χ) in the solution.
P = KH
where KH is the Henry’s law constant.
Different gases have different KH values at the same temperature. Thus, KH is a function of the nature of the gas.

Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.

The solubility of gases increase with decrease of temperature. Therefore, aquatic species are more comfortable in cold waters rather than in hot waters.

Applications of Henry’s law
1. To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
2. To avoid bends (a medical condition which is painful and dangerous to life caused by the formation of bubbles of N2 in the blood) the tanks used by scuba divers are filled with air diluted with He (11.7% He, 56.2% N2 and 32.1% O2).
3. At high altitudes, low pressure leads to low concentrations of O2 in blood. It causes climbers to become weak and unable to think clearly (anoxia).

Plus Two Chemistry Notes Chapter 2 Solutions

Effect of temperature :
Dissoloution of gases in liquids is an exothermic process. Hence, according to Le Chatelier’s principle solubility of gases in liquids decreases with rise in temperature.

Vapour Pressure of Liquid Solutions

Vapour Pressure of Liquid-Liquid Solutions:
Consider the two volatile liquids denoted as ‘A’ and ‘B’. When both liquids are taken in a closed vessel, both components would evaporate and an equilibrium would be established between liquid and vapour phase.
Let, PA– Partial vapour pressure of component A’
PB – Partial vapour pressure of component ‘B’
χA Mole fraction of A
χB Moiefraction of B

Raoult’s Law :
The law states that fora solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
For component ‘A’
PA ∝ χA.
PA= P°A χA
where P°A is the vapour pressure of pure component ‘A’ at the same temperature.
Similarly, for component ‘B’
PB ∝ χB
PB= P°B χB
where PB° is the vapour pressure of pure component ‘B’. Rauolt’s law also states that, at a given temperature for a solution of volatile liquids, the partial vapour pressure of each component is equal to the product of the vapour pressure of pure component and its mole fraction.

According to Dalton’s law of partial pressures,
Total pressure, P[Total] = PA + PB
Plus Two Chemistry Notes Chapter 2 Solutions 9

A plot of PA or PB versus the mole fractions χA and χB for a solution gives a linear plot as shown in the figure.
Plus Two Chemistry Notes Chapter 2 Solutions 10

Raoult’s Law as a special case of Henry’s Law:
According to Raoult’s law, the vapour pressure of volatile liquid in a solution is proportional to its mole fraction, i.e., Pi = Pi° χi

According to Henry’s law, the vapour pressure of a gas in a liquid is proportional to its mole fraction, i. e., p=KHχ

Thus, Raoult’s law becomes a special case of Henry ’s law in which KH becomes equal to Pi°.

Vapour Pressure of Solution of Solids in Liquids:
If a non-volatile solute is added to a solvent to give a solution, the surface of solution has both solute and solvent molecules; thereby the fraction of surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecules escaping from the surface is reduced. Hence, the vapour pressure of solution is lower than vapour pressure of pure solvent.
Plus Two Chemistry Notes Chapter 2 Solutions 11

General form of Raoult’s Law:
For any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

In a binary solution, let us denote the solvent by ‘A’ and solute by ‘B’.
According to Raoult’s law,
PA ∝ χA
PA = PA° χA
Total pressure, P = PA Here, PB = 0
(∵ solute is non-volatile)
P = PA° χA
For binary solution,
χA + χB = 1
χA = 1 – χB
Thus, the above equation becomes,
Plus Two Chemistry Notes Chapter 2 Solutions 12
lowering of vapour pressure.

Ideal and Non-ideal Solutions :
Ideal Solutions:
The solutions which obey Raoult’s law over the entire range of concentrations.

Important properties of Ideal Solutions
i. PA = P°A χA ; PB = P°B χB
ii. Enthalpy of mixing is zero (∆mixH = 0)
iii. Volume of mixing is zero (∆mixV = 0)

If the intermolecular attractive forces between A – A and B – Bare nearly equal to those between A – B, it leads to the formation of ideal solution.

Plus Two Chemistry Notes Chapter 2 Solutions

Examples:

  1. Solution of n-hexane and n-heptane
  2. Solution of bromoethane and chloroethane
  3. Solution of benzene and toluene

Non-ideal Solutions :
solutions which do not obey Raoult’s law overthe entire range of concentration. The vapour pressure of such solutions is either higher or lower than that predicted by Raoult’s law.

If the vapour pressure is higher, it exhibits positive deviation and if the vapour pressure is lower it exhibits negative deviation from the Raoult’s law.

Solutions showing positive deviation :
the intermolecular attractive forces between the solute- solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. Thus, in such solutions molecules will find it easier to escape than in pure state. This will increase the vapour pressure and results in the positive deviation.
Plus Two Chemistry Notes Chapter 2 Solutions 13
(dotted line represents graph for ideal solution).
Examples:
Ethanol + Water, Ethanol + Acetone, CCl4 + Chloroform, C6H6 + Acetone , n-Hexane + Ethanol

Solution showing negative deviation:
In the case of negative deviation, the intermolecular attractive forces between solvent-solute molecules are greater than those between solvent-solvent and solute-solute molecules and leads to decrease in the vapour pressure.
Plus Two Chemistry Notes Chapter 2 Solutions 14

Examples:
1. Mixture of phenol and aniline – In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules.
2. Mixture of acetone and chloroform – Here chloroform molecule is able to form hydrogen bond with acetone molecule.
Plus Two Chemistry Notes Chapter 2 Solutions 15

3. H2O + HCl, (4) H2O + HNO3, (5) CHCl3 + (C2H5)2O

Azeotropes:
binary mixtures having same composition in liquid and vapour phase and boil at a constant temperature. It is not possible to separate the components of azeotropes by fractional distillation.

Plus Two Chemistry Notes Chapter 2 Solutions

Solutions which show large positve deviation from Raoult’s law form minimum boiling azeotrope at a specific composition. For example, ethanol-water mixture forms a minimum boiling azeotrope (b.p. 351.1 K) when approximately 95% by volume of ethanol is reached.

The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. For example, nitric acid and water form a maximum boiling azeotrope (b.p. 393.5 K) at the approximate composition, 68% nitric acid and 32% water by mass.

Colligative Properties :
properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. These are,
i. Relative lowering of vapour pressure of the solvent \(\left(\frac{\Delta p_{1}}{p_{1}^{0}}\right)\)
ii. Elevation of boiling point of the solvent (∆Tb)
iii. Depression of freezing point of the solvent (∆Tf)
iv. Osmotic pressure of the solution (π)

Relative Lowering of Vapour Pressure:
When a non-volatile solute (B) is dissolved in a liquid solvent (A), the vapour pressure of the solvent is lowered. This phenomenon is called lowering of vapour pressure. It depends only on the concentration of the solute particles and it is independent of their identity. The relation between vapour pressure of solution, mole fraction and vapour pressure of the solvent is given as,
PA = χAA ……………(1)
The lowering of vapour pressure of solvent ∆ PA is given as,
∆ PA = P°A – PA ……………(2)
Substitute the equation (1) in (2)
∆ PA = P°A – P°AχA
= P°A(1 – χA)
∆ PA = P°AχB …………..(3) ∵ (1 – χA) = χB
The relative lowering of vapour pressure is given as,
Plus Two Chemistry Notes Chapter 2 Solutions 16
of vapour pressure and is equal to the mole fraction of solute.
From equation (4),
Plus Two Chemistry Notes Chapter 2 Solutions 17
For dilute solutions nB < < nA, hence neglecting nB In the denominator, the above equation becomes,
Plus Two Chemistry Notes Chapter 2 Solutions 18
where wA and wB are the masses and MA and MB are the molar masses of solvent and solute respectively.
Plus Two Chemistry Notes Chapter 2 Solutions 19

Elevation of boiling point (∆Tb):
The boiling point of a solution is higher than that of the pure solvent. The elevation in the boiling point depends ‘ on the number of solute molecules rather than on their nature.
Plus Two Chemistry Notes Chapter 2 Solutions 20

Let T°b be the boiling point of pure solvent and Tb be the boiling point of solution. The increase in the boiling point ∆Tb = Tb – T°b is known as elevation of boiling point.

Plus Two Chemistry Notes Chapter 2 Solutions

For a dilute solution, the elevation of boiling point ( ∆Tb) is directly proportional to the molal concentration of the solute in a solution (i.e., molality).
∆Tb ∝ m
∆Tb = Kbm …………(1)

where, m → molality and Kb → Boiling Point Elevation Constant/Molal Elevation Constant/ Ebullioscopic Constant.
Unit of Kb is K kg mol-1 Or K m-1
Plus Two Chemistry Notes Chapter 2 Solutions 21
Substituting the value of‘m’ in equation (1),
Plus Two Chemistry Notes Chapter 2 Solutions 22

Depression of Freezing point (∆Tf) :
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent.
Plus Two Chemistry Notes Chapter 2 Solutions 23
Let T°f be the freezing point of pure solvent and Tf be the freezing point of solution.
Depression in freezing point ∆Tf= T°f – Tf
For a dilute solution, depression of freezing point (∆Tf) is directly proportioned to molality (m) of the solution. Thus,
∆Tf ∝ m
∆Tf = Kfm ………………(1)
where, Kf – Freezing Point Depression Constant/ Molal Depression Constant/Cryoscopic Constant.
Unit of Kf is K kg mol-1 Or K m-1
Plus Two Chemistry Notes Chapter 2 Solutions 24
[Note: The values of Kb and Kf, depend upon the nature of the solvent. They Can be ascertained from the following equations:
Plus Two Chemistry Notes Chapter 2 Solutions 25
where,
R → Gas constant, MA → Molar mass of solvent
Tb → Boiling point of pure solvent of kelvin
Tf → Freezing point of pure solvent in kelvin
fusH → Enthalpy of fusion, ∆vapH → enthalpy of vapourisation.
For water, Kb = 0.52 K kg mol-1 and Kf = 1.86 K kg mol-1]

Osmosis and Osmotic Pressure:
The process of flow of the solvent molecules from pure solvent to the solution through semipermeable membrane (SPM) is called osmosis.

Semi Permeable Membrane :
The membrane which allows the passage of solvent molecules but ’ not the solute molecule is called SPM.

Example:
Parchment paper, Pig’s bladder, Cell wall, Film of cupric ferrocyanide.

Plus Two Chemistry Notes Chapter 2 Solutions

Osmotic Pressure (π):
the excess pressure which must be applied to a solution to prevent osmosis or the pressure that just stops the flow of solvent.

Osmotic pressure (π) is proportional to the molarity (C) of the solution at a given temperature (T K).
π = CRT, where R is the gas constant.
π = \(\frac{n_{B}}{V}\)RT, where nc is the number of moles of the solute and V is the volume of the solution in litres.
π = nBRT
π V= \(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}}\)RT , where wB is the mass of the solute and MB is the molar mass of the solute.
Or MB = \(\frac{\mathbf{w}_{\mathrm{B}} \mathrm{RT}}{\pi \mathrm{V}}\)

Osmotic pressure measurement is widely used to determine molar mass of proteins, polymers and other macro molecules.

Advantages of osmotic pressure method:
i) pressure measurement is around the room temperature
ii) molarity of the solution is used instead of molality
iii) the magnitude of osmotic pressure is large compared to other colligative properties even for very dilute solutions.

Isotonic Solution :
Two solutions having same (equal) osmotic pressure at a given temperature. A 0.9% solution of NaCI (normal saline solution) is isotonic with human blood, and it is safe to inject intravenously.

Hypertonic Solution :
A solution having higher osmotic pressure than another solution.
Hypotonic Solution :
A solution having lower osmotic pressure than another solution.

Reverse Osmosis:
flow of the pure solvent from solution side to solvent side through semipermeable membrane when a pressure larger than the osmotic pressure is applied to the solution side.

Uses of reverse osmosis:
Desalination of sea water, Purification of water.

Plus Two Chemistry Notes Chapter 2 Solutions

Abnormal Molar Mass :
In some cases, the molar mass determined by colligative properties do not agree with the theoretical values. This is due to association ordissociation of the solute particles in the solution.

Association of Solute Particles :
When solute particles undergo association the number of the solute particles in the solution decreases. Consequently, the experimental values of colligative properties are less than the expected values, e.g. Molecules of ethanoic acid (acetic acid) dimerise in benzene due to intermolecular hydrogen bonding.
Plus Two Chemistry Notes Chapter 2 Solutions 26
Similarly, benzoic acid undergo dimerisation when dissolved in benzene.

Dissociation of Solute Particles :
When the solute particles dissociate or ionise in the solvent, the number of particles in solution increases and so the experimental values of the colligative properties are higher than the calculated values.
e.g. KCl in water ionises as
KCl → K+ + C
Molar mass either lower or higher than the expected or normal value is called as abnormal molar mass.

van’t Hoff factor (i):
It accounts for the extent of association or dissociation.
Plus Two Chemistry Notes Chapter 2 Solutions 27

Significance of van’t Hoff factor.
i > 1 ⇒ there is dissociation of solute particles.
i < 1 ⇒ there is association of solute particles.
i < 1 ⇒ there is no dissociation and association of solute particles.

Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows:
Relative lowering of vapour pressure of solvent,
Plus Two Chemistry Notes Chapter 2 Solutions 28

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Students can Download Chapter 5 Surface Chemistry Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Surface chemistry deals with phenomena that occurs at the surfaces or interfaces.

Adsorption:
accumulation of molecular species at the surface rather than in the bulk of a solid or liquid, it is a surface phenomenon, e.g. Moisture gets adsorbed on silica gel.

Adsorbate:
molecular species or substance, which accumulates at the surface.

Adsorbent:
material on the surface of which adsorption takes place, e.g. Charcoal, Silica gel, etc.

Desorption:
process of removing adsorbed substance from the surface of adsorbent.

Difference between Adsorption and Absorption:
Adsorption –
the substance is concentrated only at the surface and does not penetrate to the bulk of the adsorbent.

Absorption –
the substance is uniformly distributed throughout the bulk of the solid, e.g. Moisture gets absorbed on anhydrous CaCl2 while adsorbed on silical gel.

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Sorption:
term used when both adsorption and absorption take place simultaneously.

Mechanism of Adsorption :
The unbalanced or residual attractive forces are responsible for attracting the adsorbate particle on adsorbent surface. During adsorption energy decreases, therefore adsorption is exothermic process, i.e., ∆H of adsorption (heat of adsorption) is always negative. The entropy of the system also decreases (∆S = – ve).

Types of Adsorption:
1. Physical Adsorption (Physisorption):
Here the adsorbed molecules are held on the surface of the adsorbent by physical forces such as van der Waals’ forces. It is reversed by reducing pressure or by heating.

Characteristics:
Lack of specificity, easily liquifiable gases readily adsorbed, reversible in nature, extent of adsorption increases with increase in surface area of adsorbent, enthalpy of adsorption quite low (20 – 40 kJ mol’ ).

2. Chemical Adsorption (Chemisorption):
the forces of interaction between the adsorbent and adsorbate are chemical in nature. It cannot be easily reversed.

Characteristics:
High specificity, irreversibility, increases with increase in surface area, enthalpy of adsorption is high (80 -240 kJ mol”1).

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Sometimes physisorption and chemisorption occur simultaneously and it is not easy to ascertain the type of adsorption. A physisorption at low temperature may pass into chemisorption as the temperature is increased. For example, dihydrogen is first adsorbed on Ni by van der Waals’forces. Molecules of hydrogen then dissociate to form hydrogen atoms which are held on the surface by chemisorption.

Comparison of Physisorption and Chemisorption

PhysisorptionChemisorption
1) Arises because of van der Waals’ force1) Caused by chemical bond formation
2) Not specific2) Highly specific
3) Reversable3) Irreversible
4) More easily liquefiable gases are adsorbed readily.4) Gases which can react with the adsorbent show chemisorption.
5) Enthalpy of adsorption is low (20-40 kJ mol’1)5) Enthalpy of adsorption is high (80-240 kJ mol-1)
6) Low temperature is favourable. It decreases with increase of temperature6) Hig temperature is favourable. It increases with increase of temperature
7) No appreciable activation energy is needed.7) High activation energy is sometimes needed.
8) Increases with an increase of surface area.8) Increases with an increase of surface area.
9) Results into multimolecular layers on adsorbent surface under high pressure.9) Results into unimolecular layer

Adsorption Isotherms:
The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve termed as adsorption isotherm.

Freundlich Adsorption isotherm:
empirical relation between the quantity of gas adsorbed by unit mass of the solid adsorbant and pressure at a particular temperature.
x/m = k.P1/n (n > 1)
x → mass of the gas adsorbed
m → mass of adsorbent
‘k’ and ‘n’ are constants which depend on the nature of the adsorbent and the gas at a particular temperature.
OR log x/m = log k + \(\frac{1}{n}\) log P
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 1

Adsorption from Solution Phase:
Freundlich’s equation approximately describes the behaviour of adsorption from solution.
\(\frac{x}{m}\) = k.C1/n m
C – equilibrium concentration
log x/m = log k + \(\frac{1}{n}\) log C
Plotting log x/m vs log C a straight line is obtained

Applications of Adsorption:
Production of high vacuum, in Gas masks – activated charcoal is filled in gas mask to adsorb poisonous gases, for removal of colouring matter from solution in heterogeneous catalysis, in chromatographies analysis, in froth floatation process.

How to find Kp with the entropy and enthalpy amounts and with Gibb’s Free Energy.

Catalysis :
The process of altering the rate of chemical reaction by the addition of a foreign substance (catalyst) is called catalysis, e.g. MnO2 acts as a catalyst in the thermal decomposition of KClO3.

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Promoters:
substances that enhance the activity of a catalyst, e.g. In Haber’s process, iron is used as catalyst and molybdenum acts as a promoter.
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 2

Poisons:
substances which decrease the activity of a catalyst.

Homogeneous and Heterogeneous Catalysis
a) Homogeneous Catalysis:
When the reactants and catalyst are in the same phase, the process is said to be homogeneous catalysis.
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 3
Here both the reactants and the catalyst are in the liquid phase.

Heterogeneous Catalysis:
If the reactants and the catalyst are in different phase, the catalysis known as heterogeneous catalysis.
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 4
Here reactants are gaseous state while the catalysts are in the solid state.

Important Features of Solid Catalysts
a) Activity:
ability of catalysts to accelerate a chemical reaction.
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 5
But pure mixture of H2 and O2 does not react at all in the absence of a catalyst.

b) Selectivity:
ability of a catalyst to direct a reaction to yield a particular product.

e.g. CO and H2 combine to form different products by using different catalysts.
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 6

Shape Selective Catalysis by Zeolites:
The catalytic reaction that depends upon the pore structure of the catalyst and size of the reactant and the product molecules.

Zeolites are good shape-selective catalysts because of their honey comb-like structures. Zeolites are widely used in petrochemical industries for cracking and isomerisation of hydrocarbon.
e.g. ZSM – 5 – which convert alcohols into petrol.

Enzyme Catalysis:
Enzymes are biological catalysts. They catalyse biological reaction in animals and plants to maintain life. e.g.

  1. Invertase – Cane sugar into glucose and fructose
  2. Zymase – Glucose into alcohols
  3. Maltase – Maltose into glucose
  4. Diastase – Starch into maltose
  5. Cellulase – Cellulose into glucose
  6. Urease – Urea into NH3 and CO2

Characteristics:
Highly efficient, highly specific in nature, highly active under optimum temperature, highly active under optimum pH

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Mechanism (Lock and key model)
The molecules of the reactant (substrate), which have complementary shape, fit into the cavities on the surface of enzyme particles just like a key fits into a lock. The enzyme catalysed reactions proceeds in two steps:
Step -1 :
Binding of enzyme to sutbstrate to form an activated complex.
E + S → ES*
Step-2 :
Decomposition of the activated complex to form product.
ES* → E + P

Catalysts in Industry

  1. Finely divided iron with molybdenum as promoter in Haber’s process. (New catalyst: a mixture of iron oxide, potassium oxide and alumina)
  2. Platinised asbestos in Ostwald’s process
  3. Platinised asbestos or V205 in Contact process

Colloids:
Heterogeneous system in which one substance is dispersed (dispersed phase) as very fine particles in another substance called dispersion medium, e.g. Starch, Gelatin. In colloids the particle size (diameter) is between 1nm and 1000 nm.

Classification of Colloids:
i) Based on physical state of dispersed phase and dispersion medium:
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 7

ii) Based on Nature of Interaction between Dispersed Phase and Dispersion Medium:
1. Lyophilic (solvent attracting) Colloids:
there is strong interaction between the dispersed phase and dispersion medium. They are reversible sols. e.g. Starch, gelatin, albumin etc.

2. Lyophobic (solvent repelling) Colloids:
there is little or no interaction between the dispersed phase and dispersion medium. They are also irreversible colloids and are not stable.

iii) Based on Types of Particle of the Dispersed Phase
a) Multimolecular Colloids :
the individual particles consist of an aggregate of atoms or small molecules with molecular size less than 1 nm, the particles are held together by van der Waals’ forces, e.g. Sulphur sol, Gold sol etc.

b) Macromolecular Colloids :
the particles of dispersed phase are sufficiently big in size, maybe in the colloidal range, e.g. Starch, cellulose, proteins.

c) Associated Colloids (Micelles):
colloids which behave as normal strong electrolytes at low concentration but get associated at higher concentrations and behaves as colloidal solutions. The associated particle formed are called micelles.
e.g. Soap, detergents etc.

The formation of micelles take place only above a particular temperature called Kraft temperature (Tk.) and above a particular concentration called Critical Micelle Concentration(CMC).

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Mechanism of micelle formation –
In soaps, the RCOO ions are present on the surface with their COO groups in water and R staying away from it and remain at the surface. At CMC, the anions are pulled into the bulk of the solution and aggregate to form ‘ionic micelle’ having spherical shape with R pointing towards the centre of the sphere and COO part remaining outward on the surface of the sphere.

Preparation of Colloids
a) Chemical Methods
Some examples:
Plus Two Chemistry Notes Chapter 5 Surface Chemistry 8

b) Electrical Disintegration or Bredig’s Arc Method
Metallic sols can be prepared by striking an arc between two electrodes of the metal, immersed in the dispersion medium. The metal is vapourised by the arc which then condenses to form particles of colloidal size. e.g. Gold sol, Platinum sol, Silver sol etc.

c) Peptization:
process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte (peptizing agent), e.g. Freshly prepared Fe(OH)3 is peptized by adding small quantity of FeCI3 solution (peptizing agent).

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Mechanism of peptization –
During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid.

Purification of Colloids:
process of reducing the amount of impurities to a requisite minimum from the colloids.
i) Dialysis:
process of removing a dissolved substance from a colloid by means of diffusion through a suitable membrane.

ii) Electro-dialysis:
process of dialysis in presence of an applied electric field. It is faster and is applicable if the dissolved substance in the impure colloid is only an electrolyte. The ions present in the colloid migrate out to the oppositely charged electrodes.

iii) Ultrafilteration:
process of separating the colloidal particles from the solvent and soluble solutes present in the colloid by ultra filters. The ultra filter paper is prepared by soaking the filter paper in a colloidion solution (4% solution of nitro cellulose in a mixture of alcohol and ether). It is then hardened by formaldehyde and finally dried.

Properties of Colloids
1) Colligative Properties:
values of colligative properties as smaller due to smaller number of particles.

2) Tyndall Effect (Optical Property):
phenomenon of the scattering of light by colloidal particles.

Conditions for observing Tyndall effect:
1. The diameter of the dispersed particles is not much smaller than the wavelength of the light used; and

2. The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude. The ultramicroscope used to observe the light scattered by colloidal particles is based on Tyndall effect.

The colour of the sky can be explained by Tyndall effect. The dust and other colloids present in the atmosphere scatter the light. Only blue light reaches to our eyes.

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

3) Colour:
It depends on the wavelength of lighty scattered by the dispersed particles which in turn depends on the size and nature of the particles and changes with the manner in which the observer receives the light, e.g. a mixture of milk and water appears blue when viewed by the reflected light and red when viewed by transmitted light.

4) Brownian Movement:
The constant zig-zag movement of the colloidal particles. It is due to the unbalanced bombardment of the particles by the molecules of the dispersion medium. It does not permit the particles to settle and is responsible for the stability of sols. ,

5) Charge on Colloidal Particles:
Colloidal particles carry an electric charge.
Positive charged sols: Al2O3. xH2O, CrO3.xH20, basic dye stuffs, blood (Haemoglobin) etc.

Negatively charged sols:
Metal sols (Cu, Ag, Au), metallic sulphides, acid dyes stuffs, starch, gelatin.

Reason for charge:
It is due to
i) electron capture by sol particles during electrodispersion of metals,
ii) preferential adsorption of ions from solution and/ or
iii) formulation of electrical double layer.

Helmholtz Electrical Double Layer:
combination of two layers of opposite charges around the colloidal particle. The first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed as diffused layer.

Electrokinetic Potential or Zeta Potential:
It is the potential difference between the fixed layer and the diffused layer of opposite changes in the electrical ‘ double layer.

Significance of Charge on Colloidal Particles:
provides stability to the colloid because the repulsive forces between charged particles having same charge prevent them from coalescing or aggregating when they come closer to one another.

6) Electrophoresis:
lled anaphoresis and that of cathode is called cataphoresis.

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Coagulation/Flocculation/Precipitation:
process of settling of colloidal particles by the addition of electrolyte.

Coagulation of lyophobic sols can be carried out by the following ways:
Electrophoresis, mutual coagulation (mixing two oppositely charged sols), boiling, persistent dialysis, addition of electrolytes, etc.

Addition of electrolytes –
Colloids interact with ion carrying charge opposite to that present on themselves. This causes neutralisation leading to their coagulation.

Hardy – Schulze Rule:
the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.

The ion having opposite charge to sol particles (coagulating ion) cause coagulation.

In the coagulation of negative sol, the flocculating power is in the order: Al3+ > Ba2+ > Na+

In the coagulation of positive sol, the flocculating power in the order:
[Fe(CN)6]4- > PO43- > SO42-> Cl

Protective Colloids:
the lyophilic sol used for protection of lyophobic sol.

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Emulsions:
liquidin liquid colloidal systems i.e., the dispersion of finely divided droplets in another liquid. There are two types of emulsions.
1) Oil dispersed in water (O/W type):
water acts as dispersion medium, e.g. Milk, Vanishing cream.

2) Water dispersed in oil (W/O type):
oil, acts as dispersion medium.e.g. Butter, Creams, Cod liveroil Emulsification – process of making an emulsion. Emulsion may be obtained by vigourously agitating a mixture of both liquids.

Emulsifying agent or emulsifier –
substance used to stabilise an emulsion. It forms an interfacial film between suspended particles and the medium, e.g.

Emulsifying agents for O/W emulsions :
Proteins, gums, natural and synthetic soaps etc.

Emulsifying agents for W/O emulsions:
Heavy metal salts of fatty acids, long chain alcohols, lampblack etc.

Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Colloids Around Us :
Fog, mist and rain; food materials, blood, soils, formation of delta.

Application of Colloids
I) In Medicine:
Colloidal medicines are more effective because they have large surface area and are, therefore, easily assimilated, e.g. Colloidal silver (Argyrol) – as eye lotion, Colloidal antimony – in curing Kalaazar, Colloidal gold – for intramuscular injection. Milk of magnesia – in stomach disorder.

II) In industries :
Electrical precipitation of smoke – by Cottrell smoke precipitator, purification of water, tanning, cleansing action of soaps and detergents (micelle formation), photographic plates and films, rubber industry and Industrial products.