Prasanna

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 4 Chemical Kinetics.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 4 Chemical Kinetics

Question 1.
The order of a chemical reaction can be zero and even a fraction but molecularity cannot be zero or a non-integer. (March – 2010)
i) What do you mean by the order of a reaction?
ii) What is the molecularity of a reaction?
iii) The conversion of molecules ‘A’ to ‘B’ follows second order kinetics. If the concentration of A is increased to three times, how will it affect the rate of formation of ‘B’?
Answer:
i) It is sum of powers of the concentration – terms of the reactants in the rate law expression.
ii) The number of reacting species in an elementary reaction.
iii) Increases by 9 times.
Rate = k[R]²
Rate’= k[3R]² = 9 [R]²; i.e., Rate = 9 Rate

Question 2.
In a class room discussion about order and molecularity of a chemical reaction, Ramu argued that “there are reactions which appear to be of higher order but actually follow first order kinetics”. (Say – 2010)
a) How far is his statement true? Give your opinion in this regard. Justify your answer using suitable example.
b) List out any three important differences between order and molecularity.
Answer:
a) The reaction appears to be of higher order but actually follows a lower order kinetics. Such reactions are called pseudo order reactions. For example, hydrolysis of ethyl acetate. This reaction appears to be of second order but actually follows first order kinetics.

Here the concentration of water does not get altered much during the course of the reaction.
Hence, [H₂O] can be taken as a constant. The equation thus becomes
Rate = k[CH₃CQOH] where k = k [H₂O]
Thus, the reaction behaves as a first-order reaction.

(b)

Question 3.
The hydrolysis of an ester in acid medium is a first-order reaction. (March – 2011)
a) What do you mean by a first-order reaction?
b) What is the relation between Rate Constant and Half-Life Period of a reaction?
c) Half-Life Period of a first-order reaction is 20 seconds. How much time will it take to complete 90% of the reaction?
Answer:
a) When the sum of the powers of the concentra¬tion terms in the rate expression is one, that reaction is called a first-order reaction.
Rate = k[A]¹

b) In the case of the first-order reaction,

Question 4.
The value of rate constant K of a reaction depends on temperature. From the values of K at two different temperatures, the Arrhenius parameters E_a, and A can be calculated. (Say – 2011)
a) The rate constant of a reaction at 600K and 900K are 0.02 s^-1 and 0.06 s^-1 respectively. Find the values of E_a and A.
b) Write the unit of rate constant ‘K’ of a reaction if the concentration is in mol L^-1 and time in s. (order of the reaction is two)
Answer:

Question 5.
Rate of a reaction is the change in concentration of any one of the reactants or any one of the products in unit time. (March – 2012)
i) Express the rate of the following reaction in terms of reactants and products:
2H I → H₂ + l₂
ii) If rate expression for the above reaction is, rate = k[H I]², What is the order of the reaction?
iii) Define order of a reaction.
iv) Whether the molecularity and the order of the above reaction are the same? Give reason.
Answer:
i) In terms of reactants

iii) Order is the sum of the powers of the concentration terms in the rate law.
rate = K [A]^x [B]^y
∴ Order = x + y

iv) Yes.
2H I = H₂ + l₂
rate = K [H I]² ∴ Order = 2
Molecularity is the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction.

∴ Here molecularity = 2

Question 6.
For a first-order reaction, the half-life period (t_1/2) is independent of initial concentration of its reacting species. (Say – 2012)
i) What is meant by half-life period of a reaction?
ii) By deriving the equation for t_1/2 of first-order reaction, prove that it is independent of initial concentration of its reacting species.
[Hint: Fora first ortler reaction, \(\left.\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\right]\)
Answer:
i) The half-life period of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration,
ii) For a first order reaction,

Thus, for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

Question 7.
a) Zero-order reaction means that the rate of a reaction is independent of the concentration of reactants. (March – 2013)
i) Write an example for a zero-order reaction.
ii) Write the integral rate expression for the zero-order reaction, R→ P.

b) The temperature dependence of the rate of a chemical reaction can be accurately explained by the Arrhenius equation. With the help of the Arrhenius equation calculate the rate constant for the first-order reaction C₂H₅l(g) → C₂H₄(g) + Hl(g) at 700K. Energy of activation (E_a) for the reaction is 209 kJ moh1 and rate constant at 600 K is 1.60 x 10^-5 S^-1 [Universal gas constant R = 8.314 JK^-1 mol^-1]
Answer:
a) i) Zero-order reaction

where ‘I is the constant of integration.
At t = 0 1 = [R]₀
[R]₀ → InitiaI concentration of the reactant
[R] → concentration at time
∴ Equation (1) becomes. [R] = ^–kt + [R]₀

Question 8.
The conversion of molecule A to B follows second order kinetics. (Say – 2013)
a) If the concentration of A is increased to 4 times, how will it affect the formation of B?
b) Indicate the order and molecularityofthe reaction given below.
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)
Answer:
a) Rate r,= k[A]²
If the concentration of A is increased by four times, the new rate, r2 = k[4A]²
or r₂ = 16 k [A]²
or r₂ = 16 r₁
i.e., rate is increased by 16 times.

b) This is a pseudo first order reaction.
Order = 1, Molecularity = 2

Question 9.
a) Consider a general reaction aA + bB → cC + dD. The rate expression for the reaction is Rate = K[A]*[B]^y (March – 2014)
i) Establish the significance of ‘(a+b)’ and ‘(x+y)’ in terms of order and molecularity.
ii) Write any two differences between order and molecularity.
b) “Reactions with zero order is possible, but zero molecularity is not”. Justify the statement.
Answer:
a) i) (a + b) – Molecularity of the reaction
(X + y) – Order of the reaction

ii)

b) The order of a reaction can be zero which means that the rate of the reaction is independent of the concentration of the reactants. But, molecularity of a reaction cannot be zero which means that there is no reacting species and hence no reaction is possible.

Question 10.
a) Unit of rate constant (K) of a reaction depends on the order of the reactions. (Say – 2014)
Values of ‘K’ of two reactions are given below. Find the order of each reaction.
i) K = 3 x 10^-2 molL^-1 s^-1
ii) K = 5 x 10^-3 mol^-1 Ls^-1
b) i) Write integrated rate equation for a first order reaction.
ii) Write the relation between half life (t^1/2) and rate constant (K) of a first order reaction.
iii) Rate constant (K) of a reaction is 5 x 10.2 s^-1.
Find the half life (t_1/2) of the reaction.
Answer:
a) i) Zero-order reaction by analysing the unit.
ii) Second-order reaction

Question 11.
The terms order and molecularity are common in chemical kinetics. (March – 2015)
a) What do you mean by order and molecularity?
b) i) Write two factors influencing rate of a reaction.
ii) WnteArrhenius equation.
Answer:
a) Order of a chemical reaction is the sum of powers of the concentration of the reactants in the rate law expression.
Molecularity of a reaction is the number of reating species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

b) i) Temperature, Nature of the reactant, Concentration of the reactant (Pressure in the case of gases). Presence of catalyst, Presence of radiation/light, Surface area etc. (any one)

Question 12.
Integrated rate expression for rate constant of first-order reaction is given by \(\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\), for a general reaction R → R (Say – 2015)
i) Derive an expression for half life period of first order reaction.
ii) A first order reaction has a rate constant 1.15 x 10^-3s^-1. How long will 5g of the reactant take to reduce 3 g?
Answer:

Question 13.
i) The molecularity of the reaction 2NO + O₂ → 2NO₂ is, (March – 2016)
a) 5
c) 2
c) 3
d) O

ii) a) What do you mean by rate oía reaction?
b) What will be the effect of temperature on rate of a reaction?
iii) A first order reaction is found to have a rate constant, k = 5.5 x 10^-14 s^-1. Find out the half-life of the reaction.
Answer:
c) 3

ii) a) The rate of a reaction is defined as the change in concentration of any one of the reactants or products ¡n unit time. i.e.,

b) The rate of most of the chemical reactions (endothermic reactions) increase with increase in temperature. For a chemical reaçtion with rise ¡n temperature by 100, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation, \(k=A e^{\frac{-E_{0}}{R T}}\)

iii) For a first order reaction, half-life period,

Question 14.
Rate of a reaction is the change in concentration of any one of the reactants or any one of the products in unit time (Say – 2016)
a) Express the rate of the following reaction in terms of reactants and products \(2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{NO}_{2(g)}\).
b) i) \(\mathrm{N}_{2} \mathrm{O}_{5(g)} \rightarrow 2 \mathrm{NO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)}\) is a first order reaction. Find the unit of K.
ii) Calculate the time required for the completion of 90% of a first order reaction. (K = 0.2303s^-1)
Answer:

Question 15.
a) Plot a graph showing variation in the concentration of reactants against time for a zero-order reaction. (March – 2017)
b) What do you mean by zero-order reaction?
C) The initial concentration of the first-order reaction, \(\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(g)}\) was 1 24 x 10^-2 mol L^-1 at 300 K. The concentration of N₂O₅ after ‘1’ hour was 0.20 x 10^-2 mol L^-1. Calculate the rate constant of the reaction at 300 K.
Answer:

b) It is a reaction for which the rate of the reaction is proportional to zero power of the concentration of reactants, i.e, order is zero.
Or, this is a reaction for which the rate of the reaction is independent of the concentration of the reactants.

Or, this is a reaction for which rate of the reaction,

Question 16.
The effect of temperature on rate of reaction is given by Arrhenius equation. (Say – 2017)
i) Write Arrtenius equation.
ii) Define activation energy (Ea).
iii) Rate constant K₂ of a reaction at 310 K is two times of its rate constant K., at 300 K. Calculate activation energy of the reaction. (1og2 0.3010, log 1=0)
Answer:
\(\begin{array}{l}
\text { I) } \mathrm{K}=\mathrm{A} \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \text { or } \\
\log \mathrm{K}=\log \mathrm{A} \frac{-\mathrm{Ea}}{2.303 \mathrm{RT}}
\end{array}\)

ii) Activation energy is the energy required to form an activated complex or It is the energy difference between the activated complex and the reactant molecules.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 3 Electrochemistry.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 3 Electrochemistry

Question 1.
From the position of elements in the electrochemical series, copper (Cu) can displace silver (Ag) from silver nitrate solution. (March – 2010)
a) Represent the cell constructed with silver and copper electrodes.
b) Write down the reaction taking place at the anode.
c) Write down the reaction taking place at the cathode.
d) Write the Nernst equation for the above cell reaction.
Answer:

Question 2.
In a class room, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday’s laws of electrolysis. (May – 2013)

a) State the Faraday’s laws of electrolysis.
b) Explain the term electrochemical equivalent.
c) Calculate the quantity of electricity required to deposit 0.09 g of Aluminium during the following electrode reaction:
Al³⁺ + 3e → Al (Atomic mass of Al = 27)
Answer:
a) First law : The amount of a substance which is deposited or liberated at any electrode during ectrolysis is directly proportional to the quantity of electricity flowing through the electrolyte.

Second law: If the same quantity of electricity is passed through different electrolytes the amount of substances formed is directly proportional to their chemical equivalent weights.

b) It is the quantity of a substance formed when one-ampere current is passed through an electrolyte for one second.

c) Quantity of electricity required to deposit 27 g of
Al = 3F = 3 x 96500 C = 289500 C
∴ the quantity of electricity required to deposit 0.09
\(g \text { of } A l=\frac{289500 \times .09}{27}=965 C\)

Question 3.
The limiting molar conductivity of an electrolyte is to Aained by adding the limiting molar conductivities of cation and anion of the electrolyte. (March – 2011)
a) Name the above law.
b) What is meant by limiting molar conductivity?
c) Explain how conductivity measurements help to determine the ionization constant of a weak electrolyte like Acetic Acid.
d)Explain the change of conductivity and molar conductivity of a solution with dilution.
Answer:
a) Kohlrauschs law.
b) It is the conductivity of an electrolyte when the concentration of the solution approaches zero (or at infinite dilution).
c) The limiting molar conductivity of acetic acid \(\left(\Lambda_{\mathrm{Gh}_{3} \mathrm{COOH}}\right)\) isdeterrnined by app’ying Kohlrauschts law \(\left(\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{0}=\Lambda_{\mathrm{CH}_{3} \mathrm{COONa}}^{0}+\Lambda_{\mathrm{HCl}}^{0}-\Lambda_{\mathrm{NaCl}}^{0}\right)\). Then, degree of dissociaijon,OE is determined using the re lation, \(a=\frac{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{C}}}{\Lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\mathrm{O}}}\)

From α, the diissociation constant can be deter mined using the relation, \(K_{a}=\frac{c a^{2}}{(1-\alpha)}\)

Substituting for CL we get,

d. Conductivity decreases with dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution. The vanation of molar conductivity with dilution is different for strong and weak electrolytes. For strong electrolytes molar conductivity in creases steadily with increase in dilution due to decrease in interionic attraction. Thus, a straight line is obtained when ∧_m is plotted against C^1/2. For weak electrolytes molar conductivity increases with dilution steadily intially and shows a steep increase, especially at lower concentration due to increase in degree of dissociation.

Thus, a plot of ∧_m against C^1/2 gives a curve,

Question 4.
The standard electrode potentials of some electrodes are given below: (May – 2011)
E°_(zn2+,zn) = – 0.76V
E°_{(cu2+, Cn)} = + 0.34V
E°_{(Ag+, Ag)} = + 080V
E°_{(H+, H2)} = 0V

a) Can CuSO₄ solution be kept in silver vessel?
b) Zinc or Copper which can displace hydrogen from dii. H₂SO₄?
c) What is the reaction taking place at SHE when its connected to Ag/Ag electrode to form a galvanic cell?
d) Find the value of K_C (equillibnum constant) in the Daniel cell at 298k.
Answer:
a) Yes.
Since the standard reduction potential of silver is more than that of copper it is less reactive than copper and hence cannot react with CuSO₄.

b) Zinc with negative standard electrode potential is more active than hydrogen and hence can displace hydrogen from dil.H₂SO₄. But copper with a positive standard electrode potential is less active than hydrogen and hence cannot displace hydrogen from dil. H₂SO₄.

c) Since silver is less active than hydrogen, when silver electrode is connected with S.H.E, silver will act as cathode and S.H.E will act as anode. At the anode H is oxidiseci to H⁺. Hence, the reaction taking place at S.H.E is

Question 5.
Daniell cell is a galvanic cell made of zinc and copper electrodes. (March – 2012)
i) Write anode and cathode reactions in Daniell cell.
ii) Nernst equation for the electrode reaction
Mⁿ⁺ + + ne- 2 M is
\(\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}=\mathrm{E}_{\left(\mathrm{M}^{n+} / \mathrm{M}\right)}^{0}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \frac{1}{\left[\mathrm{M}^{\mathrm{n}^{+}}\right]}\)
Derive Nernst equation for Daniel cell.
OR
Leclariche cell, Lead storage cell and Fuel cell are galvanic cells having different uses.

i) Among these, the Leclanche cell is a primary cell and Lead storage cell is a secondary cell. Write any two differences between primary cells and secondary cells.
ii) What is a Fuel cell?
iii) Write the overall cell reaction in H₂ – O₂ Fuel cell.
Answer:
i) Anode : Zn → Zⁿ²⁺ + 2e- (oxidation half reaction)
Cathode : Cu²⁺ + 2e^– → Cu (reduction half reaction)
Overall cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu Theceilcanberepresentedas: Znl Zn²⁺llCu²⁺lCu
Anode : Zn I ZnSO₄
Cathode : Cu I CuSO₄

which is the Nernst equation for Daniell cell.
OR

ii) Fuel cells are galvanic cells in which chemical energy from fuels like H₂, CO, CH₄ etc are converted to electrical energy.

Question 6.
Innumerable number of galvanic cells can be constructed on the pattern of Daniell cell by taking combination of different half cells. (May – 2012)
i) What is galvanic cell?
ii) Name the anode and cathode of the Daniell cell.
iii) Write the name of the half-cell represented by Pt_(S)/H_2(g)/H⁺_(aq).
iv) What is the potential of the above half-cell at all temperate res?
Answer:
i) It is a device for converting chemical energy released into electrical energy.
ii) Anode – Zn rod dipped in ZnSO₄
Cathode – Cu rod dipped on CauSO₄
iii) Standard Hydrogen Electrode (S.H.E)
iv) S.H.E is assigned a zero potential at all temperatures.

Question 7.
With decrease in concentration of an electrolytic solution, conductivity (K)decreases and molar conductivity (∧_m) increases. (March – 2013)
i) Write the equation showing the relationship between conductivity and molar conductivity.
ii) How will you account for the increase in molar conductivity with decrease in concentration?
iii) Limiting molar conductivity (L°_m) of a strong electrolyte can be determined by graphical extrapolation method. Suggest a method for the determination of limiting molar conducivity of a weak electrolyte, taking acetic acid (CH₃COOH) as example.
Answer:
i) Molar conductivity \(\left(\Lambda_{m}\right)=\frac{\kappa \times 1000}{M}\)
where M → Molanty and K → Conductivity
OR
\(\lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{C}}\) Concentration of where solution in mol/litre.

ii) Molar conductivity increase with decrease in con centration or increase in dilution as number of ions as well as mobility of ions increases with dilution.

a) For strong electrolytes, the number of ions do not increase appreciably on dilution and only mobility of ions increases due to decrease in interionic attraction.

∴ increases a little as shown in the above graph.

b) For weak electrolytes, the number of ions, as well as mobility of ions, increases on dilution. Hence, increases steeply on dilution, especially near lower concentrations as shown in graph.

iii) Using Kohlrauschs law

Thus the molar conductivity of CH3COQH at infinite dilution can be determined from the knowledge Of \(\lambda_{\mathrm{m}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)}^{0}, \lambda_{\mathrm{m}(\mathrm{HCl})}^{0}, \lambda_{\mathrm{m}(\mathrm{NaCl})}^{0}\)

Question 8.
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combi nation of different half cells. (May – 2013)
a) What is a galvanic cell?
b) Name the cathode and anode used in the Daniell cell.
c) Name the cell represented by Pt_(S), H_2(g)/H⁺_(aq).
d) According to convention what is the potential of the above cell at all temperatures?
e) Write the use of the above cell.
Answer:
a) It is a device for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as elec trical energy. e.g., Daniell cell,
b) A zinc rod dipped in 1 M solution of ZnSO₄ acts as the anode. Here oxidation takes place. A copper rod dipped in 1 M solution of CuSO₄ acts as the cathode. Here reduction takes place.
c) This represents the andard Hydrogen Electrode (S.H.E), when it acts as the anode.
d) According to convention, S.H.E is assigned a zero potential at all temperatures
e) It is used as a primary reference electrode for determining the standard electrode potential of an unknown electrode. The electrode whose standard potential is to be determined is coupled with a reference electrode of known potential i.e., S.H.E to get a galvanic cell. The potential of the resulting galvanic cell is determined experimentally. E = E – E Knowing the potential of one electrode that of the other can be calculated.

Question 9.
a) The cell reactìon in Daniell cell is Zn_(s) + CU²⁺_(aq) → Zn²⁺_(aq) + CU_(s) and Nernst equation for single electrode potential for general electrode reaction \(\mathrm{M}^{\mathrm{n}+}{ }_{(\mathrm{aq})}+\mathrm{ne}^{-} \longrightarrow \mathrm{M}_{(\mathrm{s})}\) is \(E_{M^{n+} M}=E_{M^{n+} / M}^{0}-\frac{2.303 R T}{n F} \log \frac{[M]}{\left[M^{n+}\right]}\) Derive Nernst equation for Daniell cell. (March – 2014)
b) Daniell cell is a primary cell while lead storage cell is a secondary cell. Write any one difference between primary cells and secondary cells.
Answer:
a) In Daniell cell, the electrode potential for any given concentration of Cu²⁺ and Zn²⁺ ions, we can write, For Cathode:

b) In primary cells the reaction occurs only once and after use over a period of time cell becomes dead and cannot be reused again. Here the cell reaction is irreversible. e.g., dry cell Secondary cells after use can be recharged by passing current through them in the opposite direction so that they can be used again. Here the cell reaction is reversible. e.g., Lead storage cell.

Question 10.
Fuel cells are special types of galvanic cells. (May – 2014)
a) i) Whataregalvaniccells?
ii) Write any two advantages of fuel cells.
b) Write the electrode reactions is H₂ – O₂ fuel cells.
a) i) These are devices for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as electrical energy. e.g., Daniell oeil.
ii) 1) Fuel cells are pollution free.
2) Fuel cells are highly efficient (about 70%) compared to thermal plants (about 40%)
3) Fuel cells run continuously as long as the reactants are supplied. (any two)

b) At cathode:
O₂(g) + 2H₂O(l) + 4e → 4OH(aq)
At anode:
2H₂(g) + 4OH(aq) → 4H₂O(l)

Question 11.
You are supplied with the following substances: Copper rod, Zinc rod, Salt bridge, two glass bea kers, a piece of wire, 1 M CuSO₄ solution, 1 M ZnSO₄ solution. (March – 2015)
a) Represent the cell made using the above materials.
b) i) Write the Nemst equation for the above cell.
ii) Calculate the standard EMF of the cell if
\(\begin{array}{l}
E_{\left(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\right)}=-0.76 \mathrm{~V} \\
\mathrm{E}_{\left(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\right)}=+0.34 \mathrm{~V}
\end{array}\)
Answer:
a) Zn(s)|Zn²⁺(1 M)||Cu²⁺(1 M)|Cu(s)
OR

Question 12.
a) Conductance (G), conductivity (K) and molar conductivity (∧_m) are terms used in electrolytic conduction. (May – 2015)
i) Write any two factors on which conductivity depends on.
ii) How do conductivity and molar conductivity vary with concentration of electrolytic solution?
b) Write any one difference between primary cell and secondary cell.
Answer:
a) i) 1. the nature of the electrolyte added
2. size of the ions produced and their solvation
3. the nature of the solvent and its viscosity
4. concentration of the electrolyte
5. temperature (any two factors)

ii) Conductivity always decreases with decrease in concentration both for weak and strong electrolytes. This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity, ∧_m = _kV

∧_m increases with decrease in concentration. This is because the total volume (V) of the solution containing one mole of electrolyte also increases. The decrease in K on dilution is more than compensated by increase in its volume.

In the case of strong electrolytes Am increases slowly with dilution and can be represented by the equation:
\(\Lambda_{m}=\Lambda_{m}^{0}-A c^{1 / 2}\)

where ‘c’ is the molar concentration, ‘A’ is a constant (equato to -ve of slope) and ∧_m° is the limiting molar conductivity. Here, the plot of ∧_m against ‘C^1/2‘ will be a straight line. In the case of weak electrolytes ∧_m increases steeply on dilution, especially near lower concentrations due to increase in degree of dissociation.

b) Pnmary cell – Cell in which the reaction occurs only once and after use over a period of time the cell becomes dead and cannot be reused again. Secondary cell – Cell which can be recharged after use by passing current through it in the opposite direction so that it can be used again.

Question 13.
a) Which of the following is a secondary cell? (March – 2016)
a) Dry cell
b) Leclanche cell
C) Mercury cell
d) None of these

b) What is the relationship between resistance and conductance?
c) One of the fuel cells uses the reaction of hydrogen and oxygen to form water. Write down the cell re action taking place in the anode and cathode of that fuel cell.
Answer:
d) None of these
b) Resistance is inversely proportional to conductance. Or, Conductance is the inverse of resistance.

Question 14.
Galvanic cells are classified into primary and secondary cells (May – 2016)
a) Write any two differences between primary cell and secondary cell.
b) i) What is a fuel cell?
ii) Write the overall cell reaction in H₂ – O₂ fuel cell.
Answer:
a) Primary cell
Cell reaction cannot be reversed and hence can not be recharged, cannot be reused again. e.g. Dry cell, Mercury cell

Secondary cell
Cell reaction can be reversed and hence can be recharged, can be resued again. e.g. Lead storage battery, nickel-cadmium cell

b) i) Fuel cell is a galvanic cell that is designed to convert the energy of combustion of fuels directly into electncal energy.
ii) 2H₂(g) + O₂(g) → 2H₂O(l)

Question 15.
a) Represent the galvanic cell based on the cell reaction given below (March – 2017)
\(\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})} \rightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})}\)
b) Write the half cell reaction of the above cell.
c) ∧_m⁰ for NaCI. HCI and NaAc are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively. Calculate ∧_m⁰ for HAc.
Answer:

Question 16.
a) Identify the weak electrolyte from the following: (May – 2017)
i) KCl
ii) NaCl
iii) KBr
iv) CH₃COOH

b) Kohlrausch’s law helps to determine the degree of dissociation of a weak electrolyte at a given concentration.
i) State Kohlrausch’s law.
ii) The molar conductivity ∧_m of .001 M acetic acid is 4.95 x 10^-5 S cm² mol^-1. Calculate the degree of dissociation (α) at this concentra tion if limiting molar conductivity \(\wedge_{m}^{0}\) for H+ is 340 x 10^-5 S cm² mol^-1 and for CH₃COO is 50.5 x 10^-5 S cm² mol^-1.
Answer:
a) iv) CN₃COOH
b) i) Kohlrausch’s law: Limiting molar conductivity of an electrolyte is the sum of individual contnbutions of anion and cation respectively.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 2 Solutions.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 2 Solutions

Question 1.
Colligative properties are properties of solutions which depend on the number of solute particles irrespective of their nature. (March – 2010)
a) Name the four important colligative properties.
b) What happens to the colligative properties when ethanoic acid is treated with benzene? Give reason.
Answer:
a) 1) Relating lowering of vapour pressure of the solvent
2) Depression of freezing point of the solvent
3) Elevation of Boiling point of the solvent
4) Osmotic pressure of the solution

b) Molecules of ethanoic add (acetic acid) dimenses in benzene due to hydrogen bonding. As a result of dimerisation, the actual number of solute particles in the solution is decreased. As colligative property decreases molecular mass increases.

Question 2.
a) Mr. Raju has determined the molecular masses of different solutes in different solvents by osmotic pressure measurements and presented them in the following table. Please help him to complete the table. (May – 2010)

b) The extent of deviation from ideal behaviour of a solution is explained by van’t Hoff factor, i. What is meant by van’t Hoff factor?
Answer:
a) 2B, C, 2D, E/3, F, G/5.
b) 2B and 2 D are due to association
\(\mathrm{i}=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}\)
OR
\(\mathrm{i}=\frac{\text { Observed colligative property }}{\text { Calculated colligative property }}\)

Question 3.
Colligative properties can be used to determine the molecular mass of solutes in solutions. (March – 2011)
a) What do you mean by ‘Colligative Property’?
b) Fordeterminingthe molecular mass of polymers, osmotic pressure is preferred to other properties. Why?
c) For intravenous injections only solutions with freezing point depression equal to that of 0.9% NaCI solution is used. Why?
Answer:
a) Colligative properties are those properties which depend upon no. of particles in the solution.

b) i) Pressure measurement can be done around the room temperature.
ii) Molarity of the solution is used instead of molality.
iii) Its magnitude is large compared to other colligative properties.
iv) Polymers have poor solubility.

c) This is because the osmotic pressure associated with the fulid inside the blood cell is equivalent to that of 0.9 (m/w) sodium chloride solution, i.e., blood is isotonic with this solution.

Question 4.
Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure are important colligative properties of dilute solutions. (May – 2011)
a) Relative lowering of vapour pressure of an aqueous dilute solution of glucose is 0.018. What is the mole fraction of glucose in the solution?
b) An aqueous dilute solution of a non-volatile solute boils at 373.052 K. Find the freezing point of the solution.
For water K_b = 0.52K Kg mol^-1
For water K_f = 1.86 K Kgmol^-1
Normal boiling point of water = 373K
Normal freezing point of water = 273K
Answer:
a) 0.018
Because according to Raoult’s law for solutions containing non-volatile solutes, the relaive lower-ing of vapour pressure is equal to the mole fraction of the non-volatile solute.

Question 5.
Vapour pressure of a solution is different from that of pure solvent. (March – 2012)
i) Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
ii) State the above law.
iii) Vapour pressure of chloroform (CHCI₃) and dichloro methane (CH₂CI₂) at 298 K are 200 mm and 415 mm of Hg respectively. Calculate the vapour pressure of solution prepared by mixing 24 g of chloroform and 17 g of dichloro methane at 298 K. [At. Mass : H -1, C – 12, Cl – 35.5]
Answer:
i) Raoult’s law.

ii) It states that at a given temperature for a solution of volatile liquids, the partial v.p. of each component in the solution is directly proportional to its mole fraction.

Question 6.
Colligative properties are properties of solutions which depend on the number of solute particles in the solution. (May – 2012)
i) Write the names of four important colligative properties.
ii) The value of van’t Hoff factor, ‘i’ for aqueous KCI solution is close to 2, while the value of ‘i’ for ethanoic acid in benzene is nearly 0.5. Give reason.
i) The Colligative Properties are:
1) Relative lowering of vapour pressure
2) Elevation of boiling point
3) Depression of freesing point
4) Osmotic pressure

ii) This is caused by dissociation in the case of KCI and association in the case of acetic acid.

KCI in aqueous solution undergoes dissociation as KCI → K⁺ + Cl^–

Thus, if complete ionisation occurs the number of particles in solution becomes double and hence van’t Hoff factor (i) for aqueous KCI solution is close to 2.

In the case of ethanoic acid (acetic acid) association (dimerisation) occurs in benzene through in- termolecular hydrogen bonding. Thus, if complete association occurs the number of particles in solution becomes half and hence van’t Hoff factor (i) for ethanoic acid in benzene is nearly 0.5.

Question 7.
Elevation of boiling point is a colligative property, (March – 2013)
i) What are colligative properties?
ii) Elevation of boiling point (D T_b) is directly proportional to molality (m) of solution.
Thus, DT_b = K_bm, K_b is called the molal elevation constant.
From the above relation derive an expression to obtain molar mass of the solute,
iii) The boiling point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of the solute. K_b for benzene is 2.53K kg mol^-1.
Answer:
i) Colligative properties are those properties which depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.

Question 8.
Liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law. (May – 2013)
a) State Raoult’s law.
b) What are ideal solutions?
c) Write two important properties of ideal solutions.
d) What type of deviation is shown by a mixture of chloroform and acetone? Give reason.
Answer:
a) Raoult’s law states that for a solution of volatile liquids, the partial vapour of each component in the solution is directly proportional to its mole fraction. Or The partial vapour pressure of any volatile component of a solution is equal to the product of the vapour pressure of pure component and mole fraction of that component in the solution.

b) Ideal solutions are solutions which obey Raoult’s law over the entire range of concentration.

c) i) Enthalpy of mixing of the pure components to form the solution is zero, i.e.; Δ_mixH = 0
ii) Volume of mixing is zero, i.e.; Δ_mixV = 0

d) Negative deviation. This is because chloroform molecule is able to form hydrogen bond with acetone molecule.

Question 9.
Osmotic pressure is a colligative property and it is proportional to the molarity of the solution. (March – 2014)
a) What is osmotic pressure?
b) Molecular mass of NaCI determined by osmotic pressure measurement is found to be half of the actual value. Account for it.
c) Calculate the osmotic pressure exerted by a solution prepared by dissolving 1.5 g of a polymer of molar mass 185000 in 500 mL of water at 37°C. [R = 0.0821 L atm K^-1 mol^-1].
Answer:
a) Osmotic pressure is the extra pressure applied on the solution to stop osmosis
b) NaCI, as a strong electrolyte dissociates to form Na+ and Ch ions. The number of particles doubles colligative property also doubles. Observed molecular mass will be half.

Question 10.
Molarity (M), molality (m) and mole fraction (x) are some methods for expressing concentrations of solutions. (May – 2014)
a) Which of these are temperature independent?
b) i) Define ‘molefraction’.
ii) A mixture contains 3.2 g methanol (molecular mass = 32 u) and 4.6 g ethanol, (molecular mass = 46 u) Find the molefraction of each ’ component)
Answer:
a) Molality and Mole fraction are temperature independent because these are mass to mass relationships. Mass is independent of temperature,
b) i) Mole fraction is defined as the ratio of the number of moles of that component to the total number of moles of all the components in solution.

Question 11.
a) Among the following which is not a colligative property? (March – 2015)
i) Osmotic pressure
ii) Elevation of boiling point
iii) Vapour pressure
iv) Depression of freezing point

b) i) 200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of solution at 300 K is found to be 8.3 x 10^-2 bar. Calculate the molar mass of protein. (R = 0.083 L bar K^-1 mol^-1)
ii) What is the significance of van’t Hoff factor?
Answer:
a) iii) Vapour pressure
b) i) Osmotic pressure, n = 8.3 x 10^-2 bar
Volume of the solution, V = 200 cm³ = 0.200 L
Temperature, T = 300 K
R = 0.083 L bar mol^-1 K^-1

Thus, in the case of association, the value of ‘i’ is less than unity while for dissociation ‘i’ is greater than unity. If i =1 it means that there is no association or dissociation of the solute particles in solution.

Question 12.
a) Draw a vapour pressure curve, by plotting vapour pressure against mole fraction of an ideal solution of two volatile components A and B (not to scale). Indicate partial vapour pressure of A and B (P_A and P_B) and total vapour pressure (P_total) (May – 2015)
b) What is an ideal solution?
c) Modify the above plot for non-ideal solution showing positive deviation. (Draw the above plot once again and modify).
Answer:

b) A solution which obeys Raoult’s law over the entire range of concentration is known as an ideal solution.

Question 13.
a) Number of moles of the solute per kilogram of the solvent is (March – 2016)
a) Mole fraction
b) Molality
c) Molarity
d) Molar mass

b) The extent to which a solute is dissociated or associated can be expressed by Van’t Hoff factor.’ Substantiate the statement.
c) The vapour pressure of pure benzene at a certain temperature is0.850 bar. Anon volatile, non-electrolyte solid weighing 0.5 g when added to 39 g of benzene (molar mass 78 g mol^-1), the vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance?
Answer:
a) b) Molality
b) b) The van’t Holf factor,

When i = 1 ⇒ there is no association or dissociation of solute particles.
When i < 1 ⇒ there is association of solute particles. When i > 1 ⇒ there is dissociation of solute particles.

Question 15.
Osmotic pressure is a colligative property. (May – 2016)
a) What is osmotic pressure?
b) 1.00 g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40 k. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of solute.
Answer:
a) Osmotic pressure is the extra pressure applied on the solution side to just stop osmosis i.e., the flow of the solvent from its side to solution side through the semipermeable membrane,

Question 16.
a) Henry’s law is related to solubility of a gas in liquid. (March – 2017)
i) State Henry’s law.
ii) Write any two applications of Henry’s law.
b) 1000cm³ of an aqueous solution of a protein contains 1.26 gm of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 x 10^-3 bar. Calculate molar mass of the protein. (R = 0.083 L bar mol^-1 K^-1)
Answer:
a) i) Henry’s law states that at constant tempera-ture, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. p = K_H x where, p is the partial pressure of the gas in vapour phase, K_H is the Henry’s law constant and x is the mole fraction of the gas in the solution.

ii) 1. To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

2. To avoid bends and the toxic effects of high concenration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium.

3. Low partial pressure of oxygen at high altitudes leads to low concentration of oxygen in the blood and tissues of people living at high altitudes or climbers and causes anoxia. (Any two applications)

Question 17.
a) The mole fraction of water in a mixture containing equal number of moles of water and ethanol is (May – 2017)
i) 1
ii) 0.5
iii) 2
iv) 0.25

b) The following are the vapour pressure curves of a pure solvent and a solution of a non-volatile solute in it.

Based on the above curves answer the following questions:
i) What do the curves A and B indicate?
ii) Explain why the value of TB is greater than that of Tb⁰.
Answer:
a) ii) 0.5
b) i) A-Vapour pressure curve of solvent Vapour pressure curve of solution
ii) Due to the presence of a non-volatile solute vapour pressure of solution is less than solvent and the boiling point is increased.

Kerala State Board New Syllabus Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 1 The Solid State.

Kerala Plus Two Chemistry Chapter Wise Previous Questions Chapter 1 The Solid State

Plus Two Chemistry The Solid State 4 Marks Important Questions

Question 1
a) Schottky defects and Frenkel defects are two stoichiometric defects shown by crystals. (March – 2010)
i) Classify the following crystals into those showing Schottky defects and Frenkel defects:
NaCI, AgCI, CsCI, CdCI₂
ii) Name a crystal showing both Schottky defect and Frenkel defect.
b) Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.
i) ↑↓↑↓↑↓↑↓
ii) ↑↑↓↑↓↑↑↑
iii) ↑↑↑↑↑↑↑↑
Answer:
a) i) Schottchydefect – NaCI, CsCI Frenkel defect – AgCI, CdCI₂
ii) AgBr

b) i) Antifero magnetism
ii) Ferrimagnetism
iii) Ferromagnetism

Question 2.
Based on the nature of order present in the arrangement of the constituent particles, solids are classified into two, crystalline and amorphous. (May – 2010)
a) List out any four points of difference between crystalline and amorphous solids.
b) A list of solids are given below:
Quartz, glass, iodine, ice.
From this, identify crystal (s)
i) having sharp melting point.
ii) which is/are isotropic
Answer:

b) i) Quartz, Iodine, Ice
ii) Glass

Question 3.
Cristal defects give rise to certain special properties in the solids. (March – 2011)
a) What is meant by Frenkel Defect?
b) Why does LiCI not exhibit Frenkel Defect?
c) Explain the pink colour of LiCI when heated in . the vapours of Li.
Answer:
a) The dislocation of a cation from its original site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.
b) The size of the cation is bigger than the void.
c) Due to F – centre. It is an electron trapped anion vacancy.

Question 4.
A cubic unit cell is characterized by a = b = c and α = β = γ = 90° (May – 2011)
a) Name three important types of cubic unit cells and calculate the number of atoms in one unit cell in the above three cases.
b) A metal forms cubic crystals. The mass of one unit cell of it is M/NA gram, where M is the atomic mass of the metal and NA is Avogardo Number. What is the type of cubic unit cell possessed by the metal?
Answer:
a) Simple cubic unit cell or Primitive unit cell, Body – centred cubic unit cell (bcc) and Face – centred cubic unit cell (fee).
b) Primitive cubic unit cell: This unit cell has atoms only at its comers. There are 8 corners for a cube.
Contribution by atom at the corner = 1/8
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

Body – centred cubic unit cell:
This unit cell has an atom at each of its corners and also one atom at its body centre.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

Face – centred cubic unit cell:
This unit cell contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners \(\times \frac{1}{8}\) per corner atom \(=8 \times \frac{1}{8}=1\) atom Contribution by an atom at the face centre = \(\frac{1}{2}\)

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms

∴ Total number of atoms per unit cell = 4 atoms

c) Mass of one unit cell = \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}} \mathrm{g}\)

Mass of 1 mole of unit cells \(\mathrm{N}_{\mathrm{A}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) = M gram = Gram atomic mass It means that 1 unit cell contains one atom of the metal. Hence, the type of unit cell is primitive cubic or simple cubic.

Question 5.
Solids can be classified into three types on the basis of their electrical conductivities. (March – 2012)
i) Name three types of solids classified on the basis of electrical conductivities.
ii) How will you explain such classification based on Band theory?
Answer:
i) Conductors, insulators & semiconductors
ii) In conductors, the valance band overlaps with Ir e conduction band or no energy gap exists between the valance band and conduction band.

∴ The electrons can easily go into the conduction band and hence metals are good conductors. In insulators, the energy gap between valance band and conduction band is very large. Hence electrons from valance band cannot move into the conduction band.

Semi conductors have energy gap between conductors and insulators. At room temperature, these are not good conductors. But with an in crease in temperature electrons acquire sufficient energy to move from valance band into conduction band resulting in an increase in conductivity.

Question 6.
Schottky and Erenkel defects are stoichiometric defects. (May – 2012)
i) Write any two differences between Schottky defect and Frenkel defect.
ii) When pure NaCI (Sodium Chloride) crystal is heated in an atmosphere of sodium vapours, it turns yellow. Give reason.
Answer:
i)

ii) It is caused by metal excess defect due to anion vacancies. When crystals of NaCI are heated in a atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl- ions diffuse to the surface of the crystal and combine with Na atoms to give NaCI. The electrons released from Na atoms diffuse into the crystal and occupy anionic sites to form Fcentres, which impart yellow colourto the crystal. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 7.
a) NaCI has fcc structure. Calculate the number of NaCI units in a unit cell of NaCI. (March – 2013)
b) Calculate the density of NaCl, if edge length
of NaCI unit cells is 564pm. [Molar mass of
NaCI =58.5g/mol].
Answer:
a) Number of Na ions = 12 (at edge centres) x \(\frac{1}{4}\) +
= 1(at body centre) x 1
= 3 + 1 = 4
Number of Cl- ions = 8 (at the corners) x \(\frac{1}{8}\) +
= 6 (at face centres) x \(\frac{1}{2}\)
= 1 + 3 = 4
∴ Number of NaCI units per unit cell (z) = 4

Question 8.
Unit cells can be broadly classified into 2 categories primitive and centred unit cells. (May – 2013)
a) What is a unit cell?
b) Name the three types of centred unit cells.
c) The unit cell dimension of a particular crystal system is a = b = c, α = β = γ = 90°. ldentify the crystal system.
d) Give one example for the above crystal system.
Answer:
a) The smallest repeating unit of a crystal.
b) Body centred unit cell, Face centred unit cell and End centred unit cell.
c) Cubic crystal system.
d) NaCI (Rock salt struãture)

Question 9.
a) Every substance has some magnetic properties associated with it. How will you account for the following magnetic properties? (March – 2014)
i) Paramagnetic property
ii) Ferromagnetic property

b) A compound is formed by two elements P and Q. Atoms of Q (as anions) make hep lattice and those of the element P (as cations) occupy all the tetrahedral voids. What is the formula of the compound?
Answer:
a) i) Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired . electrons, e g., O₂, Cu²⁺

ii) Ferromagnetic substances are strongly attracted by a magnetic field. They are permanantly magnetised. When a ferromagnetic substance is placed in a magnetic field all the magnetic domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced, e.g,, Fe, Co

For hep lattice, No. of particles per unit cell = 6

∴ Number of anions of Q in the unit cell = 6
Number of tetrahedral voids = 2 x N = 2 x 6 = 12
∴ Number of cations of P in the unit cell = 12
Hence, formula of the compound = P₁₂Q₂ = P₂Q

Question 10.
a) Crystalline solids are ‘anisotropic’. What is ‘anisotropy’? (May – 2014)
b) Copper crystals have fee unit cells.
i) Compute the number of atoms per unit cell of copper crystals.
ii) Calculate the mass of a unit cell of copper crystals. (Atomic mass of copper = 63.54 u)
Answer:
a) Anisotropy means physical properties shows different values along different directions, eg. refrative index electrical resistance,
b) i) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.
8 corners x \(\frac{1}{8}\) per corner atom

Question 11.
Unit cells can be divided into two categories, primitive and centred unit cells. (March – 2015)
a) Differentiate between Unit Cell and Crystal Lattice.
b) Calculate the number of atoms per unit cell in the following:
i) Body centred cubic unit cell (bcc)
ii) Face centred cubic unit cell (fee)
Answer:
a) Unit Cell – It is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.

Crystal Lattice – It is the regularthree dimensional arrangement of points in space.

b) i) Body centred cubic unit cell (bcc) – It has atoms at each of its corners and one atom at its body centre.
8 corners x \(\frac{1}{8}\) per corner atom = 8 x \(\frac{1}{8}\) = 1 atom
1 body centre atom = 1 x 1 = 1 atom
∴ Total number of atoms per unit cell = 2 atoms

ii) Face centred cubic unit cell (fee) – It contains atoms at all the corners and at the centre of all the faces of the cube.

8 corners x \(\frac{1}{8}\) per corner atom
\(=8 \times \frac{1}{8}=1\) atom

6 face – centred atoms x \(\frac{1}{2}\) atom per unit cell \(=6 \times \frac{1}{2}=3\) atoms
∴ Total number of atoms per unit cell = 4 atoms

Question 12.
a) Which of the following is not a characteristic of a crystalline solid? (May – 2015)
i) Definite heat of fusion
ii) Isotropic nature
iii) A regular ordered arrangement of constituent particles
iv) A true solid

b) Frenkel defect and Shottky defects are two stoichiometric defects found in crystalline solids.
i) What are stoichiometric defects?
ii) Write any two differences between Frenkel defect and Schottky defect.
Answer:
a) ii) Isotropic nature
b) i) Stoichiometric defects are those point defects which do not disturb the stoichiometry of the solid.

ii)

Question 13.
a) Which of the following is a molecular solid? (March – 2016)
a) Diamond
b) Graphite
c) Ice
d) Quartz

b) Unit cells can be classified into primitive and centered unit cells. Differentiate between primitive and centered unit cells.
c) Presence of excess Sodium makes NaCI crystal coloured. Explain on the basis of crystal defects.
Answer:
a) Ice
b) In primitive unit cell constituent particles are present only at the corner positions. Unit cells in which one constituent particles are present at the centres of a faces in addition to those at corners.

c) Such anionic sites occupied by unpaired electrons are called F – centres (colour centres). They impart yellow colourto the crystals of NaCI. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

Question 14.
A unit cell is a term related to crystal structure. (May – 2016)
a) What do you mean by unit cell?
b) Name any two types of cubic unit cells.
c) Calculate the number of atoms in each of the above – mentioned cubic unit cells.
d) Identify the substance which shows Frenkel defect:
i) NaCI
ii) KCI
iii) ZnS
iv) AgBr
Answer:
a) Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
b) Simple cubic unit cell, body centred cubic (bcc) unit cell, face centred cubic (fee) unit cell (any two)
c) Number of atoms per unit cell
i) Simple cubic unit cell:
Total number of atoms in one unit cell \(=8 \times \frac{1}{8}=1\) atom

ii) Body centred cubic (bcc) unit cell: Total number of atoms in one unit cell
\(=\left(8 \times \frac{1}{8}\right)+(1 \times 1)=1+1=2\) atoms

iii) Face centred cubic (fee) unit cell: Total number of atoms in one unit cell \(=\left(8 \times \frac{1}{8}\right)+\left(6 \times \frac{1}{2}\right)=1+3=4\)
(any two required)

d) ZnS or AgBr
(AgBr shows both Schottky and Frenkel defects)

Question 15.
a) Identify the non – stoichiometric defect (March – 2017)
i) Schottky defect
ii) Frenkel defect
iii) Interstitial defect
iv) Metal deficiency defect
b) What type of substance could make better permanent magnets – ferromagnetic or ferrimagnetic? Justify your answer.
c) In terms of Band theory write the differences between conductor and insulator.
Answer:
a) iv) Metal deficiency defect

b) Ferromagnetic substances could make better permanent magnets because when these substances are placed in a magnetic field all the magnetic moments (domains) get oriented in the direction of the magnetic field and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagentic substance becomes a permanent magnet.

The schematic alignment of magnetic moments in a ferromagnetic substance is as shown below:

c) In conductors the valence band is either partially filled or it is overlaped with a higher energy unoccupied conduction band so that the electrons can flow easily under an applied electric field. Whereas in insulators the energy gap between the filled valence band and the next higher unoccupied conduction band is large so that electrons cannot jump to it.

Question 16.
a) From the following choose the incorrect statement about crystalline solids. (May – 2017)
i) Melt at sharp temperature.
ii) They have definite heat of fusion.
iii) They are isotropic
iv) They have long range order.

b) Cubic unit cells are divided into primitive, bcc and fee.
i) Calculate the number of atoms in a unit cell of each of the following:
* bcc
* fcc

ii) Write two examples for covalent solids.
a) iii) They are isotropic
b) i) \(\begin{array}{l}
\text { bcc }-2\left(8 \times \frac{1}{8}+1=2\right) \\
\text { fCc }-4\left(8 \times \frac{1}{8}+6 \times \frac{1}{2}=4\right)
\end{array}\)
ii) Graphite, Diamond

Kerala State Board New Syllabus Plus Two Botany Previous Year Question Papers and Answers.

Kerala Plus Two Botany Previous Year Question Paper Say 2018 with Answers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

(Questions 1 to 3) : Answer all the questions. Each question carries 1 score. (3 × 1 = 3)

Question 1.
Identify the freshwater fish from the following:
a) Sardine
b) Mackerel
c) Rohu
d) Hilsa
Answer:
c) Rohu

Question 2.
In Gel electrophoresis the separated DNA fragments can be visualized after staining. Name the stain used for it.
Answer:
Ethidium bromide

Question 3.
In a forest ecosystem different plant species are occupied in different vertical levels. Name such vertical arrangement.
Answer:
Stratification

(Questions 4 to 14. Answer any 9 questions. Each question carries 2 scores. (9 × 2 = 18)

Question 4.
Primate and non-primate female mammals exhibit cyclic changes in the activities of ovaries and accessory ducts as well as hormones during the reproductive phase. Name the cyclic changes in these groups.
Answer:
Primates – menstrual cycle
Nonprimate – oestrous cycle

Question 5.
Bamboo species and Strobilanthus Kunthiana exhibit usual flowering phenomena. Explain their flowering characteristics.
Answer:
Bamboo – flower only after 50- 100 years
Strobilanthus kunthiana – flower once in 12 years

Question 6.
A population has certain attributes that an individual organism does not. What are they?
Answer:

• Natality
• Mortality
• Sex ratio
• Population density

Question 7.
Multiple copies of gene of interest can be synthesised in vitro. Name the technique and its requirements.
Answer:
Polymerase chain reaction
2 set of Primers and DNA Polymerase enzyme

Question 8.
Catalytic converters are used in automobiles to control air pollution. Briefly comment on its role.
Answer:
Unburnt hydrocarbons are converted into carbon dioxide and water.
Carbon monoxide and nitric oxide are changed to carbon dioxide and nitrogen gas respectively.

Question 9.
Your friend wishes to start a poultry farm. What are the important suggestions given to him for the successful management of the farm?
Answer:

1. Selection of disease-free breeds
2. Proper and safe farm conditions
3. Proper feed and water and hygiene and health care are components of poultry farm management.

Question 10.
Pollination by water is seen in Zostera and Vallisneria. Enumerate its adaptations.
Answer:

• In Vallisneria, the female flower reach the surface of water by the long stalk and the male flowers or pollen grains are released on to the surface of water. Stigma collects pollen grains and reached the water inside.
• In zostera female flowers are found inside the water and pollen grains are released in the water.
• In some species pollen grains are prevented from wetting by mucilaginous covering.

Question 11.
Parasites evolved special adaptations to live on host. What are they?
Answer:

• Loss of unnecessary sense organs
• Presence of suckers to cling on to the host
• Loss of digestive system and
• High reproductive capacity.

Question 12.
Domestic sewage and industrial effluents contain large amount of nutrients. What are the probable effects of these nutrients on water bodies?
Answer:

1. It causes excessive growth of planktonic (free-floating) algae, called an algal bloom which gives colour to the water bodies.
2. Algal blooms cause lose of water quality and fish mortality.

Question 13.
Match the Column A with Column B :

Answer:

Question 14.
Humification leads to accumulation of a dark coloured amorphous substance. Identify the substance and its peculiarities.
Answer:
The substance is humus.
It is highly resistant to microbial action and undergoes decomposition at an extremely slow rate. It serves as the reservoir of nutrients.

(Questions 15 to 18): Answer any 3 questions. Each question carries 3 scores. (3 × 3 = 9)

Question 15.
Bt cotton is a transgenic pest resistant plant.
a) How this was achieved?
b) How do this plant survive on pest attack?
Answer:
a) Cry genes are isolated from Bacillus thuringiensis and inserted into cotton by gene manipulation technique.

b) Bt toxin protein exists as inactive protoxins. It is converted into an active form in the presence of the alkaline pH of insect gut.

The activated toxin binds to the surface of midgut epithelial cells and creates pores that causes cell swelling and lysis and results in the death of insect.

Question 16.
Depending on the source of pollen, pollination can be divided into three types. What are they? Explain each.
Answer:
i) Autogamy:
It is the transfer of pollen grains from the anther to the stigma of the same flower.
ii) Geitonogamy:
Transfer of pollen grains from the anther to the stigma of another flower of the same plant.
iii) Xenogamy:
Transfer of pollen grains from anther to the stigma of a different plant.

Question 17.
Hydrach succession takes place in wetter areas and the successional series progress from ‘hydric’ to ‘mesic’ condition. List out the stages in correct sequence.
Answer:

• Phytoplankton stage.
• Submerged plant stage.
• Submerged free-floating plant stage
• Reed swamp stage
• Marsh – meadow stage.
• Scrub stage.
• Forest stage.

Question 18.
Restriction endonuclease enzymes are used to cut the DNA at specific sequence.
a) Write the name of first isolated one.
b) Write the convention for naming these enzymes.
Answer:
a) Hind II

b) First capital letter – genus name
Second two letters – species name
Next letter – the strain of bacterium from which the restriction enzyme is isolated
Last roman letter – order of isolation of enzyme.

Kerala State Board New Syllabus Plus Two Botany Previous Year Question Papers and Answers.

Kerala Plus Two Botany Previous Year Question Paper March 2019 with Answers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer All questions from 1 to 3. Each carries 1 score. (3 × 1 = 3)

Question 1.
What is the function of Restriction Endonuclease in recombinant DNA technology?
a) Linktogetherfragmentof DNA
b) Make millions of copies of DNA
c) Cut DNA into many fragments
d) Separate fragments of DNA
Answer:
Cut the DNA into many fragments

Question 2.
The government of India has introduced the concept of ……….., so as to work closely with the local communities for protecting and managing forests.
Answer:
Joint forest management (JFM)

Question 3.
Which among the following is not a greenhouse gas?
a) N₂O
b) Methane
c) Carbon dioxide
d) Ozone
Answer:
d) Ozone

Answer any 9 questions from 4 to 14. Each carries 2 scores. (9 × 2 = 18)

Question 4.
Many countries encourage the cultivation of Genetically Modified Crops (G.M. Plants). Write any two advantages of GM plants.
Answer:
Made crops more tolerant to abiotic stresses (cold, drought, salt, heat)
2 Reduce reliance on chemical pesticides (pest resistant crop)

Question 5.
Match the columns (A) and (B)

Answer:

Question 6.
Write the asexual reproductive structure given in the diagrams (a), (b), (c) and (d).

Answer:
a) zoospore
b) conidia
c) bud
d) gemmule

Question 7.
PCR and ELISA are two molecular diagnostic techniques.
a) How is PCR useful in molecular diagnosis?
b) What is the principle of ELISA?
Answer:
Early detection of disease is possible by these methods
a) PCR
i) used in the amplification of pathogenic nucleic acid.
ii) Find out cancer and other genetic diseases

b) ELISA – Antigen – antibody interaction

Question 8.
a) Identify the type of ecological pyramid given below.
b) Pyramid of energy is always upright. WHy?

Answer:
a) pyramid of biomass
b) energy decreases from one trophic level to the next or energy transfer from one trophic level to the next is 10%.

Question 9.
Deforestation is a serious issue in the present scenario. Write any two major consequences of deforestation.
Answer:

1. Leads to global warming due to excess carbon dioxide
2. Loss of biodiversity

Question 10.
Observe the flow chart given below:
a) Name the processes represented as A and B.
b) If ‘N_t’ is the population density at time t, then write down the population density equation at time t + 1.

Answer:
a) Mortality, Emigration
b) Nt + 1 = Nt + [(B + I) – (D + E)]

Question 11.
The early stages of embryo development are similar in both dicots and monocots. However, mature embryos have differences. Write two major differences between dicot embryo and monocot embryo.
Answer:

Question 12.
Given below is a flow chart showing the accumulation of DDT in different trophic levels:
a) Name the phenomenon.
b) How does it affect bird population?

Answer:
a) Biomagnification.
b) High concentrations of DDT disturb calcium metabolism in birds, which causes the thinning of eggshell and their premature breaking, causing decrease in bird populations.

Question 13.
Detritivores play a major role in decomposition.
a) What are detritivores?
b) Write an example for a detritivore.
Answer:
a) Organism which breaks up detritus into smaller particles
b) Earthworm and termite

Question 14.
Double fertilization is a characteristic feature of angiosperms.
a) Which are the events in double fertilization?
b) Name the triploid nucleus formed as a result of double fertilization.
Answer:
a) syngamy and triple fusion
b) PEN (Primary endosperm nucleus)

Answer any 3 questions from 15 to 18. Each carries 3 scores. (3 × 3 = 9)

Question 15.
Recombinant DNA technology is a complex process which involves several steps. Write down the major steps in recombinant DNA technology.
Answer:

• Isolation of the Genetic Material (DNA)
• Cutting of DNA at Specific Locations
• Amplification of Gene of Interest by using PCR
• Insertion of Recombinant DNA into the Host Cell/Organism
• Obtaining the Foreign Gene Product
• Downstream Processing

Question 16.
The discovery of Restriction Endonuclease is considered as “milestone” in the history of genetic engineering.
a) Which is the first discovered restriction endonuclease?
b) What are the criteria for naming of restriction endonuclease?
Answer:
a) Hind II

b) first capital letter – genus name
second two letters – species name
next letter – the strain of bacterium from which the restriction enzyme is isolated
last roman letter – order of isolation of enzyme

Question 17.
Observe the diagram of young anther given below.
a) Identify the parts labelled as A, B, C and D
b) Which layer nourishes the developing pollen grains?

Answer:
a) A – Epidermis
B- Endothecium
C – Middle layer
D – tapetum

b) tapetum

Question 18.
Outbreeding in animals may be outcrossing, cross- creeding and interspecific hybridisation.
a) Give an example for a progeny obtained by interspecific hybridisation.
b) How does outcrossing differs from crossbreeding?
Answer:
a) Mule

b) Out-crossing:
It is mating of animals within the same breed, but having no common ancestors on either side of their pedigree up to 4-6 generations. The offspring produced called as out-cross.

Cross-breeding:
It is the method of mating superior males of one breed with superior females of another breed.

Kerala State Board New Syllabus Plus Two Zoology Previous Year Question Papers and Answers.

Kerala Plus Two Zoology Previous Year Question Paper Say 2018 with Answers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to Candidates :

• There is a ‘cool off time’ of 10 minutes each for Botany and Zoology in addition to the writing time of 1 hour each. Further, there is ‘5 minutes’ ‘Preparatory Time’ at the end of the Botany Examination and before the commencement of Zoology Examination.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’ and ‘Preparatory time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer all questions from 1 to 3. Each question carries 1 score. (3 × 1 = 3)

Question 1.
Number of spermatids produced from 25 primary spermatocyte are ………….
a) 25
b) 50
c) 100
d) 250
Answer:
c) 100

Question 2.
Study the relationship between the first two words and fill the blank space with a suitable word.
Sterilization in male: Vasectomy
Sterilization in female: ……….
Answer:
Tubectomy

Question 3.
Identify the bacterial disease from the following:
a) Typhoid
b) Amoebiasis
c) Malaria
d) Filariasis
Answer:
a) Typhoid

Answer any 9 questions 4 – 14. Each carries 2 scores. (9 × 2 = 18)

Question 4.
The incidence of STDs are reported more among the age group between 15 – 24 years.
a) What are STDs?
b) Suggest methods to prevent STDs.
Answer:
a) Sexually transmitted disease
b) i) Avoid sex with unknown partners/multiple partners.
ii) Always use condoms during coitus.

Question 5.
Observe the following cross between heterozygous dominant progeny and homozygous recessive parent. Answer the following questions.

a) Identify the cross.
b) Mention the significance of this cross.
Answer:
a) Test cross
b) In a testcross genotype of F1 can be determined when it is crossed with recessive parent.

Question 6.
Following diagram shows amino acid sequences of a part of β chain of Haemoglobin of two individuals. Observe the amino acid sequence and answer the following questions:

a) Which among the above indicates sickle cell anaemic condition?
b) Justify your answer.
c) Describe what is single base substitution.
Answer:
a) Second amino acid chain.
b) The defect is caused by the substitution of Glutamic acid by Valine at the sixth position of the beta globin chain of the haemoglobin molecule.
c) It is due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG.

Question 7.
“Human genome project is a mega project”. Give two reasons to explain this.
Answer:

1. Determine the sequences of the 3 billion chemical base pairs.
2. Store this information in database.
3. Determine the sequences of the 3 billion chemical base pairs.

Question 8.
Observe the diagram and answer the following questions:

a) Identify the diagram.
b) Name the enzymes A, B and C.
Answer:
a) Lacoperon
b) A – beta galactosidase
B – Permease
C – Tranacetylase

Question 9.
In sewage treatment plants microbes play a significant role. Distinguish between primary and secondary treatment in sewage plants.
Answer:
Primary treatment: It is the removal of large and small particles from the sewage through filtration and sedimentation.

Secondary treatment: The primary effluent is passed into large aeration tanks that is agitated and air is pumped into it. It results the vigorous growth of useful aerobic microbes into floes.

Question 10.
Human beings can conserve and protect our eco-system and biodiversity. Prepare a handout to show different methods of Biodiversity conservation.
Answer:
In situ conservation:
Some areas are considered as ‘biodiversity hotspots’ that shows high species richness and high degree of endemism. So organisms are conserved in their natural habitat.

Ex situ Conservation:
It is the conservation of threatened animals and plants outside their natural habitat.

Question 11.
“Genetic code is universal in nature”.
a) Substantiate this statement.
b) Mention any two other salient features of genetic code.
Answer:
a) From bacteria to human being triplet codon code for the same amino acid.

b) i) The codon is triplet.
ii) One codon codes for only one amino acid, it is unambiguous.
iii) Some amino acids are coded by more than one codon, It is degenerate.
iv) The codon is read in mRNA in a contiguous fashion. There are no punctuations.

Question 12.
p² + 2pq + q² = 1 is the gene frequency of a popula¬tion showing an evolutionary priciple.
a) Name the principle
b) Enlist any three factors affecting this principle.
Answer:
a) Hardy weinberg principle.
b) Gene migration, Genetic drif, Mutation and Natural selection.

Question 13.
The blood group of a child is ‘O’. His father is with ‘A’ blood group and mother with ‘B’ blood group. Write down the genotype of the child and genotypes of parents.
Answer:
The genotype of child is O group
Father I^A i
Mother I^B i

Question 14.

Above homologous organs provide evidence for a particular type of evolution.
a) Identify the type of evolution.
b) What do you mean by homologous organs?
Answer:
a) Divergent evolution
b) Homologous organ – same structure but different function.

(Q. 15 to 18). Answer any three. Each carries 3 scores. (3 × 3 = 9)

Question 15.
Match the columns B & C with column A.

Answer:

Question 16.
Prepare a flowchart of evolution of man in descending order by choosing the names given below:
Homo sapiens, Homo erectus, Homo habilis, Australopithecines, Ramapithicus, Neanderthal
Answer:
Ramapithecus → Australopithecus → homo habilus → homo erectus → neanderthal man → Homo sapiens.

Question 17.
Classify the following barriers of innate immunity under three suitable headings:
Skin, Saliva, WBC, Monocyte, Mucus, Acid of stomach
Answer:
Physical barrier – skin and mucus
Physiological barrier – acid in stomach and saliva
Cellular barrier – WBC, Monocyte

Question 18.
Expand the following:
1) SNP
2) BAC
3) YAC
Answer:
1) SNP Single nucleotide polymorphism
2) BAC Bacterial artificial chromosome
3) YAC Yeast artificial chromosome

Kerala State Board New Syllabus Plus Two Zoology Previous Year Question Papers and Answers.

Kerala Plus Two Zoology Previous Year Question Paper March 2019 with Answers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to Candidates :

• There is a ‘cool off time’ of 10 minutes each for Botany and Zoology in addition to the writing time of 1 hour each. Further, there is ‘5 minutes’ ‘Preparatory Time’ at the end of the Botany Examination and before the commencement of Zoology Examination.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’ and ‘Preparatory time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer all questions from 1 to 3. Each carries 1 score. (3 × 1 = 3)

Question 1.
Which among the following belongs to ex-situ conservation?
Wildlife sanctuaries, Biosphere reserves, Zoological parks, National parks, Sacred groves
Answer:
Zoological park

Question 2.
The milk produced during the initial few days of lactation is called …………
Answer:
Colostrum

Question 3.
Diagrammatic representation of the central dogma given below is not correct. Make necessary corrections and redraw it:

Answer:

Answer any 9 questions from 4 to 14. Each carries 2 scores. (9 × 2 = 18)

Question 4.
Prepare a flow chart showing the evolution of modem man in the hierarchical order of their evolution using the details given below:
Homo erectus, Homo habilis, Dryopithecus, Australopithecines, Homo sapiens, Ramapithecus, Neanderthal man
Answer:
Dryopitheus Ramapithecus → Australopithecus → Homo habilus → Homo erectus → Neanderthal man →
Homo sapiens.

Question 5.
Observe the figure given below:

a) Identify the figure
b) How many histone molecules are present in the Histone core?
c) Distinguish Euchromatin and Heterochromatin
Answer:
a) Nucleosome
b) 8
c) euchromatin – Region of loosely packed chromatin Heterochromatin – Region of tightly packed chromatin.

Question 6.
Some examples of evolutionary structures are given below. Classify them under suitable heading:
a) Forelimb of Man, Cheetah, Whale, Bat
b) Wings of Butterfly, Bird
c) Thorns and tendrils of Bougainvillea and Cucurbita
d) Vertebrate hearts or brains
e) Eye of the Octopus and Mammals
f) Flippers of Penguins and Dolphins
Answer:
Homologous organs:
a) Forelimb of Man, Cheetah, Whale, Bat.
d) Vertebrate hearts or brains.
c) Thorns and tendrils of Bougainvillea and cucurbita.

Analogous organs:
e) Eye of the octopus and mammals.
f) Flippers of penguins and dolphins.
b) Wings of butterfly, bird

Question 7.
“The sex of the baby is determined by the father and not by the mother.” Do you agree with this statement? Substantiate your answer.
Answer:
yes.
In case the ovum fertilises with a sperm carrying X chromosome the zygote develops into a female (XX) and with Y-chromosome results into a male offspring. Thus the genetic makeup of the sperm that determines the sex of the child.

Question 8.
Find the odd one out. Justify your answer.
Down’s syndrome. Turner’s syndrome, Phenylketonuria, Klinefelter’s syndrome.
Answer:
Phenylketonuria, it is the mendelian disorder or metabolic disorder.

Question 9.
Observe the diagram given below showing the sectional view of the female reproductive system and name the parts labelled ‘A’, ‘B’, ‘C’ & ‘D’.

Answer:
A – endometrium
B – ovary
C – isthmus
D – ampulla

Question 10.
Microbes are useful to human beings in diverse ways. If so, name the following:
a) Microbe known as ‘‘Baker’s Yeast”.
b) Lactic acid producing bacterium.
c) Fungus which helps in the production of bio-active molecule – cyclosporine A.
d) Symbiotic nitrogen fixing bacterium.
Answer:
A – Saccharomyces
B – Lactobacillus
C – Trichoderma
D – rhizobium

Question 11.
Complete the flow chart given below:

Answer:
A – aquired immunity
B – Physiological barrier
C – Cytokine barrier
D – Cell mediated immunity

Question 12.
A wide range of contraceptive methods are presently available. If so,
a) Name one contraceptive method having least side effect.
b) Which contraceptive method is generally adviced for females as a termination method to prevent any more pregnancies?
c) List out any two possible ill-effects of the usage of contraceptive methods.
Answer:
a) Natural method
b) Tubectomy
c) Irregular menstrual bleeding and Breast cancer

Question 13.
The causes of biodiversity loss are designated as “EVIL QUARTET”. Explain the Evil Quartet in biodiversity loss.
Answer:

• Habitat loss and fragmentation:
• Over-exploitation
• Alien species invasions
• Co-extinctions

Question 14.
List of some diseases commonly occurring in man are given below. Arrange them based on causative organism in the table.
Malaria, Common cold, Filariasis, Typhoid, Ascariasis, Ringworms, Amoebiadid, Pneumonia

Answer:

Answer any 3 questions from 15 to 18. Each carries 3 scores. (3 × 3 = 9)

Question 15.
The amino acid composition of the relevant portion of β chain of two hemoglobin molecules (A & B) are shown below:

a) Which one of the polypeptide chain is abnormal?
b) Name the disorder caused by it.
c) What is the reason for this abnormality?
d) What is the effect of this abnormality in such individuals?
Answer:
a) B
b) Sickle cell anemia
c) Substitution of glutamic acid by valine.
d) The mutant hemoglobin molecule under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.

Question 16.
The diagrammatic representation of the DNA fingerprint from a crime scene and that of a suspected persons are given below:

a) What is your conclusion about the suspects based on DNA Fingerprint given?
b) What is VNTR?
c) Who developed this technique first?
Answer:
a) The DNA finger print of suspect II matches with DNA from the crime scene. So suspect II is identified as culprit.
b) variable number tandem repeats
c) Alec Jeffreys

Question 17.
a) Expand STDs.
b) Cite any two examples for STD.
c) Suggest any two methods for the prevention of STDs.
Answer:
a) sexually transmitted disease
b) Gonorrhoea, syphilis, genital herpes, chlamydiasis.
c) Avoid sex with unknown partners/multiple partners.

Question 18.
The diagrammatic representation of a process in bacteria is given below:

a) Identify the process.
b) Name the enzyme involved in this process.
c) Explain the three major steps in this process.
Answer:
a) Transcription

b) DNA dependent RNA polymerase

c) Initiation:
RNA polymerase binds to promoter and initiates transcription (Initiation).
Elongation:
It uses nucleoside triphosphates as substrate and polymerises in a template.
Termination:
Once the polymerases reaches the terminator region, the nascent RNA and RNA polymerase falls off. This results in termination of transcription.