Plus One

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 3 Chapter 4 Lathiyum Vediyundayum Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 3 Chapter 4 Lathiyum Vediyundayum

Lathiyum Vediyundayum Questions and Answers

Lathiyum Vediyundayum Summary

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 1 Kinav Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 1 Kinav

Kinav Questions and Answers

Kinav Summary

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 1 Chapter 4 Matsyam Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 1 Chapter 4 Matsyam

Matsyam Questions and Answers

Matsyam Summary

Kerala State Board New Syllabus Plus One Hindi Textbook Answers Unit 1 Chapter 1 अनुताप Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Hindi Textbook Answers Unit 1 Chapter 1 अनुताप

Anuthap Hindi Chapter प्रश्न 1.
‘उसे शाक-सा लगा’ क्यों?
अनुताप
“बाबूजी आइए…… मैं पहुँचाए देता हूँ।”
एक रिक्शेवाले ने उसके नज़दीक आकर कहा,
“असलम अब नहीं आएगा।” “क्या हुआ उसको?”
रिक्शे में बैठते हुए उसने लापरवाही से पूछा। पिछले
चार-पाँच दिनों से असलम ही उसे दफ्तर पहुंचाता रहा था।

“बाबूजी, असलम नहीं रहा…”
“क्या?”
उसे शाक-सा लगा,
“कल तो भला चंगा था।
“उसके दोनों गुर्दो में खराबी थी, डाक्टर ने रिक्शा
चलाने से मना कर रखा था,”
उसकी आवाज़ में गहरी उदासी थी,
“कल आपको दफ्तर पहुंचाकर लौटा तो पेशाब बंद हो
गया था, अस्पताल ले जाते समय उसने रास्ते में ही दम तोड़ दिया था ……”
उत्तर:

प्रश्न 2.
इनके साथ हमदर्दी जताना बेवकूफ़ी होगीयहाँ यात्री का कौनसा मनोभाव प्रकट हो रहा है?
उत्तर:
श्रमिक वर्ग के प्रति उपेक्षा का मनोभाव और सहजीव के प्रति संवेदना हीनता का मनोभाव।

प्रश्न 3.
वह किसी अपराधी की भाँति सिर झुकाए रिक्शे के साथ-साथ चल रहा था, क्यों?
उत्तर:

अनुताप अनुवर्ती कार्य

ये प्रसंग किन-किन पात्रों से संबंधित हैं?

प्रश्न 4.
i) उसे शाक-सा लगा। उसकी
ii) आवाज़ में गहरी उदासी थी।
iii) उसने रास्ते में ही दम तोड़ दिया।
iv) कल की घटना उसकी आँखों के आगे सजीव हो उठी।
v) एकबारगी उसकी इच्छा हुई कि रिक्शे से उतर जाए।
vi) किसी कार के हार्न से चौंककर वह वर्तमान में आ गया।
vii) उसके लिए यह चढ़ाई खास मायने नहीं रखती थी।
viii) वह अपराधी की भाँति सिर झुकाए चल रहा था।
उत्तर:
i) यात्री को शाक-सा लगा।
ii) रिक्शेवाले की आवाज़ में गहरी उदासी थी।
iii) असलम ने रास्ते में ही दम तोड़ दिया।
iv) कल की घटना यात्री की आँखों के आगे सजीव हो उठी।
v) एकबारगी यात्री की इच्छा हुई कि रिक्शे से उत्तर जाए।
vi) किसी कार के हार्न से चौककर यात्री वर्तमान में आ गया।
vii) रिक्शेवाले के लिए यह चढ़ाई खास मायने नहीं रखती थी।
viii) यात्री अपराधी की भाँति सिर झुकाए चल रहा था।

प्रश्न 5.
यात्री का मन संघर्ष से भरा था। वह अपना संघर्ष डायरी में लिख रहा है। वह डायरी लिखें।
उत्तर:

2014 माच 5. बुधवार

नटराज नगरः
आज मेरे लिए बड़ा मानसिक संघर्ष का दिन है। रिस्शेवाला असलम की मृत्यु की खबर सुनकर मैं व्याकुल हो गया। असलम के प्रति मुझसे हमदर्दी का अभाव हुआ। मेरा दिल पश्चाताप से उत्पन्न अनुताप से भरा है। नटराज टाकीज़ के पास की चढ़ाई पार करते समय मुझे असलम की रिक्शे से उतरना था। मैं नहीं जानता था कि असलम के गुर्यों में खराबी थी। मज़बूत कदकाठी रिक्शेवाले से हमदर्दी से मैंने ज़रूर व्यवहार किया। फिर भी मेरा आत्मसंघर्ष मैं कैसे निकालूँ?

असलम के प्रति मेरी श्रद्धांजलि…. हे भगवान! मुझे माफी दें….. भगवान मुझे अच्छी नींद दें।

Hsslive Guru Plus One Hindi Notes प्रश्न 6.
असलम की मृत्यु की खबर
उत्तर:
पत्नी : लगता है आप बड़ी परेशानी में हैं?
यात्री : हैं… हाँ… आप ने ठीक समझी।
पत्नी : क्या हुआ?
यात्री : एक रिक्शावाला…
पत्नी : रिक्शावाला?
यात्री : मैं बताता हूँ।
पत्नी : हाँ…हाँ… क्या नाम है उसका?
यात्री : असलम।
पत्नी : आप और असलम के बीच….
यात्री : असलम की मृत्यु हो गयी।
पत्नी : अरे बापरे! कैसे?
यात्री : उसके दोनों गुदों में खराबी थी।
पत्नी : हे भगवान! तो?
यात्री : मैंने यह न जानकर उससे….
पत्नी : उससे?
यात्री : रिक्शा चला कर बिना हमदर्दी से व्यवहार किया।
पत्नी : आह!
यात्री : मैं पश्चाताप विवश हूँ।
पत्नी : मैं समझ सकती हूँ।
यात्री : पश्चाताप से उत्पन्न अनुताप से…..
पत्नी : अनुताप से…
यात्री : मेरे मन ने….,
पत्नी : साफ बताईए….
यात्री : मुझे उदास बना लिया है।
पत्नी : हाँ….हाँ… मैं ने अब समझ ली आप की परेशानी का कारण।
यात्री : मैं क्या करूँ?
पत्नी : चिंता छोडिए। असलम के परिवार के लिए कुछ हम कर देंगे।
यात्री : जरूर! आप एक चाय बनाईए।
पत्नी : जी हाँ….. अब तैयार होगा।

प्रश्न 7.
असलम के प्रति अपना व्यवहार
उत्तर:
यात्री अपराधी ही है। यह इसलिए है कि असलम के प्रति यात्री द्वारा दिखाई गयी उपेक्षा के कारण असलम की मृत्यु हो गयी थी।

प्रश्न 8.
हमदर्दी का अभाव
उत्तर:
अनुताप
सालों के बाद मैं उस दिन की याद में आत्मकथा लिखता हूँ। असलम नामक एक रिक्शावाला मुझे दफ्तर ले जाता था। एक दिन दफ्तर जाते समय असलम का साथी रिक्शावाले से मैंने समझा कि असलम मर गया है। असलम के दोनों गुर्दो में खराबी थी। डॉक्टर ने रिक्शा चलाने से उसे मना कर रखा था। मुझे यह नहीं मालूम था। यह न जानकर मैंने असलम से रिक्शा चलाया। रिक्शे में बैठ कर चढ़ाई पर मैंने उसे बड़ी परेशानी दी।

रिक्शा चलाते हुए असलम धीरे-धीरे कराह रहा था। बीच-बीच में एक हाथ से पेट पकड़ लेता था। दाहिना हाथ गद्दी पर जमाकर असलम बड़ी कठिनाई और परेशानी से चकाई पर रिक्शा खींच रहा था। वह बुरी तरह हाँफ रहा था। उसके गंजे सिर पर पसीने की नन्हीं नन्हीं बूंदें दिखाई देने लगी थीं। लेकिन असलम के प्रति मेरे व्यवहार में हमदर्दी का बड़ा अभाव हुआ था। आज सालों के बाद भी मेरे मन से असलम की दयनीय अवस्था का चित्र न मिट जाता। मेरा मन पश्चाताप से उत्पन्न अनुताप से आज भी भर रहा है। असलम! आप को मेडी श्रद्धांजली क्षमायाचना के रूप में मैं समर्पित करता हूँ।

प्रश्न 9.
पश्चाताप से उत्पन्न अनुताप
उत्तर:
मित्र : अरे! आप क्यों इतना उदास हैं?
यात्री : मैं…. उदास….
मित्र : हैं… हाँ… बड़ी उदासी मैं हैं आप
यात्री : आप ने ठीक समझा।
मित्र : अरे! बापरे! क्या हुआ?
यात्री : एक रिक्शावाला…
मित्र : हाँ…हाँ… क्या नाम है उसका?
यात्री : असलम।
मित्र : आप और असलम के बीच….
यात्री : असलम की मृत्यु हो गयी।
मित्र : अरे बापरे! कैसे?
यात्री : उसके दोनों गुदों में खराबी थी।
मित्र : हे भगवान! तो?
यात्री : मैंने यह न जानकर उससे….
मित्र : उससे?
यात्री : रिक्शा चला कर बिना हमदर्दी से व्यवहार किया।
मित्र : आह!
यात्री : मैं पश्चाताप विवश’हूँ।
मित्र : मैं समझ सकता हूँ।
यात्री : पश्चाताप से उत्पन्न अनुताप से…..
मित्र : अनुताप से…
यात्री : मेरे मन ने….
मित्र : साफ बताईए….
यात्री : मुझे उदास बना लिया है।
मित्र : हॉ….हाँ… मैं ने अब समझ लिया आप की उदासी का कारण।
यात्री : मैं क्या करूं?
मित्र : चिंता छोडिए। असलम के परिवार के लिए कुछ कर दीजिए।
यात्री : जरूर।

डायरी की परख, मेरी ओर से

प्रश्न 10.
घटना की सूचना है।
उत्तर:

प्रश्न 11.
संवेदना की अनुभूति है।

प्रश्न 12.
आत्मसंघर्ष की अभिव्यक्ति है।

प्रश्न 13.
आत्मपरक शैली है।
उत्तर:
कहानी
ii) अलारक्खी क्यों हताश थी?
iii) उपर्युक्त अंश का संक्षेपण करें।
iv) अलारक्खी के उस दिन की डायरी कल्पना करके लिखिए।
v) उपर्युक्त अंश केलिए उचित शीर्षक दें।

नीचे दिए मुद्दों के आधार पर अनुताप शीर्षक की सार्थकता पर अपना विचार प्रकट करें-

प्रश्न 14.
पाठ के केंद्र भाव को सूचित करता है।
उत्तर:
अनुताप’ शीर्षक बिलकुल सार्थक है। पाठ का केन्द्रभाव यात्री का अनुताप ही है। इसको यह शीर्षक ठीक सूचित करता है। पाठ पढ़कर चरमसीमा तक पहुँचने के लिए शीर्षक हमें प्रेरित करता है। पाठ का संक्षिप्त हम शीर्षक से समझ सकते हैं। इन कारणों से अनुताप शीर्षक सार्थक और संगत है।

प्रश्न 15.
चरमसीमा तक पढ़ने को प्रेरित करता है।

प्रश्न 16.
संक्षिप्त, पर स्पष्ट है।

प्रश्न 17.
सार्थक एवं संगत है।

प्रश्न 18.
निम्नलिखित पाठभाग का अनुवाद मातृभाषा में कीजिए:
आगे वह कुछ नहीं सुन सका। एक सन्नाटे ने उसे अपने आगोश में ले लिया….। कल की घटना उसकी आँखों के आगे सजीव हो उठी। रिक्शा नटराज टाकीज़ पार कर बड़े डाकखाने की ओर जा रहा था। रिक्शा चलाते हुए असलम धीरे-धीरे कराह रहा था। बीच बीच में एक हाथ से पेट पकड़ लेता था। सामने डाक बंगले तक चढ़ाई ही चढ़ाई थी। एकबारगी उसकी इच्छा हुई थी कि रिक्शे से उतर जाए। अगले ही क्षण उसने खुद को समझाया था – रोज़ का मामला है….. कब तक उतरता रहेगा….. ये लोग नाटक भी खूब कर लेते हैं, इनके साथ हमदर्दी जताना बेवकूफी होगी….. अनाप-शनाप पैसे माँगते हैं, कुछ कहो तो सरे आम रिक्शे से उतर पड़ा था, दाहिना हाथ गद्दी पर जमाकर चढ़ाई पर रिक्शा खींच रहा था। वह बुरी तरह हाँफ रहा था, गंजे सिर पर पसीने की नन्हीं-नन्हीं बूंदे दिखाई देने लगी थीं…..।
उत्तर:

प्रश्न 19.
‘उसे शाक-सा लगा’ – क्यों?
उत्तर:
असलम की आकस्मिक मृत्यु की खबर सुनकर और जीवन की क्षणिकता के बारे में सोचकर यात्री को शाक-सा लगा।

प्रश्न 20.
‘उसकी आवाज़ में गहरी उदासी थी। क्यों?
उत्तर:
अपने साथी असलम की मृत्यु के कारण और उसको नष्ट हो जाने के कारण रिकशेवाले की आवाज़ में गहरी उदासी थी।

प्रश्न 21.
‘वह किसी अपराधी की भाँति सिर झुकाए रिक्शे के साथ-साथ चल रहा था’, क्यों?
उत्तर:
अपने सहजीव के प्रति दिखाई गई उपेक्षा से उत्पन्न पश्चाताप के कारण।

प्रश्न 22.
ये प्रसंग किन-किन पात्रों से संबंधित हैं?

a. उसे शाक-सा लगा।
उत्तर:
यात्री से।

b. उसकी आवाज़ में गरही उदासी थी।
उत्तर:
मज़बूत कदकाठी रिक्शेवाले से।

c. उसने रास्ते में ही दम तोड़ दिया।
उत्तर:
असलम से।

d. कल की घटना उसकी आँखों के आगे सजीव हो उठी।
उत्तर:
यात्री से।

e. एकबारगी उसकी इच्छा हुई कि रिक्शे से उतर जाए।
उत्तर:
यात्री से।

f. किसी कार के हार्न से चौंककर वह वर्तमान में आ गया।
उत्तर:
यात्री से।

g. उसके लिए यह चढ़ाई खास मायने नहीं रखती थी।
उत्तर:
मज़बूत कदकाठी रिक्शेवाले से।

प्रश्न 23.
वह अपराधी की भाँति सिर झुकाए चल रहा था।
उत्तरः
यात्री से।

प्रश्न 24.
‘वह किसी अपराधी की भाँति सिर झुकाए रिकशे के साथ चल रहा था।’ अपराधी की भाँति कौन चल रहा था?
उत्तर:
यात्री।

प्रश्न 25.
यात्री सिर झुकाए रिकशे के साथ अपराधी जैसे क्यों चल रहा था?
उत्तर:
असलम के प्रति दिखाई गयी उपेक्षा से उत्पन्न पश्चाताप के कारण।

प्रश्न 26.
यात्री के मनोभाव के साथ ‘अनुताप’ लघुकथा के शीर्षक का कोई संबंध है?
उत्तर:
‘अनुताप’ शीर्षक से बिल्कुल संबंध है। यात्री द्वारा असलम के प्रति दिखाई गयी उपेक्षा के कारण असलम की मृत्यु हो गयी थी। यात्री के मन में इससे उत्पन्न पश्चाताप ‘अनुताप’ शीर्षक से संबंधित है।

प्रश्न 27.
‘वह किसी अपराधी की भाँति सिर झुकाए रिकशे के साथ चल रहा था। यात्री पश्चाताप से विवश होकर अपनी बहन को पत्र लिखता है। प्रस्तुत पत्र तैयार करें।
उत्तर:
स्थान,
तारीख,

प्रिय बहन रमा,
तुम कैसी हो? ठीक हो न? मैं यहाँ पर ठीक हूँ। फिर भी, दो दिनों से मेरा मन बहुत दुःखित है। मेरे परिचय का एक रिक्शवाला था। वह मुझे रोज दफ्तर ले चलता था। उसका नाम असलम है। कल असलम की आकस्मिक मृत्यु हो गयी। उसकी मृत्यु में मेरा भी दायित्व है। उसके दोनों गुों में खराबी थी। लेकिन उसके प्रति मेरी ओर से बड़ी उपेक्षा हो गयी। उसकी मृत्यु केलिए यह भी एक कारण बना। उसके प्रति मुझसे दिखाई गयी उपेक्षा से उत्पन्न पश्चाताप और अनुताप से मेरा मन विवश हो रहा है। असलम के प्रति मेरी श्रद्धांजलि जरूर है। फिर भी, रमा मैं विवश हूँ।

मुझे जवाब देकर सान्तवना देना।

(हस्ताक्षर)
तुम्हारा भाई

सेवा में,
रमा,
गाँधी नगर,
कोच्ची

प्रश्न 28.
सूचनाः यह गद्यांश पढ़कर नीचे दिए प्रश्नों का उत्तर लिखें।
राम और श्याम अनाथ बालक थे। दिन भर काम करके वे जहाँ आश्रय मिलते वहाँ सो जाते थे। वे पढ़े-लिखे नहीं थे। बच्चे स्कूल जाते वक्त वे दोनों बडी इच्छा से देखते थे। एक दिन स्कूल जानेवाले एक बच्चे से उन्होंने अपने पढ़ने का आग्रह बताया। बच्चे ने स्कूल जाकर अपने अध्यापक से सारी बातें बताई। दूसरे दिन अध्यापक, प्रधानाध्यापक से चर्चा करके इन बालकों के पास आया। उनकी दीनता देखकर अध्यापक को बहुत दुख हुआ। उन्होने बालकों के पढ़ने का आग्रह भी समझा। वे उन दोनों को अपने घर ले गए, भोजन और कपडे दिए। स्कूल में भर्ती करवा दिया और रहने का आयोजन भी किया।

i) राम और श्याम के मन में क्या आग्रह था?
उत्तर:
पढ़ने का आग्रह था।

ii) अध्यापक को बहुत दुःख क्यों हुआ?
उत्तर:
राम और श्याम की दीनता देखकर

पढ़ने का आग्रह

अनाथ बालक राम और श्याम अनपढ़ थे। उनके मन में पढ़ने के लिए बड़ी इच्छा थी। उनकी इच्छा समझकरएक स्कूल के अध्यापक उन्हें मुफ्त में पढ़ने का प्रबंध कर दे दिया।

iv) संक्षेपण केलिए उचित शीर्षक दें।
उत्तर:
पढ़ने की इच्छा।

v) बच्चे ने स्कूल जाकर अपने अध्यापक से सारी बातें बताई। बच्चा और अध्यापक के बीच का वार्तालाप तैयार कीजिए।
उत्तर:
बच्चा : अध्यापक जी….
अध्यापक : हाँ…. हाँ… क्या बात है?
बच्चा : आज मैं स्कूल आते समय…..
अध्यापक : हाँ….. आगे बोलो…
बच्चा : दो अनाथ बालकों को देखा …..
अध्यापक : ओहो ……. फिर?
बच्चा : वे हमारे स्कूल में……
अध्यापक : स्कूल में?
बच्चा : पढ़ना चाहते हैं।
अध्यापक : अरे बापरे!
बच्चा : आप कृपया इनकी सहायता कीजिए।
अध्यापक : मैं प्रधान अध्यापक से बात करूँगा।
बच्चा : धन्यवाद गुरुजी।
अध्यापक : तुम क्लास जाओ।
बच्चा : जी गुरुजी।

अनुताप Summary in Malayalam

अनुताप शब्दार्थ

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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:

Sodium	f-block
Oxygen	s-block
Uranium	d-block
Silver	p-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.

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:

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:

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.

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⁺, O^2-, Ca²⁺, Mg²⁺, F^–, N^3-, 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^– < O^2- < N^3- < 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. N^3-, O^2-, 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: N^3-, O^2-, F^–, Na⁺, Mg²⁺ and Al³⁺, nuclear charge increase in the order:
N^3-< O^2- < F^– < Na⁺ < Mg²⁺ < Al³⁺
Therefore, the ionic radii decrease in the order:
N^3- > O^2- > 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.

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⁺

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 p^H = 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 p^H of buffer is
Answer:
6

Question 12.
In the equilibrium reaction
CaCO_3(s) → CaO_(s) + CO_2(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 _________ .

Answer:
b) HSO₄^–

Question 15.
P^H 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.

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:

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), K_c at 47 °C 3.25 × 10^-9 mol per litre. What will be the value of K_p 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
K_c = 3.25 × 10^-9
K_p = K_c (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 K_p and K_c.
2. What is the relation between K_p and K_c for the reaction, N₂(g) + O₂(g) \(\rightleftharpoons \) 2NO(g)?
Answer:
1. K_p = K_c(RT)^∆n, where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants).

2. Here, ∆n = 2 – 2 = 0
K_p = K_c(RT)^∆n , K_p = K_c (RT)°
∴ K_p = K_c

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 Product	Molar Concentration
H2	0.080
l2	0.060
HI	0.490

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:

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. \(\mathrm{K}=\frac{\left[\mathrm{NH}_{3}\right] \mathrm{HCl}}{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}\)

Question 3.
1. What is meant by K_p?
2. How K_p is related to K_c?
Answer:
1. K_p 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. K_p = K_c (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:

Answer:

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 Q_c decreases on increasing pressure. As Q_c < K_c, the reaction proceeds in the forward direction.

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.

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) + SO₄^2- (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, K_sp.
3. Obtain the relation between solubility product constant (K_sp) and solubility (S), of a solid salt of general formula M_x^p+ X_y^q-.
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,

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.

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 K_p at this temperature?
Answer:
1. Henry’s law
2. K_p =K_c(RT)^∆n
∆n = 3.2 = 1
K_p = 0.027 × (0.0831 × 1000)¹ = 2.2437

Question 20.
1. For the reaction PCL \(\rightleftharpoons \) PCl₃ +Cl_c
i) Write the expression of K_c.
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 CO₃^2-

3. Common ion effect.

Question 21.
a) The pH of black coffee is 5.0. Calculate the hydrogen ion concentration.
b) The K_sp of barium sulphate is 1.5 × 10^-9. Calculate the solubility of barium sulphate in pure water.
Answer:

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.

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, Q_c?
3. Predict the direction of net reaction in the following cases:
i) Q_c < K_c
ii) Q_c > K_c
iii) Q_c = K_c
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, Q_c 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 Q_c > K_c, the reaction will proceed in the
direction of reactants (reverse reaction), i.e., net reaction goes from right to left.
ii) When Q_c < K_c, the reaction will proceed in the direction of products (forward reaction), i.e., net reaction goes from left to right.
iii) When Q_c = K_c, 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 (K_sp) 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)
K_sp = [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) + 4PO₄^3-(aq)
K_sp = [Zr⁴⁺]³[PO₄^3-]⁴ = (3S)³.(4S)⁴ = 6912S⁷

2. XY₂ is more soluble than AB.

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 K_sp = Q_sp. Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till K_sp = Q_sp.

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:

Answer:

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.

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.
K_a value is directly proportional to the acid strength, i.e., greater the K_a 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 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

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

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.

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:

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:

Homogeneous	Heterogeneous
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:

Question 8.
Calculate the mole fraction of NaOH in a solution containing 20 g of NaOH per 360 g of water.
Answer:

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:

A	B
1 amu	1.008 x 1.66 x10–24
Mass of 1 H atom	6.02 x 1023
Molar volume of O2 at STP	11.2L
Volume of 14g of N2 at STP	1.66 x 10–24
Avogadro number	22.4L

Answer:

A	B
1 amu	1.66 x 10–24
Mass of 1 H atom	1.008 x 1.6 x10–24
Molar volume of O2 at STP	22.4L
Volume of 14g of N2 at STP	11.2L
Avogadro number	6.02 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

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,

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 N2	1.5 mole N2	33600mLN2 (STP)
16g 0:	– – – – mole O2	11.2 L of O2 (STP)
….g CO2	1 mole CO2	– – – – L of C O2 (STP)
28g CO	1 mole CO	– – – – mLCO (STP)

Answer:

42g N2	1.5 Mole N2	33600mLN2 (STP)
16g O2	0.5 Mole O2	11.2 L of O2 (STP)
44 g CO2	1 mole CO2	22.4 L of CO2 (STP)
28g CO	1 mole CO	22400mLCO (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 O2	1.5 mol O2	……mL O2 (at STP)
…… g Na	2 gram atom Na	2NA Na atoms
…….g CO2	2.5 mol CO2	56 L (at STP)
8.5 g NH3	……mol NH3	11.2 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 O2	1.5 mol O2	33600 mL O2 (at STP)
46g Na	2 gram atom Na	2Na Na atoms
110 g CO2	2.5 mol CO2	56 L (at STP)
8.5 g NH3	0.5 mol NH3	11.2L(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.

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, N_A = 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:

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:

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.

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.

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.

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:

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.

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

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:
1/12^th 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) 1/12^th 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)

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 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:

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

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: