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Tamilnadu Samacheer Kalvi 10th Tamil Solutions Chapter 1.2 தமிழ்சொல் வளம்

கற்பவை கற்றபின்

Question 1.
பின்வரும் நிலவகைகளின் பெயர்களுக்கான காரணங்களைக் கேட்டறிந்து வகுப்பறையில் பகிர்க.
தரிசு, சிவல், கரிசல், முரம்பு, புறம்போக்கு, சுவல், அவல்.
Answer:
தரிசு நிலம் : பயிர் செய்யாத நிலம்
சிவல் நிலம் : செந்நிலம் அல்லது சிவந்த நிலம்
கரிசல் நிலம் : கரிய நிறமுடைய மண் கொண்ட நிலம் கரிசல் நிலம் (அ) கரிந்த பாலை நிலம்
முரம்பு நிலம் : பருக்கைக் கற்கள் கொண்ட மேட்டு நிலம்
புறம்போக்கு நிலம் : ஊர்ப்புறத்தே குடிகள் வாழ்தலில்லாத நிலம்
சுவல் நிலம் : மேட்டு நிலம்
அவல் நிலம் : ‘அவல்’ என்பதன் பொருள் ‘பள்ளம்’. ஆகவே பள்ளமான நிலப்பகுதி அவல் என அழைக்கப்படுகிறது. விளை நிலமாகவும் அமைகிறது.

Question 2.
ஒரு பொருள் தரும் பல சொற்களைப் பட்டியலிடுக.
எ.கா: சொல்லுதல் – பேசுதல், விளம்புதல், செப்புதல், உரைத்தல், கூறல், இயம்பல், மொழிதல்
Answer:
அ) மலர்தல் – அவிழ்தல், அலர்தல், நெகிழ்தல், விள்ளல், விரிதல்.
ஆ) ஞாயிறு – I சூரியன், கதிரவன், வெய்யோன், பகலவன், பரிதி.
இ) அரசன் – கோ, கொற்றவன், வேந்தன், ராஜா, கோன்.
ஈ) அழகு – அணி, வடிவு, பொலிவு, எழில்.
உ) அடி- கழல், கால், தாள், பதம், பாதம்.
ஊ) தீ – அக்கினி, நெருப்பு, தழல்.
எ) அச்சம் – பயம், பீதி, உட்கு .
ஏ) துன்பம் – இன்னல், அல்லல், இடும்பை
ஐ) அன்பு – கருணை , நேசம், ஈரம், பரிவு, பற்று.
ஓ) செய்யுள் – பா, கவிதை, யாப்பு.
ஓ) பெண் – நங்கை, வனிதை, மங்கை.
ஔ) வயல் – கழனி, பழனம், செய்.

பாடநூல் வினாக்கள்

பலவுள் தெரிக

Question 1.
‘காய்ந்த இலையும் காய்ந்த தோகையும்’ நிலத்துக்கு நல்ல உரங்கள். இத்தொடரில் அடிக்கோடிட்ட பகுதி குறிப்பிடுவது
அ) இலையும் சருகும்
ஆ) தோகையும் சண்டும்
இ) தாளும் ஓலையும்
ஈ) சருகும் சண்டும்
Answer:
ஈ) சருகும் சண்டும்

Question 2.
வேர்க்கடலை, மிளகாய் விதை, மாங்கொட்டை ஆகியவற்றைக் குறிக்கும் பயிர்வகை
அ) குலை வகை
ஆ) மணி வகை
இ) கொழுந்து வகை
ஈ) இலை வகை
Answer:
ஆ) மணி வகை

குறுவினா

Question 1.
ஒரு தாற்றில் பல சீப்பு வாழைப்பழங்கள் உள்ளன.
ஒரு சீப்பில் பல தாறு வாழைப்பழங்கள் உள்ளன.
ஒரு சீப்பில் பல வாழைப்பழங்கள் உள்ளன.

மேற்கண்ட தொடர்களில் சரியான தொடர்களைச் சுட்டிக்காட்டி, எஞ்சிய பிழையான தொடரிலுள்ள பிழைக்கான காரணத்தை எழுதுக.
Answer:

சிறுவினா

Quesiton 1.
‘புளியங்கன்று ஆழமாக நடப்பட்டுள்ளது.’
இதுபோல் இளம் பயிர் வகை ஐந்தின் பெயர்களைத் தொடர்களில் அமைக்க.
Answer:
பிள்ளை – தென்னம் பிள்ளை வாங்கி வந்தேன் .
வடலி – காட்டில் பனை வடலியைப் பார்த்தேன்.
நாற்று – நெல் நாற்று நட்டேன்.
கன்று – வாழைக்கன்று நட்டேன்.
பைங்கூழ் – சோளப் பைங்கூழ் பசுமையாக உள்ளது.

நெடுவினா

Question 1.
தமிழின் சொல்வளம் பற்றியும் புதிய சொல்லாக்கத்திற்கான தேவை குறித்தும் தமிழ் மன்றத்தில் பேசுவதற்கான உரைக் குறிப்புகளை எழுதுக.
Answer:

குறிப்புச்சட்டம் –
அறிமுகவுரை சொல்வளம் சொல்லாக்கத்திற்கான தேவை நிறைவுரை அறிமுகவுரை: வணக்கம்! அன்னைமொழியே! அழகார்ந்த செந்தமிழே! எனப் போற்றப்படும் தமிழ்மொழி பிறமொழிகளுக்கெல்லாம் தலைச்சிறந்த மொழியாகும். அம்மொழியின் சொல்வளத்தைப் பற்றிக் காண்போம்.

சொல்வளம்:

• இலக்கியச் செம்மொழிகளுக்கெல்லாம் பொதுவேனும், தமிழ்மட்டும் அதில் தலை சிறந்ததாகும்.
• தமிழ்ச்சொல் வளத்தைப் பலதுறைகளிலும் காணலாம்.
• ஒருபொருட் பல சொல் வரிசைகள் தமிழைத் தவிர வேறு எந்தத் திராவிட மொழிகளின் அகராதிகளிலும் காணப்படவில்லை .
• “பிற திராவிட மொழிகளுக்குரியனவாகக் கருதப்படும் சொற்களும் தமிழில் உள்” என்கிறார் கால்டுவெல்.

சொல்லாக்கத்திற்கான தேவை:

• சொல்லாக்கத்திற்கான தேவை என்பது அதன் பயன்பாட்டு முறையைக் கொண்டே அமைகிறது.
• இன்றைய அறிவியல் வளர்ச்சிக்கேற்ப நூல்களைப் புதிய சொல்லாக்கத்துடன் படைத்தல் வேண்டும்.
• இலக்கிய மேன்மைக்கும் மக்கள் அறிவுடையவர்களாக உயர்வதற்கும், புதிய சொல்லாக்கம் தேவை.
• மொழி என்பது உலகின் போட்டி போராட்டத்திற்கு ஒரு போர்க்கருவியாகும். அக்கருவி காலத்திற்கேற்ப மாற்றப்பட வேண்டும்.
• தமிழின் பெருமையும் மொழியின் சிறப்பும் குன்றாமல் இருக்க தமிழில் புதிய சொல்லாக்கம் தேவை.
• உலகின் பிற ஆய்வுச் சிந்தனைகளைத் தமிழ்ப்படுத்தி எழுதும் போது பிறமொழி அறியாத தமிழரும்
• அவற்றைக் குறித்து அறிந்துகொள்ள முடியும்.
• மொழியே கலாச்சாரத்தின் வழிகாட்டி. அதை நிலைநாட்ட புதிய சொல்லாக்கம் தேவை.
• மக்களிடையே பரந்த மனப்பான்மையையும், ஆளுமையையும் நிலைநாட்ட புதிய சொல்லாக்கம் தேவைப்படுகிறது.
• பிறமொழிச் சொற்கள் கலவாமல் இருக்க காலத்திற்கேற்ப புதிய கலைச்சொல்லாக்கம் ஏற்படுத்த வேண்டும்.

நிறைவுரை:

மென்மையான தமிழை மேன்மையான தமிழாக்க அறிவியல் தொழில்நுட்பச் சொற்களைத் தமிழ்ப்படுத்தி தமிழின் பெருமையை உலகிற்குக் கொண்டு செல்வோம்.

கூடுதல் வினாக்கள்

பலவுள் தெரிக

Question 1.
மரஞ்செடியினின்று பூ கீழே விழுந்த நிலையைக் குறிக்கும் சொல் எது?
அ) அரும்பு
ஆ) மலர்
இ) வீ
ஈ) செம்மல்
Answer:
இ) வீர

Question 2.
திராவிட மொழிகளின் ஒப்பியல் இலக்கணம் என்னும் நூலை எழுதியவர் யார்?
அ) பாவாணர்
ஆ) கால்டுவெல்
இ) இரா. இளங்குமரனார்
ஈ) திரு.வி.க
Answer:
ஆ) கால்டுவெல்

Question 3.
திருவள்ளுவர் தவச்சாலை அமைந்துள்ள இடம் எது?
அ) அல்லூர்
ஆ) திருவள்ளூர்
இ) கல்லூர்
ஈ) நெல்லூர்
Answer:
அ) அல்லூர்

Question 4.
குச்சியின் பிரிவு எச்சொல்லால் அழைக்கப்படுகிறது? அ) போத்து
ஆ) குச்சி
இ) இணுக்கு
ஈ) சினை
Answer:
இ) இணுக்கு

Question 5.
பொருத்துக.
1. தாள் – அ) குத்துச்செடி, புதர் முதலியவற்றின் அடி
2. தண்டு – ஆ) நெட்டி, மிளகாய்ச்செடி முதலியவற்றின் அடி
3. கோல் – இ) தண்டு, கீரை முதலியவற்றின் அடி
4. தூறு – ஈ) நெல், கேழ்வரகு முதலியவற்றின் அடி
அ) 1.ஈ 2.இ 3.ஆ 4.அ
ஆ) 1.ஆ 2.அ 3.ஈ 4-இ
இ) 1.இ 2.ஈ 3.அ 4.ஆ
ஈ) 1.அ 2.இ 3.ஆ 4.ஈ
Answer:
அ) 1.ஈ 2.இ 3.ஆ 4.அ

Question 6.
பொருத்துக.
1. தட்டு – அ) கரும்பின் அடி
2. கழி – ஆ) புளி, வேம்பு முதலியவற்றின் அடி
3. கழை – இ) கம்பு, சோளம்
முதலியவற்றின் அடி – ஈ) மூங்கிலின் அடி
அ) 1.ஆ 2.அ 3.ஈ. 4.இ
ஆ) 1.இ 2.அ 3.ஈ. 4.ஆ
இ) 1.ஈ 2.இ 3.அ 4.ஆ
ஈ) 1.இ 2.அ 3.ஆ 4.ஈ
Answer:
ஆ) 1.இ 2.அ 3.ஈ 4.ஆ

Question 7.
பொருத்துக.
1. கவை – அ) அடிமரத்தினின்று பிரியும் மாபெரும் கிளை
2. கொம்பு – ஆ) கிளையின் பிரிவு
3. சினை – இ) கவையின் பிரிவு
4. போத்து – ஈ) சினையின் பிரிவு
அ) 1.அ 2.இ 3.ஆ 4.ஈ
ஆ) 1.ஈ 2.இ 3.அ 4.ஆ
இ) 1.ஆ 2.அ 3.ஈ. 4.இ
ஈ) 1.இ 2.ஆ 3.அ 4.ஈ
Answer:
அ) 1.அ 2.இ 3.ஆ 4.ஈ

போத்து 8.
பொருந்தாதவற்றைக் கண்டறிக.
அ) தூறு
ஆ) கழி
இ) கழை
ஈ) கவை
Answer:
ஈ) கவை

Question 9.
பொருந்தாதவற்றைக் கண்டறிக.
அ) தாள்
ஆ) தண்டு
இ) கிளை
ஈ) கோல்
Answer:
இ) கிளை

Question 10.
பொருந்தாதவற்றைக் கண்டறிக.
அ) கவை
ஆ) தட்டு
இ) கொம்பு
ஈ) சினை
Answer:
ஆ) தட்டு

Question 11.
பொருந்தாத இணையைக் கண்டறிக.
அ) கவை – குச்சியின் பிரிவு
ஆ) கொம்பு – கவையின் பிரிவு
இ) போத்து – சினையின் பிரிவு
ஈ) குச்சி – போத்தின் பிரிவு
Answer:
அ) கவை-குச்சியின் பிரிவு

Question 12.
வேறுபட்ட குழுவினைக் கண்டறிக.
அ) தாள், தண்டு, கோல், தூறு
ஆ) கவை, கொம்பு, கிளை, சினை
இ) சுள்ளி, விறகு, வெங்கழி, கட்டை
ஈ) கழி, குச்சு, இணுக்கு, கழை
Answer:
ஈ) கழி, குச்சு, இணுக்கு, கழை

Question 13.
பொருத்துக.
1. இலை – அ) தென்னை , பனை முதலியவற்றின் இலை
2. தாள் – ஆ) சோளம், கம்பு முதலியவற்றின் அடி
3. தோகை – இ) புளி, வேம்பு முதலியவற்றின் இலை
4. ஓலை – ஈ) நெல், புல் முதலியவற்றின் இலை
அ) 1.ஈ 2.இ 3.அ 4.ஆ
ஆ) 1.இ 2.ஈ 3.ஆ 4.அ
இ) 1.ஆ 2.அ 3.ஈ 4.இ
ஈ) 1.இ 2.ஈ 3.அ 4.ஆ
Answer:
ஆ) 1.இ 2.ஈ 3.ஆ 4.அ

Question 14.
பொருத்தமற்ற ஒன்றைக் கண்டறிக.
அ) சண்டு
ஆ) சருகு
இ) தோகை
ஈ) கட்டை
Answer:
ஈ) கட்டை

Question 15.
பொருந்தாத இணையைக் கண்டறிக.
அ) சண்டு – காய்ந்த தாளும் தோகையும்
ஆ) சருகு – காய்ந்த இலை
இ) தாள் – புலி, வேம்பு முதலியவற்றின் இலை
ஈ) தோகை – சோளம், கம்பு முதலியவற்றின் இலை
Answer:
இ) தாள் – புலி, வேம்பு முதலியவற்றின் இலை

Question 16.
வேறுபட்ட குழுவினைக் கண்டறிக.
அ) தோகை, ஓலை, சண்டு, சருகு
ஆ) துளிர், முறி, கொழுந்து, கொழுந்தாடை
இ) பூம்பிஞ்சு, பிஞ்சு, வடு, கவ்வை
ஈ) கருக்கல், கச்சல், கொத்து, குலை
Answer:
ஈ) கருக்கல், கச்சல், கொத்து, குலை

Question 17.
தும்பி – இச்சொல்லின் பொருள்
அ) தும்பிக்கை
ஆ) வண்டு
இ) துந்துபி
ஈ) துன்பம்
Answer:
ஆ) வண்டு

Question 18.
தாவரங்களின் இலை வகைகளைக் குறிக்கும் சொற்களுள் – சரியானவற்றைச் தேர்ந்தெடு.
அ) இலை, தாள், தோகை, ஒலை, சண்டு, சருகு
ஆ) இலை, தோகை, தாள், தளிர், குருத்து, அரும்பு
இ) தாள், தோகை, தூறு, தட்டு, தண்டு, ஓலை
ஈ) இலை, தாள், ஓலை, தளிர், கொழுந்து, சண்டு
Answer:
அ) இலை, தாள், தோகை, ஒலை, சண்டு, சருகு

Question 19.
‘நாடும் மொழியும் நமதிரு கண்கள்’ என்று பாடியவர் யார்?
அ) பாரதியார்
ஆ) பாரதிதாசன் இ) பெருஞ்சித்திரனார்
ஈ) தேவநேயப் பாவாணர்
Answer:
அ) பாரதியார்

Question 20.
சொல்லாராய்ச்சியில் பாவாணரும் வியந்த பெருமகனார் யார்?
அ) தேவநேயப் பாவாணர்
ஆ) இளங்குமரனார்
இ) திரு.வி.கலியாணசுந்தரனார்
ஈ) மறைமலையடிகள்
Answer:
ஆ) இளங்குமரனார்

Question 21.
பாவாணர் நூலகம் ஒன்றை உருவாக்கியவர் யார்?
அ) தேவநேயப் பாவாணர்
ஆ) இளங்குமரனார்
இ) திரு.வி.க
ஈ) மறைமலையடிகள்
Answer:
ஆ) இளங்குமரனார்

Question 22.
தமிழ்வழித் திருமணங்களை நடத்தி வருபவர் யார்?
அ) தேவநேயப் பாவாணர்
ஆ) பெருஞ்சித்திரனார்
இ) இளங்குமரனார்
ஈ) பெருந்தேவனார்
Answer:
இ) இளங்குமரனார்

Question 23.
விழிகளை இழக்க நேரிட்டால் கூட தாய்த்தமிழினை இழந்துவிடக் கூடாது என்று எண்ணியவர் யார்?
அ) ஜி. யு. போப்
ஆ) வீரமாமுனிவர்
இ) இளங்குமரனார்
ஈ) பெருங்குமரனார்
Answer:
இ) இளங்குமரனார்

Question 24.
இளங்குமரனார் யார் போல இமைகளை மூடியபடி எழுதும் ஆற்றலைக் கற்றுக் கொண்டவர்?
அ) திரு.வி.க
ஆ) பாவாணர்
இ) மு.வ
ஈ) ஜீவா
Answer:
அ) திரு.வி.க

Question 25.
இமைகளை மூடியபடி எழுதும் ஆற்றலைக் கற்றுக்கொண்டவர்
அ) தமிழழகனார்
ஆ) அப்பாத்துரையார்
இ) தேவநேயப் பாவாணர்
ஈ) இரா.இளங்குமரனார்
Answer:
ஈ) இரா.இளங்குமரனார்

Question 26.
விழிகளை மூடியபடி எழுதும் ஆற்றல் வாய்ந்த வர்கள் ………… …………
அ) திரு.வி.க., இளங்குமரனார்
ஆ) தமிழழகனார், அப்பாத்துரையார்
இ) தேவநேயப் பாவாணர், கால்டுவெல்
ஈ) பெருஞ்சித்திரனார், சுந்தரனார்
Answer:
அ) திரு.வி.க., இளங்குமரனார்

Question 27.
‘தமிழ்த்தென்றல்’ என்று போற்றப்பட்டவர் யார்?
அ) இளங்குமரனார்
ஆ) பெருந்தேவனார்
இ) திரு.வி.க
ஈ) ம.பொ .சி
Answer:
இ) திரு.வி.க

Question 28.
உலகிலேயே ஒரு மொழிக்காக உலக மாநாடு நடத்திய முதல் நாடு எது? மாநாட்டுக்குரிய அம்முதல் மொழியும் தமிழே என்று கூறியவர் யார்?
அ) மலேசியா, க. அப்பாத்துரையார்
ஆ) சிங்கப்பூர், தேவநேயப் பாவாணர்
இ) இந்தியா, இளங்குமரனார் ஈ) கனடா, ஜி.யு. போப்
Answer:
அ) மலேசியா, க. அப்பாத்துரையார்

Question 29.
‘பன்மொழிப் புலவர்’ என்றழைக்கப்பட்டவர் யார்?
அ) க.அப்பாத்துரையார்
ஆ) தேவநேயப் பாவாணர்
இ) இளங்குமரனார்
ஈ) ஜி.யு.போப்
Answer:
அ) க.அப்பாத்துரையார்

Question 30.
சம்பா நெல்லின் உள் வகைகள் எத்தனை?
அ) 30
ஆ) 60
இ) 40
ஈ) 80
Answer:
ஆ) 60

Question 31.
‘மொழி ஞாயிறு’ என்றழைக்கப்பட்டவர் யார்? அ) க.அப்பாத்துரை
ஆ) தேவநேயப் பாவாணர்
இ) இளங்குமரனார்
ஈ) ஜி.யு.போப்
Answer:
ஆ) தேவநேயப் பாவாணர்

Question 32.
‘தமிழ்ச்சொல் வளம்’ என்னும் கட்டுரை இடம் பெற்றுள்ள நூல் எது?
அ) சொல்லாய்வுக் கட்டுரைகள், தேவநேயப் பாவாணர்
ஆ) தேவநேயம், இளங்குமரனார்
இ) மொழி மரபு, மு.வ ஈ) ஆய்வியல் நெறிமுறைகள், பொற்கோ
Answer:
அ) சொல்லாய்வுக் கட்டுரைகள், தேவநேயப் பாவாணர்

Question 33.
பொருத்திக் காட்டுக.
i) சுள்ளி – 1. காய்ந்த குச்சு (குச்சி)
iii) விறகு – 2. காய்ந்த சிறுகிளை
iii) வெங்கழி – 3. காய்ந்த கொம்பும் கவையும் அடியும்
iv) கட்டை – 4. காய்ந்த கழி
அ) 1, 2, 4, 3
ஆ) 2, 1. 3, 4
இ) 1, 3, 4, 2
ஈ) 4, 3, 2, 1
Answer:
அ) 1, 2, 4, 3

Question 34.
பொருத்திக் காட்டுக.
i) இளநீர் – 1. வாழைப்பிஞ்சு
ii) நுழாய் – 2. இளநெல்
iii) கருக்கல் – 3. இளம்பாக்கு
iv) கச்ச ல் – 4. முற்றாத தேங்காய்
அ) 4, 3, 2, 1
ஆ) 2, 3, 4, 1
இ) 1, 2, 3, 4
ஈ) 3, 4, 2, 1
Answer:
அ) 4, 3, 2, 1

Question 35.
பொருத்திக் காட்டுக.
i) சிவியல் – சூட்டினால் பழுத்த பிஞ்சு
ii) அளியல் – பதராய்ப் போன மிளகாய்
iii) சொண்டு – குளுகுளுத்த பழம்
iv) வெம்பல் – சுருங்கிய பழம்
அ) 4, 3, 2, 1
ஆ) 3, 2, 1, 4
இ) 1, 3, 4, 2
ஈ) 1, 3, 4, 2
Answer:
அ) 4, 3, 2, 1

Question 36.
ஒரு நாட்டு வளத்திற்குத் தக்கபடியே, அந்நாட்டு மக்களுக்கு எது அமைந்திருக்கும்? அ) அன்பொழுக்கம்
ஆ) அறிவொழுக்கம்
இ) களவொழுக்கம்
ஈ) கற்பொழுக்கம்
Answer:
ஆ) அறிவொழுக்கம்

Question 37.
செந்தமிழ்ச் சொற்பிறப்பியல் அகரமுதலித் திட்ட இயக்குநராகப் பணியாற்றியவர் யார்?
அ) தேவநேயப் பாவாணர்
ஆ) வெள்ளைவாரணார்
இ) இளங்குமரனார்
ஈ) பெருந்தேவனார்
Answer:
அ) தேவநேயப் பாவாணர்

Question 38.
உலகத் தமிழ்க் கழகத்தை நிறுவித் தலைவராக இருந்தவர் யார்?
அ) தனிநாயகம் அடிகள்
ஆ) தேவநேயப் பாவாணர்
இ) இளங்குமரனார்
ஈ) மு. வரதராசனார்
Answer:
ஆ) தேவநேயப் பாவாணர்

Question 39.
போர்ச்சுகீசு நாட்டின் தலைநகர் எது?
அ) லெபனான்
ஆ) லிசுபன்
இ) கெய்ரோ
ஈ) ஹராரே
Answer:
ஆ) லிசுபன்

Question 40.
இந்திய மொழிகளிலேயே மேலை நாட்டு எழுத்துருவில் முதலில் அச்சேறிய மொழி எது?
அ) இந்தி
ஆ) தமிழ்
இ) தெலுங்கு
ஈ) வங்காளம்
Answer:
ஆ) தமிழ்

Question 41.
கார்டிலா என்னும் போர்ச்சுகீசிய நூல் தமிழில் மொழிபெயர்க்கப்பட்டதைக் குறிப்பிடுவது எது?
அ) ஆறாம் உலகத் தமிழ் மாநாட்டு மலர்
ஆ) நான்காம் உலகத் தமிழ் மாநாட்டு மலர்
இ) செம்மொழி மாநாட்டு மலர்
ஈ) தமிழிலக்கிய வரலாறு மு.வ.
Answer:
அ) ஆறாம் உலகத் தமிழ் மாநாட்டு மலர்

Question 42.
கொழுந்தாடை என்பது யாது?
அ) நெல், புல் ஆகியவற்றின் கொழுந்து
ஆ) புளி, வேம்பு முதலியவற்றின் கொழுந்து
இ) தென்னை, பனை முதலியவற்றின் கொழுந்து
ஈ) கரும்பின் நுனிப்பகுதி
Answer:
ஈ) கரும்பின் நுனிப்பகுதி

குறுவினா

Question 1.
தாவரத்தின் அடிப்பகுதியைக் குறிப்பதற்கான சொற்கள் யாவை?
Answer:
தாள், தண்டு , கோல், தூறு, தட்டு, கழி, கழை, அடி.

Question 2.
தாவரங்களின் அடியிலிருந்து பிரிந்து செல்லும் பிரிவுகளுக்கு வழங்கப்படும் சொற்கள் யாவை?
Answer:
கவை, கொம்பு, கொப்பு, கிளை, சினை, போத்து, குச்சி, இணுக்கு.

Question 3.
தாவரங்களின் காய்ந்த பகுதிகளுக்கு வழங்கப்படும் சொற்கள் யாவை?
Answer:
சுள்ளி, விறகு, வெங்கழி, கட்டை.

Question 4.
தாவரங்களின் இலை வகைகளைக் குறிக்கும் சொற்கள் யாவை?
Answer:
இலை, தாள், தோகை, ஓலை, சண்டு, சருகு.

Question 5.
தாவரங்களின் நுனிப்பகுதியைக் குறிக்கும் சொற்கள் யாவை? (கொழுந்து வகை)
Answer:

• துளிர் அல்லது தளிர்
• குருத்து
• முறி அல்லது கொழுந்து
• கொழுந்துதாடை

Question 6.
பூவின் நிலைகளைக் குறிக்கும் சொற்கள் யாவை?
Answer:
அரும்பு, போது, மலர், வீ, செம்மல்.

Question 7.
தாவரங்களின் பிஞ்சு வகைகளுக்கு வழங்கப்படும் சொற்கள் யாவை?
Answer:
பூம்பிஞ்சு, பிஞ்சு, வடு, மூசு, கவ்வை, குரும்பை, முட்டுக்குரும்பை, இளநீர், நுழாய், கருக்கல், கச்சல்.

Question 8.
தாவரங்களின் குலை வகைகளைக் குறிப்பதற்கான சொற்கள் யாவை?
Answer:
கொத்து, குலை, தாறு, கதிர், அலகு அல்லது குரல், சீப்பு.

Question 9.
கெட்டுப்போன காய்க்கும் கனிக்கும் தாவரத்திற்கேற்ப வழங்கப்படும் சொற்கள் யாவை?
Answer:
சூம்பல், சிவியல், சொத்தை, வெம்பல், அளியல், அழுகல், சொண்டு, தேரைக்காய், அல்லிக்காய், ஒல்லிக்காய், கோட்டான்காய் (அ) கூகைக்காய்.

Question 10.
பழங்களின் மேற்பகுதியினைக் குறிக்க வழங்கும் சொற்கள் யாவை?
Answer:
தொலி, தோல், தோடு, ஓடு, குடுக்கை, மட்டை, உமி, கொம்மை.

Question 11.
தானியங்களுக்கு வழங்கப்படும் சொற்கள் யாவை?
Answer:
கூலம், பயறு, கடலை, விதை, காழ், முத்து, கொட்டை, தேங்காய், முதிரை.

Question 12.
தாவரங்களின் இளமைப் பெயர்களை எழுது.
Answer:
நாற்று, கன்று, குருத்து, பிள்ளை , குட்டி, பைங்கூழ், மடலி (அ) வடலி.

Question 13.
கோதுமையின் வகைகளில் சிலவற்றைக் கூறு.
Answer:
சம்பாக்கோதுமை, குண்டுக்கோதுமை, வாற்கோதுமை.

Question 14.
சம்பா நெல் வகைகளை எழுதுக.
Answer:
ஆவிரம்பூச்சம்பா, ஆனைக் கொம்பன் சம்பா, குண்டுச் சம்பா, குதிரைவாலிச் சம்பா, சிறுமணிச்சம்பா, சீரகச்சம்பா முதலிய அறுபது வகைகள் சம்பாவில் உள்ளன.

Question 15.
இரா. இளங்குமரனாரின் தமிழ்ப்பணியைத் தரம் உயர்த்திய நல்முத்துகள் யாவை?
Answer:

• இலக்கண வரலாறு
• தமிழிசை இயக்கம்
• தனித்தமிழ் இயக்கம்
• பாவாணர் வரலாறு
• குண்டலகேசி உரை
• யாப்பருங்கலம் உரை
• புறத்திரட்டு உரை
• திருக்குறள் தமிழ் மரபுரை
• காக்கைப் பாடினிய உரை
• தேவநேயம்

முதலியன இரா. இளங்குமரனாரின் தமிழ்ப்பணியைத் தரமுயர்த்திய நல்முத்துகளாகும்.

Question 16.
உலகத்தமிழ் மாநாடு குறித்து க. அப்பாத்துரையார் கூறுவன யாவை?
Answer:
“உலகத்திலேயே ஒரு மொழிக்காக உலக மாநாடு நடத்திய முதல் நாடு மலேசியா. அம்மாநாட்டுக்குரிய முதல் மொழியும் தமிழே!” என்று க. அப்பாத்துரையார் கூறுகின்றார்.

சிறுவினா

Question 1.
தேவநேயப்பாவாணர் பற்றி நீவீர் அறிந்தவற்றை எழுதுக.
Answer:
பெயர் : தேவநேயப் பாவாணர்
சிறப்புப்பெயர் : மொழிஞாயிறு
படைப்புகள் : இலக்கணக் கட்டுரைகள், மொழியாராய்ச்சிக் கட்டுரைகள், சொல்லாய்வுக் கட்டுரைகள்.
பணி : செந்தமிழ்ச் சொற்பிறப்பியல் அகர முதலித் திட்ட இயக்குநர், உலகத் தமிழ்க் கழகத் தலைவர்

Question 2.
இரா. இளங்குமரனார் குறித்து நீவீர் அறிந்தவற்றைக் கூறுக.
Answer:
பெயர் : இரா. இளங்குமரனார்
தமிழ்ப்பற்று : விழிகளை இழக்க நேரிட்டால் கூட தாய்த்தமிழினை இழந்துவிடக் கூடாது என்றார்.
திரு.வி.க வழி : தமிழ்த்தென்றல் திரு.வி.க. போல் இமைகளை மூடியபடி எழுதும் ஆற்றலைக் கொண்டவர்.
சிறந்த நூல்கள் : இலக்கண வரலாறு, தமிழிசை இயக்கம், தனித்தமிழ் இயக்கம், பாவாணர் வரலாறு, குண்டலகேசி உரை, யாப்பருங்கலம் உரை, திருக்குறள் தமிழ் மரபுரை, புறத்திரட்டு உரை, காக்கைப் பாடினிய உரை, தேவநேயம்
பிற செய்திகள் : திருவள்ளுவர் தவச்சாலை மற்றும் பாவாணர் நூலகத்தை அமைத்தார்.

Question 3.
கார்டிலா – நூல் குறிப்பு வரைக.
Answer:

• 1554-ல் போர்ச்சுகீசு நாட்டில் தமிழில் முதன்முதலில் மொழிபெயர்க்கப்பட்ட நூல்.
• ரோமன் வரிவடிவில் அச்சிடப்பட்ட நூல்.
• இதனை Carthila de lingoa Tamul e Portugues என்பர்.
• இந்திய மொழிகளுள் மேலை நாட்டு மொழிகளில் அச்சிடப்பட்ட முதல் நூல் தமிழ்மொழி நூலே.

Question 4.
எந்தெந்தத் தாவரங்களின் அடிப்பகுதி என்னென்ன பெயர்களால் அழைக்கப்படுகிறது எனப் பட்டியலிடுக.
Question

• தாள் – நெல், கேழ்வரகு முதலியவற்றின் அடி
• தண்டு – கீரை, வாழையின் அடி
• கோல் – நெட்டி, மிளகாய்ச் செடியின் அடி
• துறு – குத்துச்செடி, புதர் முதலியவற்றின் அடி
• தட்டு (அ) தட்டை – கம்பு, சோளம் முதலியவற்றின் அடி
• கழி – கரும்பின் அடி
• கழை – மூங்கிலின் அடி
• அடி – புளி, வேம்புவின் அடி

Question 5.
தாவரங்களின் அடியிலிருந்து பிரிந்து செல்லும் பிரிவுகளுக்கு வழங்கும் சொற்களையும் அவை
தாவரங்களின் எப்பகுதிக்குப் பொருந்தும் என்பதையும் எழுது.
Answer:

• கவை – மரத்தினின்று பிரியும் மாபெரும் கிளை
• கொம்பு (அ) கொப்பு – கவையின் பிரிவு
• கிளை – கொம்பின் பிரிவு
• சினை – கிளையின் பிரிவு
• போத்து – சினையின் பிரிவு
• குச்சி – போத்தின் பிரிவு
• இணுக்கு – குச்சியின் பிரிவு

Question 6.
எந்தெந்தத் தாவரங்களின் இலைகள் என்னென்ன பெயர்களால் அழைக்கப்படுகின்றன என்பதைப்
பட்டியலிடுக.
Answer:

• புளி, வேம்பு – இலை
• தென்னை, பனை – ஓலை
• நெல், புல் – தாள்
• காய்ந்த இலை – சருகு
• சோளம், கரும்பு – தோகை

Question 7.
தாவரங்களின் பிஞ்சு வகைகளுக்கு வழங்கப்படும் பெயர்களையும் அவை எத்தாவரப் பிஞ்சுகளுக்குப்
பொருந்தும் என்பதையும் கூறு.
Answer:
(பூம்பிஞ்சு -பூவோடு கூடிய இளம்பிஞ்சு, பிஞ்சு – இளம்காய்)

• வடு – மாம்பிஞ்சு
• இளநீர் – முற்றாத தேங்காய்
• மூசு – பலாப்பிஞ்சு
• நுழாய் – இளம்பாக்கு
• கவ்வை – எள் பிஞ்சு
• கருக்கல் – இளநெல்
• குரும்பை – தென்னை, பனை பிஞ்சு
• கச்சல் – வாழைப்பிஞ்சு

Question 8.
எந்தெந்தத் தாவரங்களின் குலைகள் என்னென்ன பெயர்களால் அழைக்கப்படுகின்றன எனப் பட்டியலிடுக.
Answer:

• கொத்து – அவரை, துவரை
• கதிர் – கேழ்வரகு, சோளக் கதிர்
• குலை – கொடி முந்திரி
• அலகு (அ) குரல் – நெல், தினைக் கதிர்
• தாறு – வாழைக்குலை
• சீப்பு – வாழைத் தாற்றின் பகுதி

Question 9.
கெட்டுப்போன காய்க்கும் கனிக்கும் தாவரத்திற்கேற்ப வழங்கப்படும் பெயர்கள் யாவை?
Answer:

• சூம்பல் – நுனியில் சுருங்கிய காய்
• சிவியல் – சுருங்கிய பழம்
• சொத்தை – புழுபூச்சி அரித்த காய் (அ) கனி
• வெம்பல் – சூட்டினால் பழுத்த பிஞ்சு
• அளியல் – குளுகுளுத்த பழம்
• அழுகல் – குளுகுளுத்து நாறிய பழம் (அ) காய்
• சொண்டு – பதறாய்ப் போன மிளகாய்
• கோடான்காய் (அ) கூகைக்காய் – கோட்டான் அமர்ந்ததினால் கெட்ட காய்
• தேரைக்காய் – தேரை அமர்ந்ததினால் கெட்டகாய்
• அல்லிக்காய் – தேரை அமாந்ததினால் கெட்ட தேங்காய்
• ஒல்லிக்காய் – தென்னையில் கெட்ட காய்

Question 10.
பழங்களின் மேற்பகுதியினைக் குறிக்க அவற்றின் தன்மைக்கேற்ப தாவரங்களுக்கு வழங்கப்படும் சொற்களை எழுதுக.
Answer:

• தொலி – மிக மெல்லியது
• குடுக்கை – சுரையின் ஓடு
• தோல் – திண்ணமானது
• மட்டை – தேங்காய் நெற்றின் மேற்பகுதி
• தோடு – வன்மையானது
• உமி – நெல், கம்பு ஆகியவற்றின் மூடி
• ஓடு – மிக வன்மையானது
• கொம்மை – வரகு, கேழ்வரகு உமி

Question 11.
தானியங்களுக்குத் தாவரத்திற்கேற்ப வழங்கப்படும் சொற்கள் யாவை?
Answer:

• கூலம் – நெல், புல் தானியங்கள்
• பயறு – அவரை, உளுந்து
• கடலை – வேர்க்கடலை
• விதை – கத்தரி, மிளகாய் வித்து, கொண்டைக் கடலை
• காழ் – புளி, காஞ்சிரை வித்து
• முத்து – வேம்பு, ஆமணக்கு வித்து
• கொட்டை – மா, பனை வித்து
• தேங்காய் – தென்னை வித்து
• முதிரை – அவரை, துவரை பயறுகள்

Question 12.
தாவரங்களின் இளம் பருவத்திற்கான சொற்களையும் அதற்குரிய தாவரங்களின் பெயர்களையும் எழுது.
Answer:

• நாற்று – நெல், கத்திரியின் இளநிலை
• கன்று – மா, புளி, வாழையின் இளநிலை
• குருத்து – வாழையின் இளநிலை
• பிள்ளை – தென்னையின் இளநிலை
• குட்டி – விளாவின் இளநிலை .
• மடலி (அ) வடலி – பனையின் இளநிலை
• பைங்கூழ் – நெல், சோளத்தின் இளநிலை

Question 13.
தமிழ்ச் சொல் வளம் குறித்து கால்டுவெல் குறிப்பிடும் கருத்தை எழுதுக.
Answer:
தமிழல்லாத திராவிட மொழிகளின் அகராதிகளை ஆராயும்போது, தமிழிலுள்ள ஒருபொருட் பலசொல் வரிசைகள் அவற்றில் இல்லாக்குறை எந்தத் தமிழறிஞர்க்கும் மிகத் தெளிவாகத் தோன்றும்.

தமிழில் மட்டும் பயன்படுத்தப்பட்டுத் தமிழுக்கே சிறப்பாக உரியனவாகக் கருதப்படும் சொற்கள் மட்டுமன்றித் தெலுங்கு, கன்னடம் முதலிய பிற திராவிட மொழிகளுக்குரியனவாகக் கருதப்படும் சொற்களும் தமிழில் உள என்கிறார் கால்டுவெல்.

Question 14.
எவற்றை அமைத்துக்கொள்வது, சிறந்த மதிநுட்பமும் பண்பாடும் உடைய மக்கட்கே இயலும்?
Answer:
திருந்திய மக்களை மற்ற உயிரினின்றும் பிரித்துக் காட்டுவது மொழியாதலின், அதுவே ஒரு நாட்டாரின் அல்லது இனத்தாரின் நாகரிகத்தை அளந்தறிவதற்கும் சிறந்த வழியாகும்.

பொருளைக் கூர்ந்து நோக்கி நுண்பாகுபாடு செய்து அவற்றிற்கேற்பப் பருப்பொருட் சொற்களும், நுண்பொருட்சொற்களும் அமைத்துக்கொள்வது, சிறந்த மதிநுட்பமும் பண்பாடும் உடைய மக்கட்கே இயலும்.

Question 15.
பழங்காலத்தில் விளைந்த அளவு இக்காலத்தில் விளையாதவை எவை? விளைந்து வருபவை எவை?
Answer:
பழங்காலத்தில் விளைந்த அளவு பொன்னும் மணியும் முத்தும் பவளமும் இன்று விளையாவிடினும் அருமையான கூலங்களும் சிறுகூலங்களும் இன்றும் தென்றமிழ் நாட்டில் விளைந்து வருகின்றன.

நெடுவினா

Question 1.
தமிழ், சொல் வளமுடையதென்றும் தமிழ்நாடு பொருள் வளமுடையதென்றும் தெளிவாக விளங்குவதற்கான காரணங்கள் சிலவற்றைத் தொகுத்தெழுதுக.
Answer:

முன்னுரை:

தமிழ், சொல் வளமுடையது, தமிழ்நாடு பொருள் வளமுடையது என்பதைப் பாவாணர், சொல்லாய்வுக் கட்டுரைகள் என்ற நூலில், ‘தமிழ்ச் சொல்வளம்’ என்னும் கட்டுரையில் எடுத்துரைக்கிறார்.

தாவரங்களின் அடிப்பகுதிப் பெயர்:

• தாள் – நெல், கேழ்வரகு முதலியவற்றின் அடி
• தண்டு – கீரை, வாழையின் அடி
• கோல் – நெட்டி, மிளகாய்ச் செடியின் அடி
• தூறு- குத்துச்செடி, புதர் முதலியவற்றின் அடி
• தட்டு (அ) தட்டை – கம்பு, சோளம் முதலியவற்றின் அடி
• கழி – கரும்பின் அடி
• கழை – மூங்கிலின்
• அடி – புளி, வேம்புவின் அடி

அடிப்பகுதிபிரிவு பெயர்:

• கவை – மரத்தினின்று பிரியும் மாபெரும் கிளை
• கொம்பு (அ) கொப்பு – கவையின் பிரிவு
• கிளை – கொம்பின் பிரிவு
• சினை – கிளையின் பிரிவு
• போத்து – சினையின் பிரிவு
• குச்சி – போத்தின் பிரிவு
• இணுக்கு – குச்சியின் பிரிவு

தாவர இலைப் பெயர்:

• புளி, வேம்பு – இலை
• தென்னை – பனை
• நெல், புல் – தா ள்
• காய்ந்த இலை – சருகு
• சோளம், கரும்பு – தோகை

தாவரங்களின் பிஞ்சுவகைப் பெயர்:
(பூம்பிஞ்சு -பூவோடு கூடிய இளம்பிஞ்சு, பிஞ்சு – இளம்காய்)

• வடு – மாம்பிஞ்சு
• இளநீர் – முற்றாத தேங்காய்
• மூசு – பலாப்பிஞ்சு
• நுழாய் – இளம்பாக்கு
• கவ்வை – எள் பிஞ்சு
• கருக்கல் – இளநெல்
• குரும்பை – தென்னை, பனை பிஞ்சு
• கச்சல் – வாழைப்பிஞ்சு

தாவரங்களின் குலைப் பெயர்:

• கொத்து – அவரை, துவரை
• கதிர் – கேழ்வரகு, சோளக் கதிர்
• குலை – கொடி முந்திரி
• அலகு (அ) குரல் – நெல், தினைக் கதிர்
• தாறு – வாழைக்குலை
• சீப்பு – வாழைத் தாற்றின் பகுதி

கெட்டுப்போன காய், கனிப்பெயர்:

• சூம்பல் – நுனியில் சுருங்கிய காய்
• சிவியல் – சுருங்கிய பழம்
• சொத்தை – புழுபூச்சி அரித்த காய் (அ) கனி
• வெம்பல் – சூட்டினால் பழுத்த பிஞ்சு
• அளியல் – குளுகுளுத்த பழம்
• அழுகல் – குளுகுளுத்து நாறிய பழம் (அ) காய்
• சொண்டு – பதறாய்ப் போன மிளகாய்
• கோடான்காய் (அ) கூகைக்காய் – கோட்டான் அமர்ந்ததினால் கெட்ட காய்
• தேரைக்காய் – தேரை அமர்ந்ததினால் கெட்டகாய்
• அல்லிக்காய் – தேரை அமாந்ததினால் கெட்ட தேங்காய்
• ஒல்லிக்காய் – தென்னையில் கெட்ட காய்

முடிவுரை:

மேற்குறித்த பெயர்கள் மூலம், தமிழின் சொல்வளத்தையும் தமிழ்நாட்டின் பொருள் வளத்தையும் நன்கு அறிய முடிகின்றது.

దుర్గా సప్తశతీ ప్రధానిక రహస్యం

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 8 Ionic Equilibrium Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 8 Ionic Equilibrium

12th Chemistry Guide Ionic Equilibrium Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 x 10^-4 mol L^-1 solubility product of Ag₂C₂O₄ is ………………
(a) 2.42 x 10^-8 mol³ L^-3
(b) 2.66 x 10^-12 12 mol³ L^-3
(c) 45 x 10^-11 mol³ L^-3
(d) 5.619 x 10^-12 mol³ L^-3
Answer:
(d) 5.619 x 10^-12 mol³ L^-3
Solution:
Ag₂C₂O₄ 2Ag⁺ + C₂ O₄^2-
[Ag⁺] = 2.24 x 10^-4 mol L^-1
[C₂ O₄^2-] = \(\frac{2.24 \times 10^{-4}}{2}\) mol L^-1
= 1.12 x 10^-4 mol L^-1
K_sp = [Ag⁺]² [C₂O₄^2-]
=(2.24 x 10^-4 mol L^-1)² (1.12 x 10^-4 mol L^-1)
=5.619 x 10^-12 mol³ L^-3

Question 2.
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.
(i) 60 mL \(\frac { M }{ 10 }\) HCI + 40 mL \(\frac { M }{ 10 }\) NaOH
(ii) 55 mL \(\frac { M }{ 10 }\) HCl + 45 mL \(\frac { M }{ 10 }\) NaOH
(iii) 75 mL \(\frac { M }{ 5 }\) HCI +25 mL \(\frac { M }{ 5 }\) MNaOH
(iv) 100 mL \(\frac { M }{ 10 }\) HCI+ 100 mL \(\frac { M }{ 10 }\) NaOH

pH of which one of them wilt be equal to 1?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Answer:
(d) (iii) 75 mL \(\frac { M }{ 5 }\) HCI + 25 mL \(\frac { M }{ 5 }\) NaOH
No of moles of HCl = 0.2 x 75 x 10^-3 = 15 x 10^-3
No of moles of NaOH = 0.2 x 25 x 10^-3 = 5 x 1o^-3
No of moles of HCl after mixing = 15 x 10^-3 – 5 x 10^-3
∴ Concentration of HCl
\(\frac{\text { No of moles of }}{\mathrm{HCl} \text { Vol inlitre }}\)
for (iii) solution, pH of 0.1 M HCI = – 1og₁₀ (0.1) = 1.

Question 3.
The solubility of BaSO₄ in water is 2.42 x 10^-3 gL^-1 at 298K. The value of its solubility product
(K_sp) will be …………………..
(Given molar mass of BaSO₄ = 233g mol^-1)
(a) 1.08 x 10^-14 mol²L²
(b) 1.08 x 10^-12 mol²L²
(c) 1.08 x 10^-10 mol² L²
(d) 1.08 x 10^-8 mol²L^-2
Answer:
(c) 1.08 x 10^-10 mol² L²
Solution:
BaSO₄ \(\rightleftharpoons\) Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 x 10^-3g L^-1)²
= \(\left(\frac{2.42 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}}{233 \mathrm{~g} \mathrm{~mol}^{-1}}\right)^{2}\)
= (0.01038 x 1o^-3)²
= (1.038 x 10^-5)²
= 1.077 x 10^-10
= 1.08 x 10^-10 mol² L²

Question 4.
pH of a saturated solution of Ca(OH)₂ is 9. The Solubility product (K_sp) of Ca(OH)₂
(a) 0.5 x 10^-15
(b) 0.25 x 10^-10
(c) 0.125 x 10^-15
(d) 0.5 x 10^-10
Answer:
(a) 0.5 x 10^-15
Solution:
Ca(OH)₂ \(\rightleftharpoons\) Ca²⁺ + 2OH^–
Given that pH = 9
pOH = 14 – 9 = 5
[p0K = – 1og₁₀[OH^–]]
[OH^–] = 10^-pOH
[OH^–] =10^-5M
K_sp = [Ca²⁺] [OH^–]²
= \(\frac{10^{-5}}{2} \times\left(10^{-5}\right)^{2}\) = 0.5 x 10^-15

Question 5.
Conjugate base for bronsted acids H₂O and HF are ………………
(a) OH^– and H₂FH⁺, respectively
(b) H₃O⁺ and F^–, respectively
(c) OH^– and F^–, respectively
(d) H₃O⁺ and H₂F⁺, respectively
Answer:
(c) OH^– and F^–, respectively
Solution:

∴ Conjugate bases are OH^– and F^– respectively

Question 6.
Which will make basic buffer?
(a) 50 mL of 0.1M NaOH + 25mL of 01M CH₃COOH
(b) 100 mL of 0.1M CH₃COOH + 100 mL of 0.1M NH₄OH
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH₄OH
(d) 100 mL of 0.1M HCI + 100 mL of O.1 M NaOH
Answer:
(c) 100 mL of 0.1M HCI + 200 mL of 0.1M NH₄OH
Solution:
Basic buffer is the solution which has weak base and its salt

Question 7.
Which of the following fluro – compounds is most likely to behave as a Lewis base?
(a) BF₃
(b) PF₃
(c) CF₄
(d) SiF₄
Answer:
(b) PF₃
Solution:
BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CF₄ → neutral → neither lewis acid nor base
SiF₄ → neutral → neither lewis acid nor base

Question 8.
Which of these is not likely to act as lewis base?
(a) BF₃
(b) PF₃
(c) CO
(d) F^–
Answer:
(a) BF₃
Solution:
BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
F^– → unshared pair of electron → lewis base

Question 9.
What is the decreasing order of strength of bases?
OH^–, NH₂^–, H – C ≡ C^– and CH₃ – CH₂^–
(a) OH^– > NH₂^– > H – C ≡ C > CH₃ – CH₂^–
(b) NH₂^– > OH^– > CH₃ – CH₂^– > H – C ≡ C^–
(c) CH₃ – CH₂^–, > NH₂^– > H – C ≡ C^– > OH^–
(d) OH^– > H – C ≡ C^– > CH₃ – CH₂^– > NH₂^–
Answer:
(c) CH₃ – CH₂^–, > NH₂^– > H – C ≡ C^– > OH^–
Solution:
Acid strength decreases in the order
HOH > CH ≡ CH > NH₃ > CH₃CH₃
Its conjucate bases arc in the reverse order
CH₃ – CH₂^– > NH₂^– > H – C ≡ C^– > OH^–

Question 10.
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
(a) acidic, acidic, basic
(b) basic, acidic, basic
(c) basic, neutral, basic
(d) none of these
Answer:
(b) basic, acidic, basic
Solution:

Question 11.
The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅NH) in a 0.10M aqueous pyridine solution (Kb for C₅H₅N = 1.7 x 10^-9) iS ……………..
(a) 0.006%
(b) 0.013%
(c) 0.77%
(d) 1.6%
Answer:
(b) 0.013%

Percentage of dissociation
= \(\sqrt { 1.7 }\) x 10^-4 x 100 = 1.3 x 10^-2 = 0.013%

Question 12.
Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
(a) 37 x 10^-2
(b) 10^-6
(c) 0.111
(d) none of these
Answer:
(a) 3.7 x 10^-2
pH = – log₁₀ [H⁺]
[H⁺] = 10^-pH
Let the volume be x mL
V₁M₁ + V₂M₂ + V₃M₃ = VM
x mL of 10^-1M + x mL of 10^-2M + x mL of 10^-3 M
= 3 x mL of [H⁺]
= 3 x mL of [H⁺]

= 0.037 = 3.7 x 10^-2

Question 13.
The solubility of AgCl (s) with solubility product 1.6 x 10^-10 in 0. 1 M NaCl solution would be ………….
(a) 1.26 x 10^-5 M
(b) 1.6 x 10^-9 M
(c) 1.6 x 10^-11 M
(d) Zero
Answer:
(b) 1.6 x 10^-9 M
AgCl (s) \(\rightleftharpoons\) Ag⁺(aq) + Cl^–(aq)

Question 14.
If the solubility product of lead iodide is 3.2 x 10^-8, its solubility will be …………..
(a) 2 x 10^-3M
(b) 4 x 10^-4 M
(c) 1.6 x 10^-5 M
(d) 1.8 x 10^-5 M
Answer:
(a) 2 x 10^-3M
Solution:
PbI₂ (s) \(\rightleftharpoons\) Pb²⁺ (aq) + 2I^– (aq)
K_sp = (s) (2s)²
3.2 x 10^-8 = 4s³
S = \(\left(\frac{3.2 \times 10^{-8}}{4}\right)^{1 / 3}\)
= (8 x 10^-9)^1/3 = 2 x 10^-3MNaOH

Question 15.
Using Gibb’s free energy change, ∆G⁰ = 57.34 KJ mol^-1, for the reaction, X₂Y_(s) \(\rightleftharpoons\) 2X⁺ + Y^2-(aq), calculate the solubility product of X₂Y in water at 300K (R = 8.3 J K^-1 Mol^-1)
(a) 10^-10
(b) 10^-12
(c) 10^-14
(d) can not be calculated from the given data
Answer:
(a) 10^-10
Solution:

Question 16.
MY and NY₃, are insoluble salts and have the same K_sp values of 6.2 x 10^-13 at room temperature. Which statement would be true with regard to MY and NY₃?
(a) The salts MY and NY₃ are more soluble in O.5 M KY than in pure water
(b) The addition of the salt of KY to the suspension of MY and NY₃ will have no effect on
(c) The molar solubities of MY and NY₃ in water are identical
(d) The molar solubility of MY in water is less than that of NY₃
Answer:
(d) The molar solubility of MY in water is less than that of NY₃
Solution:
Addition of salt KY (having a common ion Y^–) decreases the solubility of MY and NY₃ due to common ion effect. Option (a) and (b) are wrong.
For salt MY, MY \(\rightleftharpoons\) M⁺ + Y^–
K_sp = (s) (s)
6.2 x 10^-13 = s²
therefore \(\mathrm{s}=\sqrt{6.2 \times 10^{-13}} \simeq 10^{-7}\)
for salt NY₃,
NY₃ ⇌ N³⁺ + 3Y^–
K_sp = S (3S)³
K_sp = 27s⁴
\(S=\left(\frac{6.2 \times 10^{-13}}{27}\right)^{1 / 4} \mathrm{~s} \simeq 10^{-4}\)
The molar solubility of MY in water is less than of NY₃

Question 17.
What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
(a) 2.0
(b) 3
(c) 7.0
(d) 12.65
Answer:
(d) 12.65
Solution:
x ml of 0.1 m NaOH + x ml of 0.01 M HCI
No. of moles of NaOH = 0.1 x x x 10^-3 = 0.l x x 10^-3
No. of moles of HCl = 0.01 x x x 10^-3 = 0.01 x x 10^-3
No. of moles of NaOH after mixing = 0.1 x x 10^-3 – 0.01x  x 10^-3
= 0.09x x 10^-3
Concentration of NaOH = \(\frac{0.09 x \times 10^{-3}}{2 x \times 10^{-3}}=0.045\)
[OH^–] = 0.045
p^OH = – log (4.5 x 10^-2)
= 2 – log 4.5
= 2 – 0.65 = 1.35
pH = 14 – 1.35 = 12.65

Question 18.
The dissociation constant of a weak acid is 1 x 10^-3 . In order to prepare a buffer solution with a pH =4, the [Acid] / [Salt] ratio should be ………………..
(a) 4:3
(b) 3:4
(c) 10:1
(d) 1:10
Answer:
(d) 1:10
Solution:
K_a = 1 x 10^-3 ; pH = 4

Question 19.
The pH of 10^-5 M KOH solution will be …………..
(a) 9
(b) 5
(c) 19
(d) none of these
Answer:
(a) 9
Solution:

[OH^–] = 10^-5M.
pH = 14 – pOH .
pH = 14 – ( – log [OH^–])
= 14 + log [OH^–]
= 14 + log 10^-5
= 14 – 5 = 9

Question 20.
H₂PO₄^– the conjugate base of …………….
(a) PO₄^3-
(b) P₂O₅
(c) H₃PO₄
(d) HPO₄^2-
Answer:
(c) H₃PO₄
Solution:

Question 21.
Which of the following can act as lowery – Bronsted acid well as base?
(a) HCl
(b) SO₄^2-
(c) HPO₄^2-
(d) Br^–
Answer:
(c) HPO₄^2-
Solution:
HPO₄^2- can have the ability to accept a proton to form HPO₄^–
It can also have the ability to donate a proton to form PO₄^-3.

Question 22.
The pH of an aqueous solution is Zero. The solution is ……………..
(a) slightly acidic
(b) strongly acidic
(c) neutral
(d) basic
Answer:
(b) strongly acidic
Solution:
pH = – log₁₀[H⁺]
[H⁺] =10^-pH
= 10⁰ = 1
[H⁺] = 1 M
The, solution is strongly acidic

Question 23.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by ………………

Answer:
a
Solution:
According to Henderson equation

Question 24.
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?

Answer:
c) h = \(\sqrt{\frac{K_{h}}{K_{a} \cdot K_{b}}}\)
Solution:
h = \(\sqrt{\frac{K_{h}}{K_{a} \cdot K_{b}}}\)

Question 25.
Dissociation constant of NH₄OH is 1.8 x 10^-5 the hydrolysis constant of NH₄Cl would be …………….
(a) 1.8 x 10^-19
(b) 5.55 x 10^-10
(c) 5.55 x 10^-5
(d) 1.80 x 10^-5
Answer:
(b) 5.55 x 10¹⁰
Solution:
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{W}}}{\mathrm{K}_{\mathrm{b}}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\)
= 0.55 x 10^-9
= 5.5 x 10^-10

II. Answer the following questions

Question 1.
What are lewis acids and bases? Give two example for each.
Answer:

Lewis acids:	Lewis bases:
1. Electron pair acceptors	Electron pair donors
2. EX: BF3 , AlCl3	EX: NH3 ,H2O

Question 2.
Discuss the Lowry – Bronsted concept of acids and bases.
Answer:

• Lowry Bronsted acids – proton donors
• Lowry Bronsted bases – proton acceptors
• An acid₁ after donating a proton becomes a base₁ (conjugate base)
• A base₂ after accepting a proton becomes an acid₂ (conjugate acid)
• In general Lowry – Bronsted acid — base reaction is
  
• In the above reaction
  
• A conjugate acid – base pair differs by a proton.

Limitation : BF₃, AlCl₃ etc., that do not donate protons are known to behave as acid s.

Question 3.
Indentify the conjugate acid base pair for the following reaction in aqueous solution.
i) HS^– (aq) + HF \(\rightleftharpoons\) F^– (aq) + H₂S (aq)
ii) HPO₄^2- + SO₃^2- \(\rightleftharpoons\) PO₄^3- + HSO₃^–
iii) NH₄⁺ + CO₃^2- \(\rightleftharpoons\) NH₃ + HCO₃^–
Answer:

Question 4.
Account for the acidic nature of HClO₄. In terms of Bronsted – Lowry theory, identify its conjugate base.
Answer:
HClO₄ \(\rightleftharpoons\) H⁺ + ClO₄^–
1. According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

2. Let us consider the stabilities of the conjugate bases ClO₄^– , ClO₃^–, CIO₂^– and ClO^– formed from these acid HClO₄, HClO₃, HCIO₂, HOCI respectively.

These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently, the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

3. The charge stabilization mercases in the order, ClO^– < ClO₂^– < ClO₃^– < ClO₄^– .

This means ClO₄^– will have maximum stability and therefore will have a minimum attraction for W. Thus CIO₄^– will be weakest base and its conjugate acid HCIO₄ is the strongest acid.

4. CIO₄^– is the conjugate base of the acid HClO₄.

Question 5.
When aqueous ammonia is added to CuSO₄ solution, the solution turns deep blue due to the formation of tetrammine copper (II) complex, [Cu(H₂O)₆]²⁺(aq) + 4NH₃ (aq) \(\rightleftharpoons\) [Cu(NH₃)₄]²⁺ (aq), among H₂O and NH₃ which is stronger Lewis base.
Answer:

• [Cu(H₂O)₆]²⁺(aq) + 4NH₃ (aq) \(\rightleftharpoons\) [Cu(NH₃)₄]²⁺ (aq) + H₂
• Nitrogen less electronegative than oxygen and donates its lone pair of electrons readily. Hence NH₃ is a stronger Lewis base.
• If a better Lewis base (ligand) is available, a Lewis acid (central metal ion) will react (Ligand exchange reaction)
• In this reaction, H₂O is exchanged with NH₃.
• The Lewis acid Cu²⁺ exchanges the Lewis base H with better Lewis base N-H₃ to form [Cu(NH₃)₄]²⁺
• Hence NH₃ is a stronger Lewis base than H₂O in this reaction.

Question 6.
The concentration of hydroxide ion in a water sample is found to be 2.5 x 10^-6 M. Identify the nature of the solution.
Answer:
The concentration of OH ion in a water sample is found to be 2.5 x 10 M
pOH = – log₁₀ [OH^– ]
pOH = – 1og₁₀ [2.5 x 10^-6]
= – log₁₀ [2.5] – log₁₀ [10^-6]
= – 0.3979 – ( – 6)
= – 0.3979 + 6
pOH = 5.6

We know that,
pH + pOH = 14
pH + 5.6 = 14
pH = 14 – 5.6
pH = 8.4
pH = 8.4, shows the nature of the solution is basic.

Question 7.
A lab assistant prepared a solution by adding a calculated quantity of HCl gas 25°C to get a solution with [H₃O⁺] = 4 x 10⁵ M. Is the solution neutral (or) acidic (or) basic.
Answer:
[H₃O⁺] = 4 x 10^-5M
pH = – log₁₀ [H₃O⁺]
pH = – 1og₁₀[4 x 10⁵]
pH = – log₁₀ [4] – log₁₀ [10^-5]
pH = – 0.6020 – ( – 5) = – 0.6020 + 5
pH = 4.398
Therefore, the solution is acidic.
Since pH is less than 7, the solution is acidic

Question 8.
Calculate the pH of 0.04 M HNO₃ solution.
Answer:
Concentration of HNO₃ = 0.04M
[H₃O⁺] = 0.04 mol dm^-3
pH = – 1og[H₃O⁺]
= – log (0.04)
= – log(4 x 10^-2)
= 2 – log4 = 2 – 0.6021
= 1.3979
= 1.40

Identify the degree and leading coefficient calculator of polynomial functions.

Question 9.
Define solubility product.
Answer:
Solubility product:
It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric coefficient in a balanced equilibrium equation.

Question 10.
Define ionic product of water. Give its value at room temperature.
Answer:
1. The product of the concentration of H⁺ and OH^– ions in water at a particular temperature is known as ionic product.
2. The ionic product of water at room temperature (25°C) is,
K_w = [H⁺] [OH⁺] (or)
K_w= [H₃O⁺] [OH⁺]
K_w =(1 x 10^-7) (1 x 10^-7)
K_w= 1 x 10^-14 mol² dm^-6

Question 11.
Explain common ion effect with an example.
Answer:
Common ion Effect:
When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists.
CH₃COOH (aq) \(\rightleftharpoons\) H⁺(aq)+ CH₃COO^– (aq)

However, the added salt, sodium acetate, completely dissociates to produce Na⁺ and CH₃COO ion.
CH₃COONa (aq) → Na⁺ (aq) + CH₃COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier’s priñciple that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH₃COO^– ions combines with H ions to produce much more unionized CH₃COOH i.e.,

the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 12.
Derive an expression for Ostwald’s dilution law.
Answer:

• Ostwald’s dilution law relates the dissociation constant of the weak acid (Ka) with its degree of dissociation (a) and the concentration ( C)
• Ostwald’s dilution law states that, when dilution increases, the degree of dissociation of weak electrolyte also increases
• Degree of dissociation
  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
• Consider the dissociation of a weak acid (CH₃COOH)

Question 13.
Define pH.
Answer:
pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.
pH = – log₁₀ [H₃O⁺] (or) pH = – log₁₀ [H⁺]

Question 14.
Calculate the pH of 1.5 x 10^-3 M solution of Ba(OH)₂
Answer:
Acidity of Ba(OH)₂ in 2
∴ Normality = Molarity x acidity
= 15 x 10^-3 x 2
= 3 x 10^-3
So, [OH^–] = Normality = 3 x 10^-3
pOH = – log [OH^–]
= -log3 x 10^-3
= -log3 – log10^-3
= log 3 +3 log 10
= 3 – log3
= 3 – 0.4771
pOH = 2.5229
pH+ pOH = 14
pH = 14 – pOH
=14 – 2.5229
= 11.4771
pH = 11.48

Question 15.
50 ml of 0.05 M HNO₃ is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.
Solution:
Number of millinioles VrnI x Molarity
Millimoles of HNO₃ = 50 x 0.05 = 2.5
Millimoles of KOH = 50 x 0.025 = 1.25
After mixing millimoles of
HNO₃ remaining = 2.5 – 1.25 = 1.25
Total volume = 50 + 50 = 100 ml
Molarity = \(\frac{\text { Number of millimoles }}{\mathrm{Vml}}\)
= 1.25/100
Molarity = 1.25 x 10^-2
HNO₃ is mono basic.
∴ Normality Molarity x basicity
= 1.25 x 10^-2 x 1
= 1.25 x 10^-2
∴[H₃O⁺] = 1.25 x 10^-2
pH = – log[H₃O⁺]
= – log 1.25 x 10^-2
= – log 1.25 – log 10^-2
= – log 1.25 + 2 log 10
=2-log 1.25 = 2-0.0969
pH = 1.9031

Question 16.
The K_a value for HCN is 10^-9. What is the pH of 0.4 M HCN solution?
Answer:
K_a =10^-9
c = O.4M
pH = – log₁₀ [H₃O⁺]

∴ pH = – log(2 x 10^-5)
= – log 2 – log (10^-5)
= – 0.3010 + 5
pH = 4.699

Question17.
Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that.
K_a = K_b = 1.8 x 10^-5
Solution:
K_a = K_b = 1.8 x 10^-5

Question 18.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.
Answer:
Consider the reaction between the strong acid HCl and the weak base NH4OH to form the salt NH₄Cl and water.
\(\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

In aqueous solution NH4C1 is completely dissociated as follows:
\(\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \rightarrow \mathrm{NH}_{4}{ }^{+}{ }_{(\mathrm{aq})}+\mathrm{Cl}^{-}{ }_{(\mathrm{aq})}\)

NH₄⁺ is the strong conjugate acid of the weak base NH₄OH.
Hence it has a tendency to react with 0H from water to form unionised NH4OH.
\(\mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}+\mathrm{H}^{+}(\mathrm{aq})\)
No such tendency is shown by Cl^– the weak conjugate base of the strong acid HCI.

In the above reaction 11+ ion is formed.
∴ [H⁺] > [OH^–] and the solution is acidic and the pH is less than 7.
Equilibrium constant (hydrolysis constant) for the above equilibrium is

Question 19.
Solubility product of Ag₂CrO₄ is 1 x 10^-12. What is the solubility of Ag₂CrO₄ in 0.01 M AgNO₃ solution?
Answer:


[Ag+] = 2 S + 0.01
0.01 > > 2 S, 2 S can be neglected.
∵ [Ag⁺] = 0.01 = 1 x 10^-2
[CrO₄^–] = S; K_sp= 1 x 10-12
∴ K_sp = [Ag⁺]² [CrO₄^2-]
1 x 10^-12 = (1 10^-2)² x s
∴ S = \(\frac{1 \times 10^{-12}}{1 \times 10^{-4}}\) ⇒ S = 1 x 10^-8M

Question 20.
Write the expression for the solubility product of Ca₃(PO₄)₂
Answer:
Ca₃(PO₄)₂ (s) \(\rightleftharpoons\) 3Ca²⁺ (3s) + 2PO₄^3- (2s)
Solubility of Ca₃(PO₄)₂ is,
K_sp = [Ca²⁺]³ . [PO₄^3-]²
K_sp = (3s)³ . (2s)²
K_sp= 27 s³ . 4s²
K_sp = 108s⁵.

Question 21.
A saturated solution, prepared by dissolving CaF₂(s) in water, has [Ca²⁺] = 3.3 x 10^-4 M. What is the K_spof CaF₂?
Answer:
CaF_{2 (s)} \(\rightleftharpoons\) Ca²⁺_(aq) + 2F^–_(aq)
[F^–] = 2 [Ca²⁺] = 2 x 33 x 10^-4 M
= 6.6 x 10^-4 M
= [Ca²⁺] [F^–]²
= (3.3 x 10^-4) (6.6 x 10^-4)²
= 1.44 x 10^-10

Question 22.
K_sp of AgCl is 1.8 x 10^-10. Calculate molar solubility in 1 M AgNO₃
Answer:

[Cl^–] = S
K_sp = [Ag⁺] [Cl^–]
1.8 x 10^-10 = (1) (S)
S = 1.8 x 10^-10M

Question 23.
A particular saturated solution of silver chromate Ag₂CrO₄ has [Ag⁺] = 5 x 10^-5 and [CrO₄]^2- = 4.4 x 10 M. What is the value of for Ag₂CrO₄?
Answer:
Ag₂CrO₄ (s) \(\rightleftharpoons\) 2Ag⁺(aq) + CrO₄^2-(aq)
K_sp = [Ag⁺]² [CrO₄^2-]
= (5 x 10^-5)² (4.4 x 10^-4)
= 1.1 x 10^-12

Question 24.
Write the expression for the solubility product of Hg₂CI₂.
Answer:

Question 25.
K_sp of Ag₂CrO₄ is 1.1 x 10^-12 What is the solubility of Ag₂CrO₄ in 0.1M K₂CrO₄.
Answer:

Question 26.
Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO₃)₂ and 0.100 L of 0.2 M NaCl are mixed? (PbCI₂) = 1.2 x 10^-5.
Answer:
Total volume = 0.150 + 0.100 = 0.250L

Number of moles of Pb2 = V x M
= 0.150 x 0.1

Number of moles of Cl^– = V x M = 0.100 x 0.2

Ionic Product = [Pb²⁺] [Cl^–]²
= (0,06) (0.08)²
= 3.84 x 10^-4
Given K_sp= 1.2 x 10^-5
Ionic product > K_sp
∴ Precipitation of PbCl₂ will occur

Question 27.
of Al(OH)₃ is 1 x 10^-15 M. At what pH does 1.0 x 10^-13 M AI³⁺ precipitate on the addition of buffer of NH₄CI and NH₄OH solution?
Answer:
Al(OH)₃ Al³⁺ (aq) + 3OH^– (aq)
K_sp = [Al³⁺] [OH^–]³
Al(OH)₃ precipitates when
[Al³⁺] [OH^–]³ > K_sp
(1 x 10^-3)[OH^–]³ > K_sp
[OH^–]³ > 1 x 10^-12
[OH^–] > 1 x 10^-4M
[OH^–] = l x 10^-4 M
pOH = – 1og₁₀[OH^–] = – log (1 x 10^-4) = 4
pH = 14 – 4 = 10
Thus, Al (OH)₃ precipitates at a pH of 10

III. Evaluate Yourself

Question 1.
Classify the following as acid (or) base using Arrhenius concept

1. HNO₃
2. Ba(OH)₂
3. H₃PO₄
4. CH₃COOH

Answer:
1. HNO₃:
Nitric acid, dissociates to give hydrogen ions in water.
HNO₃ is acid.

2. Ba(OH)₂:
Barium hydroxide dissociates to give hydroxyl ions in water.
Ba(OH)₂ is base.

3. H₃PO₄:
Orthophosphoric acid dissociates to give hydrogen ions in water.
H₃PO₄ is acid.

4. CH₃COOH:
Acetic acid dissociates to give hydrogen ions in water.
CH₃COOH is acid.

Question 2.
Write a balanced equation for the dissociation of the following in water and identify the conjugate acid-base pairs.
i) NH₄
ii) H₂SO₄
iii) CH₃COOH.
Evaluate.
Answer:

Question 3.
Identify the Lewis acid and the Lewis base in the following reactions.
(i). CaO + CO₂ → CaCO₃
ii) CH₃ – O – CH₃ – AlCl₃ →

Answer:
(i). CaO + CO₂ → CaCO₃
CaO – Lewis base – All metals oxides are Lewis bases
CO₂ – Lewis acid – CO₂ contains a polar double bond.

ii)

Question 4.
H₃BO₃ accepts hydroxide ion from water as shown below
Answer:
H₃BO₃ (aq) + H₂O(l) = B(OH)₄^– + H⁺
Predict the nature of H₃BO₃ using Lewis concept.

Question 5.
At a particular temperature, the K_w of a neutral solution was equal to 4 x 10^-14. Calculate the concentration of [H₃O⁺] and [OH^–].
Answer:
Given solution is neutral
[H₃O⁺] = [OH^–]
Let [H₃O⁺] = x ; then [OH^–] = x
K_w = [H₃O⁺] [OH^–]
4 x 10^-14 = x . x
x² = 4 x 10^-14
X = \(\sqrt{4 \times 10^{-14}}=2 \times 10^{-7}\)

Question 6.
a) Calculate pH of 10^-8 M H₂SO₄
b) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4
c) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCI with 50 ml 0.1 M NaOH
Answer:
a.

In this case the concentration of H₂SO₄ is very low and hence [H₃O] from water cannot be neglected
∴ [H₃O⁺] = 2 x 10^-8 (from H₂SO₄) + 10^-7 (from water)
= 10^-8(2+ 10)
= 12 x 10^-8 = 1.2 x 10^-7
pH = – log₁₀[H₃O⁺]
= – log₁₀( 1.2 x 10^-7)
= 7 – log₁₀1.2
= 7 – 0.0791 = 6.9209

b. pH of the solution = 5.4
[H₃O⁺] = antilog of (- pH)
= antilog of (- 5.4)
= antilog of (-6 + 0.6) = \(\overline{6} .6\)
= 3.981 x 10^-6
i.e., 3.98 x 10^-6 mol dm^-3

c. No. of moles of HCl = 0.2 x 50 x 10^-3 = 10 x 10^-3
No. of moles of NaOH =0.1 x 50 x 10^-3 = 5 x 10^-3
No. of moles of HCl after mixing = 10 x 10^-3 – 5 x 10^-3
= 5 x 10^-3
after mixing total volume = 100 mL
∴ Concentration of HCl in moles per litre
\(=\frac{5 \cdot 10^{-3} \mathrm{~mole}}{100 \cdot 10^{-3} \mathrm{~L}} ;\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-2} \mathrm{M}\)
[H₃O⁺] = 5 x 10^-2 M
pH = – log ( 5 x 10^-2)
= 2 – log 5
= 2 – 0.6990
= 1.30

Question 7.
K_b for NH₄OH is 1.8 x 10^-5 Calculate the percentage of ionisation of 0.06 M ammonium hydroxide solution.
Answer:

Question 8.
1. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.
2. Calculate the pH of a buffer solution consisting of 0.4M CH₃COOH and 0.4 M CH₃COONa. What is the change in the pH after adding 0.01 mol of HCI to 500m1 of the above buffer solution.
Assume that the addition of HCI causes negligible change In the volume. Given: (K = 1.8 x 105).
Answer:
1. Dissociation of buffer components
NH₄OH (aq) \(\rightleftharpoons\) NH₄⁺ (aq) + OH^– (aq)
NH₄CI → NH₄⁺ + Cl^–
Addition of OH^–
The added H⁺ ions are neutralized by NH₄OH and there is no appreciable decrease in pH.
NH₄OH(aq) + H⁺ \(\rightleftharpoons\) NH₄⁺(aq) + H₂O (1)
Addition of
NH₄^– (aq) + OH^– (aq) → NH₄OH (aq)
The added OH ions react with NH₄ to produce unionized NH₄OH. Since NH₄OH is a weak base, there is no appreciable increase in pH.

2. pH of buffer

Addition of 0.01 mol HCI to 500 ml of buffer
Added [H⁺]
= \(\frac{0.01 \mathrm{~mol}}{500 \mathrm{~mL}}=\frac{0.01 \mathrm{~mol}}{\frac{1}{2} \mathrm{~L}}\)
= 0.02

∴ pH = – log (1.99 x 10^-5)
= 5 – log 1.99
= 5 – 0.30
= 4.70

Question 9.
1. How can you prepare a buffer solution of pH9. You are provided with 0.1 M NH₄OH solution and ammonium chloride crystals. (Given: pKb for NH₄OH is 4.7 at 25°C)

2. What volume of 0.6 M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid. (Given: pK_a for formic acid is 3.75.)
Answer:
1. We know that pH + pOH = 14
9 + pOH = 14
= pOH = 14 – 9 = 5

[NH₄Cl] = 0.1 M x 1.995
= 0. 1995 M
=0.2 M
Amount of NH₄CI required to prepare 1 litre 0.2 M solution = Strength of NH₄CI x molar
mass of NH₄CI
= 0.2 x 535
= 10.70 g
10.70 g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2 M solution. On mixing equal volume of the given NH₄OH solution and the prepared NH₄CI solution will give a buffer solution with the required pH value (pH = 9).

2.

[Sodium formate] = number of moles of HCOONa
= 0.6 x V x 10^-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10^-3
[formic acid] = number of moles of HCOOH
= 0.8 x 100 x 10^-3
= 80 x 10^-3
4 = 3.75 + log \(\frac { 0.6V }{ 80 }\)
0.25 = log \(\frac { 0.6V }{ 80 }\)
antilog of 0.25 = \(\frac { 0.6V }{ 80 }\)
0.6V = 1.778 x 80
= 1.78 x 80
= 142.4
V = \(\frac { 142.4 mL }{ 0.6 }\) = 237.33 mL

Question 10.
Calculate the
i) hydrolysis constant
ii) degree of hydrolysis and
iii) pH of 0.05M sodium carbonate solution pK_a for HCO₃^– is 10.26.
Answer:
Sodium carbonate is a salt of weak acid, H₂CO₃ and a strong base, NaOH and hence the solution is alkaline due to hydrolysis.

Given
K_w = 1 x 10^-14
c = 0.05 M
PK_a = 10.26
K_a = – log K_a
K_a = antilog of (- pK_a)
K_a = antilog of (- 10.26)
K_a = 5.49 x 10^-11

h = 6.034 x 10^-2

12th Chemistry Guide Ionic Equilibrium Additional Questions and Answers

I. Choose the best answer

Question 1.
The Latin word acidus means
a) bitter
b) sour
c) sweet
d) salty
Answer:
b) sour

Question 2.
According to Arrhenius concept, an acid is
a) hydrogen ion donor
b) hydrogen ion acceptor
c) hydroxyl ion donor
d) electron donor
Answer:
a) hydrogen ion donor

Question 3.
According to Arrhenius concept, a base is
a) hydrogen ion donor
b) hydroxyl ion acceptor
c) hydroxyl ion donor
d) electron acceptor
Answer:
c) hydroxyl ion donor

Question 4.
Arrhenius theory does not explain the behaviour of acids and bases in
a) aqueous solvents
b) non-aqueous solvents
c) water
d) None of the above
Answer:
b) non-aqueous solvents

Question 5.
According to Lowry-Bronsted theory, an acid is
a) proton donor
b) proton acceptor
c) hydroxyl ion donor
d) electron donor
Answer:
a) proton donor

Question 6.
According to Lowry-Bronsted theory, a base is
a) proton donor
b) proton acceptor
c) hydroxyl ion acceptor
d) electron donor
Answer:
b) proton acceptor

Question 7.
According to Lowry-Bronsted theory, an equilibrium exists between an acid and its
a) base
b) conjugate acid
c) conjugate base
d) water
Answer:
c) conjugate base

Question 8.
A conjugate acid-base pair differs only by
a) an electron
b) a proton
c) a hydroxyl ion
d) none of the above
Answer:
b) a proton

Question 9.
Lowry-Bronsted theory could not explain the acidic behaviour of
a) BF₃
b) AlCl₃
c) SO₂
d) all the above
Answer:
d) all the above

Question 10.
According to Lewis concept an acid is
a) proton donor
b) proton acceptor
c) electron pair acceptor
d) electron-pair donor
Answer:
c) electron pair acceptor

Question 11.
According to Lewis concept a base is
a) hydroxyl ion donor
b) hydroxyl ion acceptor
c) electron pair acceptor
d) electron-pair donor
Answer:
d) electron-pair donor

Question 12.
In a coordination complex the central metal ion acts as
a) Arrhenius acid
b) Lewis base
c) Lowry-Bronsted acid
d) Lewis acid
Answer:
d) Lewis acid

Question 13.
Which of the following is true for acidic solutions?
a) [H₃O⁺] > [OH^–]
b) [H₃O⁺] < [OH^–]
c) [H₃O⁺] = [OH^–]
d) [H₃O⁺] < [OH^–]
Answer:
a) [H₃O⁺] > [OH^–]

Question 14.
Electron deficient molecules can act as
a) Arrhenius base
b) Lewis base
c) Lowry-Bronsted base
d) Lewis acid
Answer:
d) Lewis add

Question 15.
Molecules with one or more lone pairs of electrons can act as
a) Arrhenius acid
b) Lewis base
c) Lowry-Bronsted acid
d) Lewis acid
Answer:
b) Lewis base

Question 16.
Which among the following is a Lewis base?
a) BF₃
b) SO₃
C) SF₄
d) CaO
Answer:
d) CaO

Question 17.
Which among the following is not a Lewis base?
a) MgO
b) CO₂
c) H₂O
d) Na₂O
Answer:
b) CO₂

Question 18.
Which among the following is a Lewis acid?
a) NH₃
b) CaO
c) RNH₂
d) FeCl₃
Answer:
d) FeCl₃

Question 19.
Which among the following is not a Lewis acid?
a) SiF₄
b) CH₂=CH₂
c) BeF₂
d) Fe³⁺
Answer:
b) CH₂=CH₂

Question 20.
The conjugate acid of H2O is
a) HCl
b) OH^–
c) H₃O⁺
d) HSO₄^–
Answer:
c) H₃O⁺

Question 21.
The conjugate base of H2O is
a) HCl
b) OH^–
c) H₃O⁺
d) HSO₄^–
Answer:
b) OH^–

Question 22.
Which of the following can act both as Bronsted acid and Bronsted base?
a)Cl^–
b) H₃O⁺
c) HCO₃^–
d) CO₃^2-
Answer:
c) HCO₃^–

Question 23.
In which of the following cases, the sparingly soluble salt solution is unsaturated?
a) Ionic product > solubility product (Ksp)
b) Ionic product < solubility product (Ksp)
c) Ionic product = solubility product (KcD)
d) Both (a) and (b)
Answer:
b) Ionic product > solubility product (Ksp)

Question 24.
Which of the following is the strongest conjugate base?
a) Cl^–
b) SO₄^2-
c) CH₃COO^–
d) NO₃^–
Answer:
c) CH₃COO^–
Reason: The conjugate base of a weak acid is a stronger base.

Question 25.
Among the following the Ka value for a strongest acid is
a) 1.8 x 10^-4
b) 1.8 x 10^-5
c) 1.8 x 10⁵
d) 1.8 x 10⁴
Answer:
c) 1.8 x 10⁵

Question 26.
Which among the following is the strongest base?
a) HSO₄^–
b) H₂O
C) F^–
d) H^–
Answer:
d) H^–
H₂ is the weakest acid and its conjugate base
H^– is the strongest base.

Question 27.
Which of the following salts do not undergo salt hydrolysis?
a) Sodium acetate
b) Ammonium acetate
c) Ammonium chloride
d) Sodium nitrate
Answer:
d) Sodium nitrate

Question 28.
If the hydrogen ion concentration of the solution is 10^-5M, its hydroxyl ion concentration is
a) 10^-5 M
b) 10^-9M
c) 10^-14M
d) 10^-7M
Ans:
b) 10^-9M

Question 29.
If the hydrogen ion concentration of a solution is 10~SM, its pOH is
a) 5
b) 9
c) 14
d) 7
Answer:
b) 9 Reason: pH = -log[H₃0⁺] = -log 10^-5 = 5
pH + pOH = 14; pOH = 14 – pH = 14 – 5 = 9

Question 30.
If the pH of a solution is 9, the solution is
a) acidic
b) neutral
c) basic
d) strongly acidic
Answer:
c) basic
Reason:.pH < 7; pOH > 7 = Acidic
pH > 7; pOH < 7 = Basic

Question 31.
If the pH of a solution is zero the solution is
a) acidic
b) neutral
c) basic
d) strongly acidic
Answer:
d) strongly addic

Question 32.
The pOH of IN HCI is
a) 0
b) 1
c) 7
d) 14
Answer:
d) 14
Reason: pH = -log[H₃O⁺] = -log 1 = 0
pOH = 14 – pH = 14 – 0 = 14

Question 33.
As [H₃O⁺] of a solution increases, its pH
a) increases
b) deposes
c) remains the same
d) becomes
Answer:
b) decreases

Question 34.
The concentration of a solution of acetic acid changes from 10^-4 M to 10^-2 M, its degree of dissociation
a) increases
b) decreases
c) remains the same
d) becomes zero
Answer:
b) decreases
Reason: As dilution increases (Concentration decreases) degree of dissociation increases. Here concentration increases. Hence degree of dissociation decreases.

Question 35.
Which of the following represents Ostwald’s dilution law for a binary electrolyte whose degree of disscociation is a and concentration C

Answer:
c

Question 36.
NH₄OH is a weak base because
a) It has a low vapour pressure
b) It is completely ionised
c) it is only partially ionised
d) It has low density
Answer:
c) it is only partially ionised

Question 37.
The concentration of acetic acid changes from 10^-2M to 10^-4 M its degree of dissociation
a) increases
b) decreases
c) remains the same
d) becomes zero
Answer:
a) increases

Question 38.
For a weak acid the hydrogen ion concentration is given as

Answer:
d) both (a) & (b)

Question 39.
When ammonium chloride is added to ammonium hydroxide, the degree of dissociation of ammonium hydroxide
a) increases
b) decreases
c) remains the same
d) becomes zero
Answer:
b) decreases
Reason: Common ion effect

Question 40.
The relationship between the solubility product (Ksp) and molar solubility (S) for Ag2 (OO4) is
a) K_sp = S³
b) K_sp = S²
c) K_sp = 4S³
d) K_sp = 3S²
Answer:
c) K_sp = 4S³

Question 41.
The decrease in degree of dissociation of HF on addition of NaF is known as
a) buffer action
b) neutralization
c) common ion effect
d) hydrolysis
Answer:
c) common ion effect

Question 42.
Which is not a buffer solution?
a) CH₃COOH + CH₃COONa
b) HCl + NaCl
c) NH₄OH + NH₄Cl
d) H₂CO₃+NaHCO₃
Answer:
b) HCl + NaCl
Reason: A strong acid and its salt is not a buffer solution. A weak acid or base and its salt is a buffer solution.

Question 43.
Which among the following does not undergo hydrolysis?
a) NH₄Cl
b) CH₃COONa
c) NaCl
d) CH₃COONH₄
Answer:
c) NaCl
Reason: NaCt is a salt of strong acid and strong base.

Question 44.
Hydrolysis of CH3COONa gives
a) acidic solution
b) basic solution
c) neutral solution
d) No solution
Answer:
b) basic solution
Reason: CH₃COONa is a salt of weak acid and strong base. Hence CH₃COO^– undergoes hydrolysis giving a basic solution.

Question 45.
The pH of the solution resulting from the hydrolysis of NH₄Cl is
a) 7
b) greater than 7
c) less than 7
d) 14
Answer:
c) less than 7
Reason: NH₄Cl is a salt of strong acid and weak base. NH₄C⁺ undergoes hydrolysis giving an acidic solution. Hence pH is less than 7.

Question 46.
For the hydrolysis of salt of weak acid and weak base which among the following is true?
a) K_h K_a = K_w
b) K_h K_b = k_w
c) K_a K_b K_h = K_w
d) K_h= \(\frac{K_{1} K_{b}}{K_{w}}\)
Answer:
c) K_a K_b K_h = K_w

Question 47.
The pH of the solution obtained from the hydrolysis of salt of strong acid and weak base

Answer:
b

Question 48.
\(X_{m} Y_{n(s)} \stackrel{H_{2} O}{\rightleftharpoons} m X_{(a q)}^{n+}+n Y_{(a q)}^{m-}\)
The solubility product of the salt X_mY_n is given as
a) K_sp = [X^m+]ⁿ[Y^n-]^m
b) K_sp = m[Xⁿ⁺] n[Y^m-]
c) K_sp = [Xⁿ⁺]^m [Y^m-]ⁿ
d) K_sp =[X_mY_n]^m+n
Answer:
c) K_sp = [Xⁿ⁺]^m [Y^m-]ⁿ

Question 49.
The aqueous solution sdium acetate, ammonium chloflde, sodi maitrate are respectively. PTA -61
a) Neutral,acidic,basic
b) acidic, basic, neutral
c) basic, acidic, neutral
d) basic, aidk, basic
Answer:
d) basic, acidic, basic

Question 50.
The relationship between the solubility product and molar solubility of Al₂(SO₄)₃ is
a) S²
b) 4S³
c) 108S⁵
d) 27S⁵
Answer:
c) 108S⁵
Reason:

II. Pick out the correct statements

Question 1.
i) Ka measures the strength of an acid
ii) Larger the pK_a value, stronger is the acid.
iii) Weak acids undergo complete ionisation
iv) Acids with K_a value greater than 10 are considered as strong acids.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statement:
ii) Larger the pK_a value, weaker is the acid, (or) Lower the pKa value, stronger is the acid.
iii) Weak acids undergo partial ionisation, (or) strong acids undergo complete ionisation.

Question 2.
i) Degree of dissociation
\(=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
ii) dilution law is applicable to strong acids.
iii) Strong acids undergo complete dissociation
iv) In a weak acid when dilution increases by 100 times the dissociation increases by 10 times.
a) (i) & (ii) b) (ii) & (iii)
c) (ii) & (iv) d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statement:
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement:
ii) Ostwald’s dilution law is applicable to strong acids.
iv) In a weak acid when dilution increases by 100 times the dissociation increases by 10 times.

Question 3.
i) A buffer solution is a mixture of a weak acid and its conjugate base
ii) Common ion effect is based on Le Chatelier’s principle.
iii) Blood contains a buffer solution of HCl and NaCl
iv) For a buffer solution
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
\(\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{acid}]}{[\mathrm{salt}]}\)
Answer:
a) (i) & (ii)
Correct statement:
iii) Blood contains a buffer solution of H₂CO₃ and NaHCO₃
iv) For a buffer solution
pH = pKa + log\(\frac{[\text { acid }]}{[\text { salt }]}\)

Question 4.
i) The solution obtained from the hydrolysis of CH₃COONa is acidic
ii) The solution obtained from the hydrolysis of NaCl is neutral
iii) The nature of the solution obtained from the hydrolysis of a salt of weak acid and weak base depends on the strength of acid or base.
iv) The solution obtained from the hydrolysis of NH4C/ is basic
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
i) The solution obtained from the hydrolysis of CH₃COONa is basic
iv) The solution obtained from the hydrolysis of NH₄Cl is acidic

III. Pic out the correct statements

Question 1.
For the equilibrium
\(\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{ClO}_{4}^{-}\)
Which are the incorrect statements?
i) HClO₄ is the conjugate acid of H₂O
ii) H₂O is the conjugate base of H₃O⁺
iii) H₃O⁺ is the conjugate base of H₂O
iv) \(\mathrm{ClO}_{4}^{-}\) is the conjugate base of HClO₄
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement:
i) HClO₄ is the conjugate acid of \(\mathrm{ClO}_{4}^{-}\)
iii) H₃O⁺ is the conjugate acid of H₂O

Question 2.
i) All metal ions are Lewis acids.
ii) All metal oxides are Lewis acids.
iii) All anions are Lewis bases
iv) Molecules that contain polar double bond
are Lewis bases
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (ìv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statement:
ii) All metal oxides are Lewis bases.
iv) Molecules that contain polar double bond are Lewis acids.

Question 3.
i) The conjugate base of OH^– is O^2-
ii) The conjugate acid of \(\mathbf{N H}_{2}^{-}\) is NH₃
iii) The conjugate acid of H₂SO₄ is \(\mathbf{H S O}_{4}^{-}\)
iv) The conjugate base of H₂O is H₃O⁺
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement:
iii) The conjugate acid of H₂SO₄ is \(\mathbf{H S O}_{4}^{-}\)
iv) The conjugate base of H₂O is H₃O⁺

Question 4.
i) If the hydrogen ion concentration of a solution is 10^-9 M, it is basic
ii) 1f the pOH of a solution is 14, it is strongly basic.
iii)As pH of a solution increases, its pOH also increases.
iv) At 25°C the sum of pH and pOH of a solution
is 14.
a) (i) & (ii) b) (ii) & (iii) c) (iii) & (iv)
d) (j) & (iv) Ans: b) (ii) & (iii)
Correct statement:
ii) If the pOH of a solution is 14 it is strongly acidic. (or) 1f the pH of a solution is 14, it is strongly basic
iii) As pH of a solution increases, its pOH decreases. (or) As pH of a solution decreases, its pOH increases.

IV. Assertion and reason

Question 1.
Assertion (A): Ionic product of water K_w increases with increase in temperature
Reason (R): Dissociation of water is an exothermic reaction.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong.
Correct R: Dissociation of water is an endothermic reaction.

Question 2.
Assertion (A): Carbanion is a Lewis base
Reason (R): Carbanion can donate a pair of electron.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A& R are correct, R explains A Correct statement:

Question 3.
Assertion (A): An aqueous solution of IICI is acidic
Reason (R): An aqueous solution of HCl
contains less H30* than OH- ions.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong.
Correct R: An aqueous solution of HCl contains more H₃O⁺ than 0H ions.

Question 4.
Assertion (A): When sodium acetate is added to acetic acid, the dissociation of acetic acid increases.
Reason (R): The addition of CH₃COO^– ion shifts the equilibrium of dissociation of acetic acid to the left.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
d) A is wrong, but R is correct.
Correct A: When sodium acetate is added to acetic acid, the dissociation of acetic acid decreases.

V. Match the following:

Question 1.

Answer:
i) c
ii) d
iii) a
iv) b

Question 2.

Answer:
i) c
ii) d
iii) b
iv) a

VII. Two Mark Questions

Question 1.
What are Arrhenius acids and base? Give examples.
Answer:

Arrhenius acids	Arrhenius bases
1. Hydrogen ion donor in water	Hydroxyl ion donor in water
Ex. HCl, H2SO4	Ex. NaOH, Ca(OH)2

Question 2.
What are the limitations of Arrhenius concept of acids and bases?
Answer:

• Does not explain the behaviour of acids and bases in non-aqueous solvents such as acetone, THF etc.,
• Does not account for the basic nature of the substances like ammonia which do not contain a hydroxyl group.

Question 3.
How does BF₃ act as Lewis acid?
Answer:

Boron has a vacant 2p-orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond. Hence BF₃ acts as a Lewis acid.

Question 4.
Arrange HCl, HCOOH and CH3COOH in their increasing order of acid strength if their K_a values at 25°C are 2 x 10⁶, 1.8 x 10^-4 and 1.8 x 10^-5 respectively.
Answer:

• As K_a value increases, the acid strength increases.
• Increasing order of Ka value is 1.8 x 10^-5 < 1.8 x 10^-4 < 2 x 10⁶
• The increasing order of acid strength is CH₃COOH < HCOOH < HCl.

Question 5.
What is neutralization?
Answer:
The reaction in which an acid reacts with a base to form salt and water is called neutralization.

Question 6.
What do yo mean by salt hydrolysis?
Answer:
Salt hydrolysis is the reaction of the cation or the anion or both the ions of the salt with water to produce either acidic or basic solution.

Question 7.
What is Buffer index (β) ?
Answer:
It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pFl by unity.
\(\beta=\frac{\mathrm{dB}}{\mathrm{d}(\mathrm{pH})}\)
dB = number of gram equivalents of acid/base added to one litre of buffer solution. d(pH) = The change in the pH after the addition of acid / base.

Question 8.
How solubility product is determined by molar solubility?
Answer:
Solubility product can be calculated from the molar solubility i.e,, the maximum number of moles of solute that can be dissolved in one litre of the solution.

Question 9.
Define pOH
Answer:
pOH of a solution is defined as the negative logarithm of base 10 of the molar concentration of hydroxyl ions present in the solution. pOH = -log₁₀[OH^–]

Question 10.
Write the pH value of the following substances.
(a) Vinegar
(b) Black coffee
(c) Baking soda
(d) Soapy water
Answer:

Substance	pH
Vinegar	2
Black coffee	5
Baking soda	9
soapy water	12.

VII.Three Mark Questions

Question 1.
Derive the relation between pH and pOH
Answer:

Question 2.
What are buffer solutions? Explain their types with examples.
Answer:
Solutions which resist drastic changes in its pH upon the addition of small amount of acids or bases are called buffer solutions.
A buffer solution consists of a mixture of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt)

Types:
Acidic buffer: Solution of acetic acid and sodium, acetate
Basic buffer: Solution of ammonium hydroxide and ammonium chloride.

Question 3.
What are conjugate acid-base pairs? Give example.
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
The species that remains after the donation of
a proton is a base (Base}) and is called the conjugate base of the Bronsted acid (Acid}). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs.
Conjugate acid – base pair

VIII. Five Mark Questions.

Question 1.
Write a note on Lewis’s concepts of acids and bases ?
Answer:

Question 2.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of weak acid and strong base. Answer:
Consider the reaction between the weak acid CH₃COOH and the strong base NaOH to form the salt CH₃COONa and water.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aqq}}+\mathrm{NaOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{I})}\)

In aqueous solution CH3COONa is completely dissociated as follows,
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}+\mathrm{Na}_{(l)}^{+}\)
CH₃ COO^– is the strong conjugate base of the weak acid CH₃COOH
\(\mathrm{CH}_{3} \mathrm{COO}_{\text {(aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_{\text {(ач) }}^{-}\)
No such tendency is shown by Na+, the weak conjugate acid of the strong base NaOH.
In the above reaction OH^– ion is formed.
[OH^–] > [H⁺] and the solution is basic and the pH is greater than 7.
Equilibrium constant (hydrolysis constant) for the above

Dissociation of the weak acid is CH3COOH(aq) \(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}{ }_{(\mathrm{aq})}+\mathrm{H}^{+}{ }_{(a q)}\)
Dissociation constant of the weak acid

Question 3.
Explain buffer action of acidic buffer?
Answer:

• The chemical reaction which is responsible to resist the pH change of a buffer solution is called its buffer action.
• To resist changes in its pH on the addition of an acid or base, the buffer solution should contain both acidic as well as basic components.
• These components neutralise the effect of added acid or base.
• At the same time, these components should not consume each other.
• Consider an acidic buffer containing CH₃COOH and CH₃COONa.

When an acid is added to this buffer, the added H⁺ ions are consumed by the conjugate base CH₃COO^– to form undissociated CH₃COOH. This keeps the H+ concentration and the pH unchanged.
\(\mathrm{CH}_{3} \mathrm{COO}_{(a \mathrm{q})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{ay}}\)

When a base is added to this buffer, the added OH- ions are consumed by H30+. To maintain the equilibrium more CH3COOH is dissociated keeping the H+ Concentration and the pH unchanged

Question 4.
Derive Henderson-Hasselbalch equation.
Answer:
Consider an acidic buffer

• The weak acid is dissociated only to a small extent.
• Due to the common ion effect, the dissociation is further suppressed.
• Hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid.
• Similarly, the equilibrium concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

Equations (4) & (5) are known as
Henderson-Hasselbaich equations.

Question 5.
Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of a weak acid and weak base.
Answer:

• Consider the reaction between the weak acid CH₃COOH and the weak base NH₄OH to form the salt CH₃COONH₄ and water.
  
• \(\mathrm{NH}_{4}^{+}\) is the strong conjugate acid of the weak base NH₄OH.
• CH₃COO^– is the strong conjugate base of the weak acid CH₃COOH.
• Hence both \(\mathrm{NH}_{4}^{+}\) and CH₃COO^– have the tendency to react with water.
• The nature of the solution depends on the strength of acid or base.
• if K_a >K_b then the solution is acidic and pH > 7.
• if K_a < K_b then the solution is basic and pH > 7.1
  
• if K_a = K_b then the solution is neutral and pH = 7.

Consider the hydrolysis of both \(\mathrm{NH}_{4}^{+}\) and CH₃COO^–

Dissociation of a weak acid

Dissociation of weak base

Question 6.
0.1 M Solution of HF is a weak acid. But 5M solution of HF is a stronger acid. Why?
Answer:

• Hydrofluoric acid is a much stronger acid when it is concentrated than when it is diluted. As the concentration of hydrofluoric acid approaches 100 percent.
• It’s acidity increases because of home association, where a base and conjugate acid form a bond.
• 3HF ⇌ H₂F⁺ + HF₂^–
• The FHF-bifluoride anion is stabilized by a strong hydrogen bond between hydrogen and fluorine.
• The stated ionization constant of hydrofluoric acid, 10 – 3.15, does not reflect the true acidity of concentrated HF solutions.
• Hydrogen bonding also accounts for the higher boiling point of HF compared to other hydrogen halides.

IX. Additional problems:

Problems based on pH.

Question 1.
What is the pH of 0.001M NaOH?
Solution:
[OH^–] = Normality = Molarity x acidity
= 0.001 x 1 = 1 x 10^-3
pOH = -log 10 [OH^–] = -log 1 x 10^-3
= -log 1 – log 10^-3 [log 1 = 0]
= 0 + 3log10 [log10 = 1]
pOH = 3
pH + pOH = 14
pH = 14 – pOH
= 14-3
pH = 11

Question 2.
What is the pH of a solution with hydronium ion concentration 6.2 x 10^-9 M.
Solution:
[H₃O⁺] = 6.2 x 10^-9
pH = -log10[H₃O⁺]
= – log 6.2 x 10^-9
= – log 6.2 – log 10^-9
= – log 6.2 + 9 log 10
= 9 – log 6.2
= 9 -0 .7924
pH = 8.2076

Question 3.
What is the pH of 10^-2 M Ca(OH)₂?
Solution:
[OH^–] = Normality = Molarity x acidity
= 10^-2 x 2
[OH^–] = 2 x 10^-2
pOH = -log10[OH^– ]
= -log 2 x 10^-2
= -log 2 – log 10^-2
= -log 2 + 21ogl0
= 2 – log 2
= 2 – 0.3010
pOH = 1.6990
pH+pOH = 14
pH = 14 -pOH
= 14 -1.6990
pH = 12.3010

Question 4.
Calculate the hydroxyl ion concentration of a solution with pH = 5
Solution:
[H₃O⁺] = 10^pH
[H₃O⁺] = 10^-5
[H₃O⁺][OH^–] = 10^-14
[OH^–] = \(\frac{10^{-14}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\)
= \(\frac{10^{-14}}{10^{-5}}\)
[OH^–] = 10^-9

Question 5.
Calculate the pH of 10^-8M HNO₃
[H₃O⁺] = 10-8 which is less than 10 7
[H₃O⁺] from water should be added.
[H₃O⁺] = 10 “(from water) + 10-8(from HN03)
[H₃O⁺] = 10-7 (1+10-1)
= 10-7 (1+0.1)
[H₃O⁺] = 1.1 x 10^-7
pH = – log10 [H₃O⁺]
= – log 1.1 x 10^-7
= – log 1.1 – log 10^-7
= – log 1.1 + 7 log 10
= 7 – log 1.1
= 7 – 0.0414
pH = 6.9586

Question 6.
The dissociation constant of 0.1 M weak acid is 1 x 10^-5. Calculate its pH.
Answer:

Question 7.
If a solution with pH = 5 is diluted 100 times. What will be the pH of the resulting solution?
Solution:
pH = 5; [H₃O⁺] = 10-pH
[H₃O⁺] = 10^-5
After dilution
[H₃O⁺] = \(\frac{10^{-5}}{100}=10^{-7} \mathrm{M}\)
As [H₃O⁺] = 10^-7 M, the [H₃O⁺] contributed
by water should be added.
Final [H₃O⁺] = 10^-7 (from water) + 10^-7 (from the solution
= 10^-7(1+1)
[H₃O⁺] = 2 x 10^-7M
= – log [H₃O⁺]
= – log 2 x 10^-7
= – log 2 – log 10^-7
= – log 2 + 7 log 10
= 7 – log 2
= 7 – 0.3010
= 6.6990

Question 8.
Calculate the hydrogen ion concentration in human blood. (pH of human blood is 7.4)
Solution:
pH = 7.4
pH = – logio [H₃O⁺]
log [H₃O⁺] = -pH
[H30+] = Antilog of (-pH)
= Antilog of (-7.4)
= Antilog of (-7-0.4)
= Antilog of (-7-0.4+1-1)
= Antilog of (-8+0.6)
= Antilog of (8.6)
[H₃O⁺] = 3.98 x 10^-8

Question 9.
Calculate the pH of a solution obtained by mixing 50ml of 0.4 N HC1 and 50ml of 0.2N NaOH.
Solution:
Milli equivalent of 50 ml, 0.4N HCl
= V_m1 x N = 50 x 0.4 = 20
Milli equivalent of 50 ml, 0.2N NaOH
= V_m1 x N = 50 x 0.2 = 10
Remaining milli equivalent of Hcl = 20-10 =10

Question 10.
Calculate the pH of 10^-7M HCl.
Answer:
If we do not consider [H₃O⁺] from the ionisation of H₂O, then [H₃O⁺] = [HCl]= 10^-7M)
In this case the concentration of the acid is very low (10^-7M). Hence, the [H₃O⁺] (10^-7M) formed due to the auto ionisation of water cannot be neglected.
[H₃O⁺] = 10^-7 (from HCl) + 10^-7 (from H₂O)
= 10^-7 (1+1) = 2 x 10^-7
pH = -log₁₀ [H₃O⁺]
= log₁₀[2 x 10^-7] = – log 2 + log 10^-7
= -log 2 – (-7) log10¹⁰
= 7 – log2 .
= 7 – 0.3010=6.6990
= 6.70

Problems based on Henderson – Hasselbaich equations

Question 1.
A buffer solution contains equal volumes of 0.2 M NH₄OH and 0.02 M NH₄Cl. Kb of NH₄OH is 1 x 10^-5. Calculate the pH of the buffer.
Solution:
[Base] = 0.2 M; [salt] = 0.02 M
K_b = 1 x 10^-5
pK_b = -log K_b = – log 1 x 10^-5 = 5
For a basic buffer
\(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\text { Base }]}\)
= 5+log 0.02/0.2
= 5+1og10^-1
= 5 – 1
pOH = 4
pH + pOH = 14
pH = 14 – pOH
= 14 -4
pH = 10

Question 2.
Find the pH of a buffer solution containing 0.20 mole per litre CH₃COONa and 0.15 mole per litre CH₃COOH, K_a for acetic acid is 1.8 x 10^-5
Solution:
[acid] = 0.15 M; [salt] = 0.20 M
K_a = 1.8 x 10^-5
PK_a = -log Ka = – log L8 x 10^-5
= – log 1.8 – log10^-5
= – log 1.8 + 5 log 10
= 5 – log 1.8
= 5 – 0.2553 pKa
PK_a = 4.7447
For an acidic buffer
\(\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)
= 4.7447 + log 0.20/0.15
= 4.7447 + log 4/3
= 4.7447 + log 4 – log 3
= 4.7447+0.6021-0.4771
pH = 4.8697

Question 3.
Calculate the pKb of NH₄OH, if the pH of a buffer solution containing 0.1N NH₄OH and 0.1 N NH₄Cl is 9.25.
Solution;)
pH of basic buffer = 9.25
∴ pOH = 14 – pH
= 14 – 9.25
pOH = 4.75
[Base] = 0.1N ; [Salt] = 0.1 N
\(\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)
\(\mathrm{pK}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\text { Salt }]}{[\text { Base }]}\)
PK_b = 4.75 – log M = 4.75 – log 0.1/0.1 = 4.75 – 0 = 4.75

Question 4.
Calculate the pH of the solution by mixing 1.5 mole of HCN and 0.15 mole of KCN in water and making up the total volume to 0.5 litre.
Solution:
K_a = 5 x 10^-10;
fAcid] = 1.5/0 5 – 3.0
[Salt] = 0.15/0.5 = 0.3
pK_a = – log K_a
= – log 5 x 10^-10
= – log 5 – log 10^-10
= – log 5 + 10 log^-10
= 10 – log 5
= 10 – 0.6990
pK^-10 = 9.3010
\(\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\)
= 9.3010 + log 0.3/3.0
= 9.3010 + log 10^-10
= 9.3010 -1 log 10
pH = 8.3010

Problems based on Ostwald’s dilution law

Question 1.
Calculate the degree of dissociation of 0.1 N CH₃COOH if Ka for CH₃COOH is 1 x 10-5.
Solution:

Question 2.
What is the dissociation constant of 0.1 M HCN which is 0.01% ionised?
Solution:
C = 0.1M ; α = 0.01% = \(\frac{0.01}{100}\) = 1 x 10 ^-4
Ka = ?
Ka = C α² = 0.1 x (1 x 10^-4)² = 1 x 10^-9

Question 3.
A weak monobasic acid is 1% ionised in 0.1M solution at 25°C. What is the percentage of ionisation in its 0.025M Solution?
Solution:
C =0.1 M; α = 1% = 1/100 = 10^-2 Ka = ?
C =0.025 M % a = ?
Ka = C α² = 0.1 x (10^-2)² = 10^-5
\(\alpha=\sqrt{\frac{K a}{C}}=\sqrt{\frac{10^{-5}}{0.025}}=0.02\)
% α =100 x α = 100 x 0.02 = 2%

Question 4.
A mono basic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionisation in this solution?
Solution:
C = 0.005 M
pH = 5
∴ [H₃O⁺] = 10^-ph = 10^-5
HA → H⁺ + A^–
ka = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
[H⁺] = [A^–] = 10^-5
[HA] = 0.005M

α = 2 x 10^-3
% α = 100 x α
= 100 x 2 x 10^-3 = 0.2%

Problems based on Solubility Product

Question 1.
Calculate the solubility product of AgCl if the solubility of AgCl is 1.3 x 10^-5 moles per litre at 25°C.
Solution:
S = 1.3 x 10^-5 M ; Ksp = ?

Question 2.
Calculate the solubility of Calcium fluoride in a saturated solution if its solubility product is 3.2 x 10^-11 M³.
Solution:
Ksp = 3.2 x 10^-11 M³; S = ?

Question 3.
The solubility of AgCl is 0.0014 g per litre at 18°C. Calculate its solubility product at 18°C. Molecular weight of AgCl is 143.5. Solubility in
Answer:

Ksp = [Ag⁺][Cl^–]
Ksp = S x S
Ksp = S²
Ksp = (9.756 x 10^-6M)²
Ksp = 95.1795 x 10^-12
Ksp = 9.518 x 10^-11 M²

Question 4.
The solubility product of a salt having general formula MX₂ in water is 4 x 10^-12. What is the concentration of M²⁺ ions in the aqueous solution of the salt?
Solution:

Question 5.
Solid Ba (N0₃)₂ is gradually dissolved in 1 x 10^-4 m Na₂CO₃ solution. At what concentration of Ba²⁺ will a precipitate begin to form?
(Ksp for BaCO₃ = 5.1 x 10^-9 )
Solution:

Ionic Product > Ksp, Precipitation will occur.
∴ After attaining 5.1 x 10^-5 M [Ba²⁺] concentration, a precipitate will begin to form.

Students can download 5th Maths Term 1 Chapter 2 Numbers InText Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 1 Chapter 2 Numbers InText Questions

Rounding to the Nearest Thousandth Calculator is an online tool that rounds off a decimal number to its nearest thousandths place.

Activity (Text Book page No.26)

Answer:

Activity (Text Book page No.27)

Answer:

Crore in many ways

Answer:

Fill in the correct numbers in the following tables.

Answer:

Activity: 1 (Text Book page No.28)

The Abacus shows the number 34,284

Number name: ________
Expanded form: 3 Ten thousands + 4 thousands +2 hundreds + 8 tens + 4 ones.
= 30,000 + _____ + 200 + _____ + _____
3 × 10,000 + 4 × ____ + 2 × 100 + 8 x ____ + ____ × 1
Answer:
Number name: Thirty four thousand two hundred and eighty four.
Expanded form: 3 Ten thousands + 4 thousands +2 hundreds + 8 tens + 4 ones.
= 30,000 + 4000 + 200 + 80 + 4
= 3 × 10,000 + 4 × 000 + 2 × 100 + 8 x 10 + 4 × 1

Try These (Text Book page No.29)

How many thousands are there in 3,45,789?
Answer:
345 thousands

Activity: 2 (Text Book page No.29)

Look at the Abacus and fill in the blanks.

Number: ________
Number name: _______
Expanded form: 5 ten lakhs + ______ lakhs + ______ ten thousands + 1 _______ + 2 ________ + 5 tens + ______ ones
= 5000000 + ____ + 40000 + ______ + 200 + 50 + 8
Answer:

Number: 52,41,258
Number Name: Fifty two lakhs forty one thousands two hundred and fifty eight.
Expanded form:
= 5 ten lakhs + 2 lakhs + 4 ten thousands + 1 x thousand + 2 hundreds + 5 tens + 8 ones
= 5000000 + 200000 + 40000 + IQOO + 200 + 50 + 8.

Activity: 3 (Text Book page No.30)

Look at the Abacus and fill in the blanks.

Given number: _________
Number Name: _______________
Expanded form: 6 crores + _____ Ten Iakhs + _____ Lakhs + 3 _______+ 5 Thousands + ______ + 1 Ten + 5 ones.
= 6,00,00,000 + 40,00,000 + ________ + ________ + 5000 + 400 + _______ + 5
= 6 × _____ + 4 × _____ + 2 × 100,000 + 3 × 10000 + 5 × _______ + _______ × 100 + 1 × 10 + _______ × 1
Answer:

Given Number: 6,42,35,415
Number Name: Six crores forty two lakhs thirty five thousands four hundred and fifteen
Expanded form: 6 Crores + 4 ten lakhs + 2 lakhs + 3 ten thousands + 5 Thousands + 4 hundreds + 1 ten + 5 ones.
= 6,00,00,000 + 40,00,000 + 2,00,000 + 30,000 + 5,000 + 400 + 10+5.
= 6 × 1,00,00,000 + 4 × 10,00,000 + 2 × 1,00,000 + 3 × 10,000 + 5 × 10,000 + 4 × 100 + 1 × 10 + 5 × 1.

Activity (Text Book page No.30)

Write the place value of 7 and 1 for the given numbers.
a. 81,70,453
b. 3,46,710
c. 1,87,13,971
Answer:
a. 81,70,453
The place value of 7 is 7 × 10,000 = 70,000
The place value of 1 is 1 × 1,00,000 = 1,00,000

b. 3,46,710
The place value of 1 is 1 × 10 = 10
The place value of 7 is 7 × 100 = 700

c. 1,87,13,971
The place value of 1 is 1 × 1 = 1
The place value of 7 is 7 × 10 = 70
The place value of 1 is 1 × 10,000 = 10,000
The place value of 7 is 7 × 1,00,000 = 7,00,000
The place value of 1 is 1 × 1,00,00,000 = 1,00,00,000

Try These (Text Book page No.34)

1. 3,002 ____ 8,002
2. 43,731 ____ 44,371
3. 43,115 ____ 43,511
4. 13,435 ____ 13,4753
Answer:
1. 3,002 ____ 8,002
= 3,002 < 8,002

2. 43,731 ____ 44,371
= 43,731 < 44,371

3. 43,115 ____ 43,511
= 43,115 < 43,511

4. 13,435 ____ 13,4753
= 13,435 < 13,4753

Activity (Text Book page No.34)

1. Form the smallest and greatest five digit numbers using the given digits once.
(a) 7, 1, 0, 5, 4
(b) 3, 4, 7, 0, 9
(c) 9, 7, 1, 6, 4
(d) 4, 5, 9, 6, 7
Answer:
(a) 7, 1, 0, 5, 4
Smallest number = 10,457
Greatest number = 75,410

(b) 3, 4, 7, 0, 9
Smallest number = 30,479
Greatest number = 97,430

(c) 9, 7, 1, 6, 4
Smallest number = 14,679
Greatest number = 97,641

(d) 4, 5, 9, 6, 7
Smallest number = 45,679
Greatest number = 97,654

2. Write the smallest numbers in the fruit and the greatest number in the flower.

(a) 45678, 145, 7829
(b) 23, 8873, 88738, 883
Answer:
(a) 45678, 145, 7829

(b) 23, 8873, 88738, 883

Try These (Text Book page No.36)

1. Arrange the following numbers in the ascending order and descending order.
(i) 33,270; 1,078; 137; 27,935
(ii) 44,918; 32,113; 23,112; 42,231
(iii) 75,343; 30,475; 43,452; 13,055
(iv) 733; 34,946; 35,945; 23,745
Answer:
(i) 33,270; 1,078; 137; 27,935
Ascending order : 137,1,078, 27,935, 33,270
Descending order 5 33,270, 27,935, 1,078, 137

(ii) 44,918; 32,113; 23,112; 42,231
Ascending order : 23,112, 32,113, 42,231, 44,918
Descending order : 44,918, 42,231, 32,113, 23,112

(iii) 75,343; 30,475; 43,452; 13,055
Ascending order : 13,055, 30,475, 42,452, 75,343
Descending order ; 75,343, 42,452, 30,475,13,055

(iv) 733; 34,946; 35,945; 23,745
Ascending order : 733, 23,745, 34,946, 35,945
Descending order ; 35,945, 34,946, 23,745, 733

Tamilnadu State Board New Syllabus Samacheer Kalvi 4th Maths Guide Pdf Term 3 Chapter 6 Fraction InText Questions Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 4th Maths Solutions Term 3 Chapter 6 Fraction InText Questions

In simple terms, this fraction and whole number calculator allows you to solve fraction problems with whole numbers and fractions form.

Activity (Text Book Page No. 51)

(The four equal parts of a circle)
A Circle which is divided into four equal parts are given to Arun, Tharun, sankar and Gopi, to colour it.
They could complete the task of colouring in two minutes, which is given below.

In the above pictures one fourth (quarter) portion was coloured by tharun.
1/2 portion was coloured by __________
Answer:
Arun

Whole portion was coloured by __________
Answer:
Sankar

Three fourth of the portion was coloured by __________
Answer:
Gopi

Can we try this: (Text Book Page No. 55)

Question 1.
Anitha’s rectangular garden was divided into ________ parts.
Answer:
4

Question 2.
Portion of brinjal planted part = ________
Answer:
\(\frac { 1 }{ 4 }\)

Question 3.
Portion of Ladies finger planted port = ________
Answer:
\(\frac { 2 }{ 1 }\)

Question 4.
Portion of pumpkin planted part = ________
Answer:
\(\frac { 1 }{ 4 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using Factor Theorem:

Question 1.
Show that \(\left| \begin{matrix} x & a & a \\ a & x & a \\ a & a & x \end{matrix} \right| \) = (x – a)² (x + 2a)
Answer:

By putting x = a , we have three rows of |A| are identical. Therefore (x – a)² is a factor of |A|
Put x = – 2a in |A|

∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)² (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.

a³ [ – 1 (1 – 1) – 1 ( – 1 – 1) + 1 (1 + 1)] = k . 4a³
a³ [o + 2 + 2 ] = 4 ka³
4a³ = 4 ka³
k = 1

 

Free handy Remainder Theorem Calculator tool displays the remainder of a difficult polynomial expression in no time.

Question 2.
Show that \(\left| \begin{matrix} b\quad +\quad c & a\quad -\quad c & a\quad -\quad b \\ b\quad -\quad c & c\quad +\quad a & b\quad -\quad a \\ c\quad -\quad b & c\quad -\quad a & a\quad +\quad b \end{matrix} \right| \) = 8 abc
Answer:


since two columns identical
= bc × 0 = 0
∴ a – 0 is a factor. That is, a is a factor.
Put b = 0 in |A|

since two columns identical
= ca × 0 = 0
∴ b – 0 is a factor. That is, a is a factor.
Put c = 0 in |A|

since two columns identical
= ab × 0 = 0
∴ c – 0 is a factor. That is, c is a factor.
The degree of the product of the factors abc is 3.
The degree of the product of leading diagonal elements (b + c) (c + a) (a + b) is 3.
∴ The other factor is the constant factor k.

Question 3.
Solve that \(\left| \begin{matrix} x\quad +\quad a & b & c \\ a & x\quad +\quad b & c \\ a & b & x\quad +\quad c \end{matrix} \right| \) = 0
Answer:

Put x = 0

x = 0 satisfies the given equation. x = 0 is a root of the given equation, since three rows are identical. x = 0 is a root of multiplicity 2. Since the degree of the product of the leading diagonal elements (x + a) (x + b) (x + c) is 3. There is one more root for the given equation.
Put x = – (a + b + c)

∴ x = – (a + b + c) satisfies the given equation.
Hence, the required roots of the given equation are x = 0, 0 , – (a + b + c)

Question 4.
Show that \(\left| \begin{matrix} b\quad +\quad c & a & { a }^{ 2 } \\ c\quad +\quad a & b & { b }^{ 2 } \\ a\quad +\quad b & c & { c }^{ 2 } \end{matrix} \right| \) = (a + b + c) (a – b) (b – c) (c – a)
Answer:

Since two rows are idenctical
|A| = 0
since two rows are idenctical
|A| = 0
∴ a – b is a factor of | A |. The given determinant is in cyclic symmetric form in a , b and c. Therefore, b – c and c – a are also factors. The degree of the product of the factors (a – b) (b – c) (c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c² is 4.
Therefore, the other factor is k (a + b + c).

5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k
5 × 6 – 24 + 6 = 12k
30 – 24 + 6 = 12k
12 = 12 ⇒ k = 1

Question 5.
Solve \(\left| \begin{matrix} 4\quad -\quad x & 4\quad +\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad -\quad x & 4\quad +\quad x \\ 4\quad +\quad x & 4\quad +\quad x & 4\quad -\quad x \end{matrix} \right| \) = 0
Answer:

Put x = 0

∴ x = 0 satisfies the given equation. Hence x = 0 is a root of the given equation. since three rows are identical, x = 0 is a root of multiplicity 2.

Since the degree of the product of the leading diagonal elements (4 – x) (4 – x) (4 – x) is 3. There is one more root for the given equation.


∴ x = – 12 is a root of the given equation.
Hence, the required roots are x = 0 , 0 , – 12

Question 6.
Show that \(\left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ { x }^{ 2 } & { y }^{ 2 } & { z }^{ 2 } \end{matrix} \right| \) = (x – y) (y – z) (z – x)
Answer:

|A| = 0 since two columns identical
∴ x – y is a factor of A. The given determinant is in the cyclic symmetric form in x, y, and z. Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y) (y – z) (z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z² is 3. Therefore, the other factor is the constant factor k.

Put x = 0, y = 1, z = -1 we get

Expanding along the first column
1 (1 + 1) = 2k
2 = 2k ⇒ k = 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Having trouble working out with the Binomial theorem? You’ve come to the right place, our Binomial Expansion Calculator is here to save the day for you.

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:

(iii) (5 + x²)^2/3
Answer:

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:

Expanding binomials calculator is a free online tool that displays the expansion of the given binomial term.

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
Answer:

(ii) e^-2x
Answer:

(iii) e^x/2
Answer:

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )

(ii) log (1 – 2x)

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)

Question 7.

Answer:

Question 8.

Answer:

Question 9.
Find the coefficient of x⁴ in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:

Question 10.
Find the value of
Answer:

Students can download Maths Chapter 7 Mensuration Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Weathering the Storm in Ersama mcq online test / important questions and answers Important MCQ Questions with Answer.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.
The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______
(1) π cm²
(2) 2π cm²
(3) 3π cm²
(4) 2 cm²
Answer:
(2) 2π cm²
Hint:
C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm² = 2π cm²

Question 2.
The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.
(1) \(\frac{3}{2} \pi h\)
(2) \(\frac{2}{3} \pi h^{2}\)
(3) \(\frac{3}{2} \pi h^{2}\)
(4) \(\frac{2}{3} \pi h\)
Answer:
(3) \(\frac{3}{2} \pi h^{2}\)
Hint:
T.S.A = 2πr(h + r)
[radius is half of the height]
= \(2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)\)
= \(=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}\) sq. units

Question 3.
Base area of a right circular cylinder is 80 cm². If its height is 5 cm, then the volume is equal to _______
(1) 400 cm³
(2) 16 cm³
(3) 200 cm³
(4) \(\frac{400}{3}\) cm³
Answer:
(1) 400 cm³
Hint:
Volume of a cylinder = πr²h cu. units
[Base area (πr²) = 80 cm² = 80 × 5 cm³ = 400 cm³

Question 4.
If the total surface area of a solid right circular cylinder is 200π cm² and its radius is 5 cm, then the sum of its height and radius is ______
(1) 20 cm
(2) 25 cm
(3) 30 cm
(4) 15 cm
Answer:
(1) 20 cm
Hint:
T.S.A of a cylinder = 200π cm²
2πr (h + r) = 200π
2 × 5 (h + r) = 200
(h + r) = 20 cm

Question 5.
The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______
(1) πa²b sq.cm
(2) 2πab sq.cm
(3) 2π sq.cm
(4) 2 sq.cm
Answer:
(2) 2πab sq.cm .
Hint:
C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.
Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm³, then the volume of the cone is equal to _______
(1) 1200 cm³
(2) 360 cm³
(3) 40 cm³
(4) 90 cm³
Answer:
(3) 40 cm³
Hint:
Volume of the cone = \(\frac{1}{3}\) × volume of the cylinder
= \(\frac{1}{3}\) × 120 cm³
= 40 cm³

Question 7.
If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 96 cm
Answer:
(1) 10 cm
Hint:
Slant height of a cone

Question 8.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______
(1) 1200π cm²
(2) 600π cm²
(3) 300π cm²
(4) 600 cm²
Answer:
(2) 600π cm²
Hint:
Circumference (2πr) = 120π cm
Slant height (l) = 10 cm;
Curved surface area of a cone = πrl sq. units
= \(\frac{120 \pi}{2}\) × 10 cm² = 600π cm²

Question 9.
If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
Answer:
(4) 12 cm
Hint:
Volume of a cone = 48π cm³
[Base area (πr²) = 12π]
\(\frac{1}{3}\) πr²h = 48π
\(\frac{1}{3}\) × 12π × h = 48π
[Substitute πr² = 12π]
h = \(\frac{48}{4}\) = 12 cm

Question 10.
If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______
(1) 240 cm³
(2) 120 cm³
(3) 80 cm³
(4) 480 cm³
Answer:
(3) 80 cm³
Hint:
Volume of a cone (V) = \(\frac{1}{3}\) πr²h sq. units
Base area (πr²) = 48 sq. cm
V = \(\frac{1}{3}\) × 48 × 5 = 80 cm³

Question 11.
The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______
(1) 4 : 1
(2) 1 : 4
(3) 2 : 1
(4) 1 : 2
Answer:
(3) 2 : 1
Hint:
h₁ : h₂ = 1 : 2
r₁ : r₂ = 2 : 1
Ratio of their volumes
= \(\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}\)
= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.
If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________
(1) 8π cm²
(2) 16 cm²
(3) 12π cm²
(4) 16π cm²
Answer:
(4) 16π cm²
Hint:
C.S.A of a sphere = 4πr² sq. units
[radius = 2 cm]
= 4 × π × 22 cm²
= 16π cm²

Question 13.
The total surface area of a solid hemisphere of diameter 2 cm is equal to _______
(1) 12 cm²
(2) 12π cm²
(3) 4π cm²
(4) 3π cm²
Answer:
(4) 3π cm²
Hint:
Radius of a hemisphere = \(\frac{2}{2}\) = 1 cm
Total surface area of a hemisphere = 3πr² sq. units = 3 × π × 12 cm² = 3π cm²

Question 14.
If the volume of a sphere is \(\frac{9}{16} \pi\) cu.cm, then its radius is ________
(1) \(\frac{4}{3}\) cm
(2) \(\frac{3}{4}\) cm
(3) \(\frac{3}{2}\) cm
(4) \(\frac{2}{3}\) cm
Answer:
(2) \(\frac{3}{4}\) cm
Hint:
Volume of the sphere = \(\frac{9}{16} \pi\)

Question 15.
The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______
(1) 81 : 625
(2) 729 : 15625
(3) 27 : 75
(4) 27 : 125
Answer:
(4) 27 : 125
Hint:
Ratio of their surface area = 9 : 25

Question 16.
The total surface area of a solid hemisphere whose radius is a units, is equal to ________
(1) 2πa² sq. units
(2) 3πa² sq. units
(3) 3πa sq. units
(4) 3a² sq. units
Answer:
(2) 3πa² sq. units
Hint:
T.S.A. of a solid hemisphere = 3πr² sq. units
= 3 × π × a × a sq.units
= 3πa² sq. units

Question 17.
If the surface area of a sphere is 100π cm², then its radius is equal to ______
(1) 25 cm
(2) 100 cm
(3) 5 cm
(4) 10 cm
Answer:
(3) 5 cm
Hint:
Surface area of a sphere = 100π cm²
4πr² = 100π
r² = 25
r = √25 = 5 cm

Question 18.
If the surface area of a sphere is 36π cm², then the volume of the sphere is equal to _______
(1) 12π cm³
(2) 36π cm³
(3) 72π cm³
(4) 108π cm³
Answer:
(2) 36π cm³
Hint:
Surface area of a sphere = 36π cm²
4πr² = 36π
r² = 9
r = 3 cm
Volume of a sphere = \(\frac{4}{3} \pi r^{3}\) cu. units
= \(\frac{4}{3} \pi\) × 3 × 3 × 3 cm³ = 36π cm³

Question 19.
If the total surface area of a solid hemisphere is 12π cm² then its curved surface area is equal to ______
(1) 6π cm²
(2) 24π cm²
(3) 36π cm²
(4) 8π cm²
Answer:
(4) 8π cm²
Hint:
T.S.A of a hemisphere = 12π cm²
3πr² = 12π
r² = 4
r = 2
Curved surface area of a hemisphere = 2πr² = 2 × π × 4 = 8π cm²

Question 20.
If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____
(1) 1 : 8
(2) 2 : 1
(3) 1 : 2
(4) 8 : 1
Answer:
(1) 1 : 8
Hint:
\(r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2\)

II. Answer the following questions:

Question 1.
Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.
Answer:
Given, Circumference of the base of a cylinder = 110 cm
2πr = 110 ……. (1)
Curved surface area = 4400 cm²
2πrh = 4400 cm² ……. (2)
From (1) & (2), \(\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}\)
Height of the cylinder (h) = 40 cm
From (1), 2πr = 110
2 × \(\frac{22}{7}\) × r = 110
r = \(\frac{35}{2}\)
We know that, diameter (d) = 2 × radius
d = 2 × \(\frac{35}{2}\) = 35 cm
Diameter of the Circular cylinder = 35 cm

Question 2.
A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.
Answer:
Given, Radius of a cylinder (r) = 50 cm = 0.5 m
Height of a cylinder (h) = 3.5 m
Curved surface area of a pillar = 2πrh sq. units
Curved surface area of 12 pillars = 12 × 2πrh
= 12 × 2 × \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 132 sq. m.
Cost for painting the lateral surface of pillars per metre = ₹ 20
Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.
The total surface area of a solid right circular cylinder is 231 cm². Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.
Answer:
Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm
Curved surface area = \(\frac{2}{3}\) × T.S.A = \(\frac{2}{3}\) × 231 = 154 cm²
2πrh = 154 cm² …… (1)
Total surface area = 231 cm²
2πr (h + r) = 231
2πrh + 2πr²= 231
154 + 2πr² = 231 [from (1)]
2πr² = 231 – 154 = 77

Radius of the cylinder = 3.5 cm
Height of the cylinder = 7 cm

Question 4.
The total surface area of a solid right circular cylinder is 1540 cm². If the height is four times the radius of the base, then find the height of the cylinder.
Answer:
Given, Let the radius of the cylinder be ‘r’
Height of a cylinder (h) = 4r (by given condition)
Total surface area = 1540 cm²
2πr(h + r) = 1540 cm²

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.
If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.
Answer:
Given, In the figure, OAB is a cone and OC ⊥ AB
∠AOC = \(\frac{60^{\circ}}{2}\) = 30°
In the right ∆OAC, tan 30° = \(\frac{\mathrm{AC}}{\mathrm{OC}}\)

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.
The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.
Answer:

Given, Radius of a sector (r) = 21 cm
The angle of the sector (θ) = 180°
Let “R” be the radius of the cone.
Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.
If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.
Answer:
Given, the Curved surface area of a solid hemisphere = 2772 cm²
2πr² = 2772

Total surface area = 3πr² sq. units
= 3 × \(\frac{22}{7}\) × 21 × 21
= 4158 sq.cm
Aliter:
C.S.A of a hemisphere = 2772 cm²
2πr² = 2772 cm²
πr² = \(\frac{2772}{2}\) = 1386 cm
T.S.A of a hemisphere = 3πr² sq.units = 3 × 1386 cm² = 4158 cm²

Question 8.
An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.
Answer:
Given, Circumference of the dome = 17.6 m
2πr = 17.6
\(r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}\)
Curved surface area of the dome = 2πr² sq. units
= 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m²
= 49.28 m²
Cost of painting for one sq.metre = ₹ 5
Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.
Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.
Answer:
Given, Height of a cylinder (h) = 4.5 cm
Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.
A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.
Answer:
After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm³

Question 11.
The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Answer:
Given, Height of the wooden solid cone (h) = 12 m
Circumference of the base = 44 m
2πr = 44
r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Volume of the wooden solid = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}\)
= 88 × 7
= 616 m³
Volume of the solid = 616 m³

Question 12.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.
Answer:
Given, Edge of the cube = 14 cm
The largest circular cone is cut out from the cube.
Radius of the cone (r) = \(\frac{14}{2}\) = 7 cm
Height of the cone (h) = 14 cm
Volume of a cone

Volume of a cone = 718.67 cm³

Question 13.
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = \(\frac{22}{7}\))
Answer:
Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm
R = r + w = 5 + 0.25 = 5.25 cm
Now, outer surface area of the bowl = 2πR²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25
= 173.25 sq. cm
Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.
Volume of a hollow sphere is \(\frac{11352}{7}\) cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = \(\frac{22}{7}\))
Answer:
Let R and r be the outer and inner radii of the hollow sphere respectively.
Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.
How many litres of water will a hemispherical tank whose diameter is 4.2 m?
Answer:
Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m
Volume of the hemisphere
= \(\frac{2}{3} \pi r^{3}\) cu.units
= \(\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}\)
= 19.404 m³
= 19.404 x 1000 lit
= 19,404 litres

III. Answer the following questions.

Question 1.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer:

For cylindrical part:
Radius (r) = 7 cm
Height (h) = 6 cm
Curved surface area = 2πrh = 2 × \(\frac{22}{7}\) × 7 × 6 cm² = 264 cm²
For hemispherical part:
Radius (r) = 7 cm
Surface area (h) = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7 cm²
= 308 cm²
Total surface area = (264 + 308) = 572 cm²

Question 2.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:

Question 3.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Answer:

For cylinderical part:
Height (h) = 2.4 cm
Diameter (d) = 1.4 cm
Radius (r) = 0.7 cm
Total surface area of the cylindrical part

For conical part:
Base area (r) = 0.7 cm
Height (h) = 2.4 cm

Question 4.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
Diameter of the cylindrical well = 7 m
Radius of the cylinder (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
Volume = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}\)
= 22 × 7 × 5 m³
Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having
Length (l) = 22 m
Breadth (b) = 14 m
Let ‘h’ be the height of the platform.
Volume of the platform = 22 × 14 × h m³
Volume of the platform = Volume of the earth taken out
22 × 14 × h = 22 × 7 × 5
\(h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}\)
Thus, the required height of the platform is 2.5 m.

Question 5.
The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]
Answer:

Let the radii of circular ends are R and r [R > r]
Perimeter of circular ends are 207.24 cm and 169.56 cm
2πR = 207.24 cm

The whole surface area of the frustum = π [(R² + r²) + (R + r) l]
Required whole surface area of the frustum
= 3.14 [33² + 27² + (33 + 27) × 10] cm²
= 3.14 [1089 + 729 + 600] cm²
= 3.14 [2418] cm²
= 7592.52 cm²

Question 6.
A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = \(\frac {22}{7}\))
Answer:

Let h₁ be the length of the pipe
Let R and r be the outer and inner radii of the pipe respectively.
Iron slab:
Volume = lbh = 55 × 40 × 15 cm³
Iron pipe:
Outer diameter, 2R = 8 cm
Outer radius, R = 4 cm
Thickness, w = 1 cm
Inner radius, r = R – w = 4 – 1 = 3 cm
Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.
Answer:
Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.
A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.
Answer:

Given, Total height of solid = 13.5 cm
Diameter of the cylinder (d) = 28 m
Height of a cylinder (h) = 3 m
Height of a conical portion = 13.5 – 3 = 10.5 m
From the diagram, Radius of a cone = Radius of a cylinder
Radius (r) = 14 m

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.2

Question 1.
Which of the following data can be represented in a histogram?
(i) The number of mountain climbers in the age group 20 to 60 in TamilNadu.
Answer:
Yes

(ii) Production of cycles in different years.
Answer:
No

(iii) The number of students in each class of a school.
Answer:
No

(iv) The number votes polled from 7 am to 6 pm in an election.
Answer:
Yes

(v) The wickets fallen from 1 over to 50th over in a one day cricket match.
Answer:
Yes

Question 2.
Fill in the blanks:
(i) The total area of the histogram is _________ to the total frequency of the given data.
Answer:
proportional

(ii) A graph that displays data that changes continuously over the periods of time is _________ .
Answer:
Histogram

(iii) Histogram is a graphical representation of_________ data.
Answer:
grouped

Question 3.
In a village, there are 570 people who have cell phones. An NGO survey their cell phone usage. Based on this survey a histogram is drawn. Answer the following questions.

(i) How many people use the cell phone for less than 3 hours?
Answer:
330 people (110 + 220)

(ii) How many of them use the cell phone for more than 5 hours?
Answer:
150 of them (100 + 50)

(iii) Are people using cell phone for less than 1 hour?
Answer:
No

Question 4.
Draw a histogram for the following data.

Answer:
The given (tata IS continuous frequency distribution taking class intervals on X axis and No. of students on Y-axis, the histogram is given below.

Question 5.
Construct a histogram from the following distribution of total marks of 40 students in a class.

Answer:
The given distribution is continuous taking marks on X axis and No. of students on Y-axis the histogram is constructed.

Question 6.
The distribution of heights (in cm ) of loo people is given below. Construct a histogram and the frequency polygon imposed on it.

Answer:
The given distribution is discontinuous.
Converting into continuous distribution we have
Lower boundary = lower limit – \(\frac { 1 }{ 2 }\) (gap between adjacent class interval)
= 125 – \(\frac { 1 }{ 2 }\) (1) = 124.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 }\) (gap between the adjacent class interval)
= 135 + \(\frac { 1 }{ 2 }\) = 135.5
∴ The new frequency table is

Question 7.
In a study of dental problem, the following data were obtained.

Represent the above data by a frequency polygon.
Answer:
Finding the midpoints of the class interval we get.

Ages	Mid point (x)	No. of patients
0 – 10	5	5
10 – 20	15	13
20 – 30	25	25
30 – 40	35	14
40 – 50	45	30
50 – 60	55	35
60 – 70	65	43
70 – 80	75	50

The points tobe plotted are A(5,5), B(15, 13), C(25, 25), D(35, 14), E(45, 30), G(65, 43.), H(75, 50) to obtain the frequency polygon ZABCDEFGHI.
Where I imagined class between 80 and 90.

Question 8.
The marks obtained by 50 students in Mathematics are given below (j) Make a frequency distribution table taking a class size of 10 marks (ii) Draw a histogram and a frequency polygon.

Answer:
Maximum marks obtained = 89
Minimum marks obtained = 08
Range = Maximum marks – Minimim marks
= 89 – 08
= 81
Taking the class size = 10, then

= \(\frac{81}{10}\) = 8.1 = 9

Now we have the continuous frequency table.

We will draw the histogram taking class interval in x-axis and frequency in y-axis as follows.

Objective Type Questions

Question 9.
Data is a collection of _________
(A) numbers
(B) words
(C) measurements
(D) all the three
Answer:
(D) all the three

Question 10.
The number of times an observation occurs in the given data is called _________
(A) tally marks
(B) data
(C) frequency
(D) none of these
Answer:
(C) frequency

This age difference calculator lets you quickly determine the age gap between two people.

Question 11.
The difference between the largest value and the smallest value of the given data is ________
(A) range
(B) frequency
(C) variable
(D) none of these
Answer:
(A) range

Question 12.
The data that can take values between a certain range is called_________
(A) ungrouped
(B) grouped
(C) frequency
(D) none of these
Answer:
(B) grouped

Question 13.
Inclusive series is a _________series.
(A) continuous
(B) discontinuous
(C) both
(D) none of these
Answer:
(B) discontinuous

Question 14.
In a class interval the upper limit of one class is the lower limit of the other class. This is _________ series.
(A) Inclusive
(B) exclusive
(C) ungrouped
(D) none of these
Answer:
(B) exclusive

Question 15.
The graphical representation of ungrouped data is ________
(A) histogram
(B) frequency polygon
(C) pie chart
(D) all the three
Answer:
(C) pie chart

Question 16.
Histogram is a graph of a ________ frequency distribution.
(A) continuous
(B) discontinuous
(C) discrete
(D) none of these
Answer:
(A) continuous

Question 17.
A _________ is a line graph for the graphical representation of the continuous frequency
distribution.
(A) frequency polygon
(B) histogram
(C) pie chart
(D) bar graph
Answer:
(A) frequency polygon

Question 18.
The graphical representation of grouped data is _________
(A) bar graph
(B) pictograph
(C) pie chart
(D) histogram
Answer:
(D) histogram

Students can download 5th Maths Term 2 Chapter 2 Numbers Additional Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 2 Chapter 2 Numbers Additional Questions

What is the factor of 50 between 11 and 20? The factors of 50 are: 1, 2, 5, 10, 25, 50 None of them is between 11 and 20.

Question 1.
Find the common factors for:
(i) 6 and 9
(ii) 15 and 20
Answer:
(i) 6 and 9
Factors of 6 is 1, 2, 3, 6
Factors of 9 is 1, 3, 9
Common factors of 6 and 9 is 1, 3

(ii) 15 and 20
Factors of 15 is 1, 3, 5, 15
Factors of 20 is 1, 2, 4, 5, 10, 20
Common factors of 15 and 20 are 1, 5.

Question 2.
Draw a picture of factor tree.
(i) 45
(ii) 100
Answer:
(i) 45

(ii) 100

Question 3.
Write down the first 5 multiple of the given numbers.
(i) 15
(ii) 25
Answer:
(i) 15
15 × 1 = 15
15 × 2 = 30
15 × 3 = 45
15 × 4 = 60
5 × 5 = 75
5, 30, 45, 60, 75

(ii) 25
25 × 1 = 25
25 × 2 = 50
25 × 3 = 75
25 × 4 = 100
25 × 5 = 125
25, 50, 75, 100, 125

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.1

Question 1.
Fill in the blanks:
(i) If 30 % of x is 150, then x is _______ .
Answer:
500
Hint:
Given 30% of x is 150
i.e \(\frac{30}{100}\) × x = 150

∴ x = 500

(ii) 2 minutes is _______ % to an hour.
Answer:
3\(\frac{1}{3}\)%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60mm
x% = \(\frac{2}{60} \times 100=\frac{200}{60}=\frac{10}{3}=3 \frac{1}{3}\)
x = 3\(\frac{1}{3}\)%

(iii) If x% of x = 25, then x = _______ .
Answer:
50
Hint:
Given that x% of x is 25
∴ \(\frac{x}{100}\) × x = 25
∴ x² = 25 × 100 = 2500
∴ x = √2500 = 50

(iv) In a school of 1400 students, there are 420 girls. The percentage of boys in the school is _______ .
Answer:
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
∴ Number of boys in school = 1400 – 420 = 980

= \(\frac{980}{14}\) = 70
% of boys = 70%

(v) 0.5252 is _______ %.
Answer:
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 × 100 = 52.52%

percent difference Calculator can be found by first finding the difference between the numbers, then finding the average of the numbers, and then dividing.

Question 2.
Rewrite each underlined part using percentage language.
(i) One half of the cake is distributed to the children.
Answer:
50% of the cake is distributed to the children
Hint:
One half is nothing but \(\frac { 1 }{ 2 }\)
as percentage, we need to multiply by 100
∴ \(\frac { 1 }{ 2 }\) × 100 = 50%

(ii) Aparna scored 7.5 points out of 10 in a competition.
Answer:
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is \(\frac{7.5}{10}\) = 0.75
For percentage, we need to multiply by 100
We get 0.75 × 100 = 75%

(iii) The statue was made of pure silver.
Answer:
The statue was made of 100% pure silver
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = \(\frac{100}{100}\)
∴ to express as percentage, \(\frac{100}{100}\) × 100% = 100%

(iv) 48 out of 50 students participated in sports.
Answer:
96% students participated in sports.
Hint:
48 out of 50 students in fraction form is \(\frac{48}{50}\)
As a percentage, we need to multiply by 100

(v) Only 2 persons out of 3 will be selected in the interview.
Answer:
Only 66\(\frac{2}{3}\)% will be selected in the interview.
Hint:
2 out of 3 in fraction form is \(\frac{2}{3}\)
to express as percentage, we need to multiply by 100
\(\frac{2}{3} \times 100=\frac{200}{3}=66 \frac{2}{3} \%\)

To convert CGPA to Percentage, we need to multiply CGPA by 9.5, which will give us the percentage.

Question 3.
48 is 32% of which number?
Answer:
Let the number required to be found be ‘x’
Given that 32% of x is 48
i.e., \(\frac{32}{100}\) × x = 48

∴ x = 150

Question 4.
What is 25% of 30% of 400?
Answer:
Required to find 25% of 30% of 400

The percentage decrease calculator determines the change from one amount to a lesser amount in terms of percent decrease.

Question 5.
If a car is sold for ₹ 2,00,000 from its original price of ₹ 3,00,000, then find the percentage of decrease in the value of the car.

Answer:
original price of car = ₹ 3,00,000
actual selling price of car = ₹ 2,00,000
Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000

Question 6.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of that number.
Answer:
Given that 75% of number less 60% of number is 82.5
Let the number be ‘x’
∴ \(\frac{75}{100}\) x x – \(\frac{60}{100}\) x x = 82.5
∴ 0.75 x – 0.60 x = 82.5
∴ 0.15 x = 82.5
∴ x = \(\frac{82.5}{0.15}=\frac{8250}{15}\) = 550
Required to find 20% of number ie 20% of x.

Question 7.
A number when increased by 18% gives 236. Find the number.
Answer:
Let the number be x. Given that when it is increased by 18%, we get 236.

Question 8.
A number when decreased by 20% gives 80. Find the number.
Answer:
Let the number be x. Given that when it is increased by 20% we get 80.

x = 100

It is important to learn to convert CGPA into percentage because both the systems are interconnected.

Question 9.
A number is increased by 25% and then decreased by 20%. Find the percentage change in that number.
Answer:
Method 1.
Let the number be x.
First it is increased by 25%
∴ It becomes x + \(\frac{25}{100}\) × x = \(\frac{125}{100}\)
Secondary it is decreased by 20%
\(\frac{125 x}{100}-\frac{20}{100} \times \frac{125}{100} x=\frac{125}{100} x \times \frac{80}{100}=x\)
Now we get back x, therefore there is no change.
Hence percentage change in that number is 0%

Method 2.
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we get

Now this 125 is decreased by 20%, we get
125 – \(\frac{25}{100}\) × 125 = 125 – 25 = 100
∴ We get back 100 ⇒ No change
Hence percentage change in that number is 0%

Question 10.
The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.
Answer:
Let number of boys be ‘b’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ \(\frac{b}{g}=\frac{5}{3}\) …… (A)
Failure in boys = 16% = \(\frac{16}{100}\) × b = \(\frac{16b}{100}\)
Failure in girls = 8% = \(\frac{8}{100}\) × g = \(\frac{8g}{100}\)
Pass in boys = 100 – 16% = 84% = \(\frac{84}{100} b\) …… (1)
Pass in girls = 100 – 8% = 92% = \(\frac{92}{100} g\) ……. (2)
From A, we have \(\frac{b}{g}=\frac{5}{3}\) , adding I on both sides, we get
\(\frac{b}{g}\) + 1 = \(\frac{5}{3}\) + 1
\(\frac{b+g}{g}=\frac{5+3}{3}=\frac{8}{3}\)
∴ g = \(\frac{3}{8}\)(b + g) ……. (3)
Similarly b = \(\frac{5}{8}\) (b + g) ……. (4)
Total pass = pass in girls + pass in boys
= (1) + (2) = \(\frac{84}{100} b+\frac{92}{100} g\)

Objective Type Questions

Question 11.
12% of 250 litre is the same as ________ of 150 litre.
(A) 10%
(B) 15%
(C) 20%
(D) 30%
Answer:
(C) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) × 250 = 30 lit.
Percentage: \(\frac{30}{150}\) × 100 = 20%

Question 12.
If three candidates A, B and C in a school election got 153,245 and 102 votes respectively, then the percentage of votes got by the winner is ________ .
(A) 48%
(B) 49%
(C) 50%
(D) 45%
Answer:
(B) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes)
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500

= \(\frac{245}{500}\) × 100 = 49%

Question 13.
15% of 25% of 10000 = ________ .
(A) 375
(B) 400
(C) 425
(D) 475
Answer:
(A) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is

Next 15% of the above is \(\frac{15}{100}\) × 2500 = 375

Question 14.
When 60 is subtracted from 60% of a number to give 60, the number is
(A) 60
(B) 100
(C) 150
(D) 200
Answer:
(D) 200
Hint:
Let the number be ‘X’
60% of the number is \(\frac{60}{100}\) × x = \(\frac{60x}{100}\)
Given that when 60 is subtracted from 60%, we get 60
i.e \(\frac{60}{100}\) x – 60 = 60
∴ \(\frac{60}{100}\) x 60 + 60 = 120
∴ x = \(\frac{120 \times 100}{60}\) = 200

Question 15.
If 48% of 48 = 64% of x, then x =
(A) 64
(B) 56
(C) 42
(D) 36
Answer:
(D) 36
Hint:
Given that 48% of 48 = 64% of x

x = 36