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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 3.1 நோயும் மருந்தும் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 3.1 நோயும் மருந்தும்

கற்பவை கற்றபின்

Question 1.
ஐம்பெருங்காப்பியங்கள், ஐஞ்சிறுகாப்பியங்கள் ஆகியவற்றின் பெயர்களைத் தொகுத்து எழுதுக.
Answer:

மாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
உடல்நலம் என்பது ……………………… இல்லாமல் வாழ்தல் ஆகும்.
அ) அணி
ஆ) பணி
இ) பிணி
ஈ) மணி
Answer:
இ) பிணி

Question 2.
நீலகேசி கூறும் நோயின் வகைகள் ……………………
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
ஆ) மூன்று

Question 3.
‘இவையுண்டார்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது_ ………………………
அ) இ + யுண்டார்
ஆ) இவ் + உண்டார்
இ) இவை + உண்டார்
ஈ) இவை + யுண்டார்
Answer:
இ) இவை + உண்டார்

Question 4.
தாம் + இனி என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………………
அ) தாம் இனி
ஆ) தாம்மினி
இ) தாமினி
ஈ) தாமனி
Answer:
இ) தாமினி

குறுவினா

Question 1.
நோயின் மூன்று வகைகள் யாவை?
Answer:

• மருந்தினால் நீங்கும் நோய்.
• எதனாலும் தீராத தன்மையுடைய நோய் மற்றொரு வகை.
• வெளியில் ஆறி உள்ளுக்குள் இருந்து துன்பம் தரும் நோய்.

Question 2.
நீலகேசியில் பிறவித் துன்பத்தைத் தீர்க்கும் மருந்துகளாகக் கூறப்படுவன யாவை?
Answer:
நல்லறிவு, நற்காட்சி, நல்லொழுக்கம் என்பவையே பிறவித் துன்பத்தைத் தீர்க்கும் மருந்துகளாக நீலகேசி கூறுகின்றது.

சிறு வினா

Question 1.
நோயின் வகைகள் அவற்றைத் தீர்க்கும் வழிகள் பற்றி நீலகேசி கூறுவன யாவை?
Answer:

• ஒளிபொருந்திய அணிகலன்களை அணிந்த பெண்ணே! நோயின் தன்மை பற்றி யார் வினவினாலும் அது மூன்று வகைப்படும் என அறிவாயாக.
• மருந்தினால் நீங்கும் நோய்.
• எதனாலும் தீராத தன்மையுடைய நோய் மற்றொரு வகை.
• வெளியில் ஆறி உள்ளுக்குள் இருந்து துன்பம் தரும் நோய்.
• அகற்றுவதற்கு அரியவை பிறவித் துன்பங்கள் ஆகும்.
• இவற்றைத் தீர்க்கும் மருந்துகள் மூன்று. நல்லறிவு, நற்காட்சி, நல்லொழுக்கம் என்பவையே அம்மருந்துகள்.
• இவற்றை ஏற்றோர் பிறவித்துன்பத்திலிருந்து நீங்கி உயரிய இன்பத்தை அடைவர்.

சிந்தனை வினா

Question 1.
துன்பமின்றி வாழ நாம் கைக்கொள்ள வேண்டிய நற்பண்புகள் யாவை?
Answer:
தருமம் செய்தல், கோபத்தைத் தணித்தல், முயற்சி செய்தல், கல்வி கற்றல், உலக நடையை அறிந்து நடத்தல், நல்ல நூல்களைப் படித்தல், பொறாமை படாமல் இருத்தல், பொய்சாட்சி சொல்லாமல் இருத்தல், இனிமையாகப் பேசுதல், பேராசையைத் தவிர்த்தல், நட்புடன் பழகுதல், பெரியோர்களை மதித்தல், ஒழுக்கம் தவறாமல் இருத்தல், நன்றியை மறவாமல் இருத்தல், காலத்தைக் கடைபிடித்தல், களவு செய்யாதிருத்தல், இழிவானதைச் செய்யாதிருத்தல், இரக்கம் கொள்ளுதல், பொய் சொல்லாதிருத்தல், ஆணவம் கொள்ளாதிருத்தல், சுறுசுறுப்புடன் இருத்தல், உடற்பயிற்சி செய்தல், அதிகாலையில் எழுந்திருத்தல் போன்றவை கைக்கொள்ள வேண்டிய நற்பண்புகள் ஆகும்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
மக்களின் உடலுக்கும் உள்ளத்திற்கும் துன்பம் தருவன …………………….
அ) நாய்கள்
ஆ) நோய்கள்
இ) பேய்கள்
ஈ) மனிதர்கள்
Answer:
ஆ) நோய்கள்

Question 2.
உள்ளத்தில் தோன்றும் தீய எண்ணங்களால் ஏற்படும் துன்பங்களையும் ……………………… என்றே நம் முன்னோர்கள் குறிப்பிட்டனர்.
அ) கவலை
ஆ) துன்பம்
இ) நோய்கள்
ஈ) பொறுமை
Answer:
இ) நோய்கள்

Question 3.
நோய்களை நீக்கும் மருந்துகளாக விளங்கும் அறக்கருத்துகளை விளக்குபவை …………………..
அ) இலக்கியங்கள்
ஆ) இலக்கணங்கள்
இ) படைப்புகள்
ஈ) முன்னோர்கள்
Answer:
அ) இலக்கியங்கள்

Question 4.
நோயைத் தீர்க்கும் மருந்துகள் ………………….
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
ஆ) மூன்று

Question 5.
………………… ஐஞ்சிறு காப்பியங்களுள் ஒன்று.
அ) சிலப்பதிகாரம்
ஆ) நீலகேசி
இ) குண்டலகேசி
ஈ) வளையாபதி
Answer:
ஆ) நீலகேசி

Question 6.
நீலகேசி ………………… சமயக் கருத்துகளைக் கூறுகிறது.
அ) சமணம்
ஆ) புத்தம்
இ) கிறித்தவம்
ஈ) இந்து
Answer:
அ) சமணம்

Question 7.
நீலகேசி, கடவுள் வாழ்த்து நீங்கலாக ……………….. சருக்கங்களைக் கொண்டது.
அ) எட்டு
ஆ) ஒன்பது
இ) ஏழு
ஈ) பத்து
Answer:
ஈ) பத்து

Question 8.
‘போலாதும்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ………………………..
அ) போ + தும்
ஆ) போல் + ஆதும்
இ) போல் + அனதும்
ஈ) போலா + தும்
Answer:
ஆ) போல்+ஆதும்

Question 9.
‘உய்ப்பனவும்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………
அ) உய் + பனவும்
ஆ) உய்ப் + பனவும்
இ) உய்ப்ப ன + உம்
ஈ) உய்ப்ப ன + அம்
Answer:
இ) உய்ப்பன+உம்

Question 10.
‘கூற்றவா’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………….
அ) கூ + அவா
ஆ) கூற்று + அவா
இ) கூற் + அவா
ஈ) கூற்று + ஆவா
Answer:
ஆ) கூற்று+அவா

Question 11.
‘ஐம்பெருங்காப்பியம்’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………..
அ) ஐந்து + காப்பியம்
ஆ) ஐந்து + பெரு + காப்பியம்
இ) ஐம்பெருங் + காப்பியம்
ஈ) ஐந்து + பெருமை + காப்பியம்
Answer:
ஈ) ஐந்து + பெருமை + காப்பியம்

Question 12.
‘அரும்பிணி’ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ………………………….
அ) அரும் + பிணி
ஆ) அரு + பிணி
இ) அருமை + பிணி
ஈ) அரும் + பணி
Answer:
இ) அருமை + பிணி

Question 13.
‘அ + பிணி’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………………….
அ) அபிணி
ஆ) அப்பிணி
இ) அப்பிணி
ஈ) அதுபிணி
Answer:
இ) அப்பிணி .

Question 14.
‘தெளிவு + ஓடு’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …………………..
அ) தெளிவுவூடு
ஆ) தெளிவோடு
இ) தெளிவுஓடு
ஈ) தெளிவாடு
Answer:
ஆ) தெளிவோடு

Question 15.
‘பிணி + உள்’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ……………….
அ) பிணியுள்
ஆ) பிணினள்
இ) பிணியாள்
ஈ) பிணிபுள்
Answer:
அ) பிணியுள்

Question 16.
‘இன்பம் + உற்றே’ என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் ………………….
அ) இன்பமுற்று
ஆ) இன்பமற்றே
இ) இன்பம் உற்றோ
ஈ) இன்பமுற்றே
Answer:
ஈ) இன்பமுற்றே

குறுவினா

Question 1.
நோய்கள் எவற்றிற்கெல்லாம் துன்பம் தருவன?
Answer:
மக்களின் உடலுக்கும் உள்ளத்திற்கும் துன்பம் தருவன நோய்கள்.

Question 2.
நம் முன்னோர்கள் எவற்றையும் நோய்கள் என்று கூறினர்?
Answer:
உள்ளத்தில் தோன்றும் தீய எண்ணங்களால் ஏற்படும் துன்பங்களையும் நோய்கள் என்றே நம் முன்னோர் கூறினர்.

Question 3.
இலக்கியங்கள் விளக்குவன யாவை?
Answer:
நோய்களை நீக்கும் மருந்துகளாக விளங்கும் அறக்கருத்துகளை இலக்கியங்கள் விளக்குகின்றன.

Question 4.
ஐம்பெருங்காப்பியங்கள் யாவை?
Answer:
சிலப்பதிகாரம், மணிமேகலை, சீவகசிந்தாமணி, வளையாபதி, குண்டலகேசி.

Question 5.
ஐஞ்சிறுகாப்பியங்கள் யாவை?
Answer:
சூளாமணி, நீலகேசி, உதயண குமார காவியம், யசோதர காவியம், நாககுமார காவியம்.

சிறுவினா

Question 1.
நீலகேசி குறித்து எழுதுக.
Answer:

• நீலகேசி என்றால் கருத்த கூந்தலை உடையவள் என்று பொருள்.
• ஆசிரியர் பெயர் அறியப்படவில்லை .
• கடவுள் வாழ்த்து நீங்கலாகப் பத்துச் சருக்கங்களைக் கொண்டது.
• 894 பாடல்களைக் கொண்டது.
• நீலகேசி தெருட்டு என்ற வேறு பெயரும் உண்டு.

சொல்லும் பொருளும்

தீர்வன – நீங்குபவை
உவசமம் – அடங்கி இருத்தல்
நிழல் இகழும் – ஒளிபொருந்திய
பேர்தற்கு அகற்றுவதற்கு
திரியோகமருந்து – மூன்று யோகமருந்து
தெளிவு – நற்காட்சி
திறத்தன – தன்மை யுடையன
கூற்றவா – பிரிவுகளாக
பூணாய் – அணிகலன்களை அணிந்தவளே
பிணி – துன்பம்
ஓர்தல் – நல்லறிவு
பிறவார் – பிறக்கமாட்டார்

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Tamilnadu Samacheer Kalvi 10th Social Science Civics Solutions Chapter 4 India’s Foreign Policy

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India’s Foreign Policy Textual Exercise

I. Choose the correct answer.

Indian Foreign Policy Class 10 Questions And Answers Question 1.
Which Minister plays a vital role in molding the foreign policy of our country?
(a) Defense Minister
(b) Prime Minister
(c) External Affairs Minister
(d) Home Minister
Answer:
(c) External Affairs Minister

Indian Foreign Policy 10th Standard Notes Question 2.
The Panchseel treaty has been signed between:
(a) India and Nepal
(b) India and Pakistan
(c) India and China
(d) India and Sri Lanka
Answer:
(c) India and China

10th Social Indian Foreign Policy Notes Question 3.
Which article of Indian constitution directs to adopt foreign policy?
(a) Article 50
(b) Article 51
(c) Article 52
(d) Article 53
Answer:
(b) Article 51

Foreign Policy Of India 10th Notes Question 4.
Apartheid is:
(a) An international association
(b) Energy diplomacy
(c) A policy of racial discrimination
(d) None of these
Answer:
(d) None of these

Indian Foreign Policy Notes 10th Class Question 5.
The Agreement signed by India and China in 1954 related to …………..
(a) Trade and Commerce
(b) Restoration of normal relations
(c) Cultural exchange programmes
(d) The Five Principles of Co-existence
Answer:
(d) The Five Principles of Co-existence

Indian Foreign Policy Class 10 Notes Question 6.
Which is not related to our foreign policy?
(a) World co-operation
(b) World peace
(c) Racial equality
(d) Colonialism
Answer:
(d) Colonialism

Indian Foreign Policy Class 10 Questions And Answers Pdf Question 7.
Which of the following country is not the founder member of NAM?
(a) Yugoslavia
(b) Indonesia
(c) Egypt
(d) Pakistan
Answer:
(d) Pakistan

Indian Foreign Policy 10th Notes Question 8.
Find the odd one:
(a) Social welfare
(b) Health care
(c) Diplomacy
(d) Domestic affairs
Answer:
(c) Diplomacy

Question 9.
Non-Alliance means ………..
(a) being neutral
(b) freedom to decide on issues independently
(c) demilitarisation
(d) none of the above
Answer:
(b) freedom to decide on issues independently

Question 10.
Non – military issues are:
(a) Energy security
(b) Water security
(c) Pandemics
(d) All the above.
Answer:
(d) All the above.

II. Fill in the blanks.

1. India conducted its first nuclear test at …………
2. At present our foreign policy acts as a means to generate ……….. for domestic growth and development.
3. ……….. is the instrument for implementing foreign policy of a state.
4. …………. was India’s policy in the face of the bipolar order of the cold war.
5. Our tradition and national ethos is to practice …………..
Answers:
1. Pokhran
2. inward investment, business and technology
3. Diplomacy
4. Non-Alignment
5. disarmament

III. Consider the following statement and tick the appropriate answer.

Question 1.
Arrange the following in the correct chronological order and choose the correct answer from the code given below.
(i) Panchsheel
(ii) Nuclear test at Pokhran
(iii) Twenty-year Treaty
(iv) First Nuclear test
(a) i, iii, iv, ii
(b) i, ii, iii, iv
(c) i, ii, iv, iii
(d) i, iii, ii, iv
Answer:
(c) i, ii, iv, iii

Question 2.
Which of the following is not about NAM?
(i) The term Non-Alignment was coined by V. Krishna Menon
(ii) It aimed to maintain national independence in foreign affairs by joining any military alliance
(iii) At present it has 120 member countries
(iv) It has transformed to an economical movement
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (ii) only
(d) (iv) only
Answer:
(c) (ii) only

Question 3.
Write true or false against each of the statement.
(a) During Cold War India tried to form a third bloc of nations in the international affairs.
(b) The Ministry of Home Affairs is responsible for the conduct of the country’s foreign relations.
(c) The nuclear test at Pokhran was done under Subterranean Nuclear Explosions Project.
Answer:
(a) True
(b) False
(c) True

Question 4.
Assertion (A): India aligned with Soviet Union by the Indo-Soviet treaty on 1971.
Reason (B): This began with a disastrous Indo-China war of 1962.
(a) A is correct and R explains A
(b) A is correct and R does not explain A
(c) A is correct and R is Wrong
(d) Both A and R are wrong
Answer:
(b) A is correct and R does not explain A

Question 5.
Assertion (A): India has formal diplomatic relations with most of the nations.
Reason (R): India is the World’s second-most populous country.
(a) A is correct and R explains A
(b) A is correct and R does not explain A
(c) A is wrong and R is correct
(d) Both are wrong
Answer:
(b) A is correct and R does not explain A

Question 6.
Avoidance of military blocs was a necessity for India after political freedom. Because India had to redeemed from
(a) acute poverty
(b) illiteracy
(c) chaotic socio-economic conditions
(d) all the above
Answer:
(d) all the above

IV. Match the following.

Answer:
1. (c)
2. (e)
3. (b)
4. (a)
5. (d)

V. Give Short Answers.

Question 1.
What is foreign policy?
Answer:
Foreign policy can be defined as a country’s policy that is conceived, designed and formulated to safeguard and promote her national interests in her external affairs in the conduct of relationships with other countries, both bilaterally and multilaterally.

Question 2.
Explain India’s nuclear policy.
Answer:
Indian nuclear programme in 1974 and 1998 is only done for strategic purposes. The two themes of India’s nuclear doctrine are
• No first use
• Credible minimum deterrence
It has decided not to use nuclear power for ‘offensive purposes’ and would never use against any non-nuclear state.

Question 3.
Highlight the contribution by Nehru to India’s foreign policy.
Answer:

1. The most idealistic phase of India’s foreign policy under the guidance of India’s first Prime Minister, Jawaharlal Nehru.
2. The new nations that got independence after the long period of colonial struggle found themselves in a very difficult situation with respect to economic development.
3. So it was necessary to align with either of the blocs – United States of America U.S.A (or) Union Soviet Socialist Republic (U.S.S.R).

Question 4.
Differentiate: Domestic policy and Foreign policy
Answer:

Question 5.
List any four guiding principles of Panchsheel.
Answer:

1. Mutual non – aggression
2. Mutual non – interference
3. Equality and co-operation for mutual benefit
4. Peaceful co-existence

Question 6.
What was the reason for India to choose the path of Non-Alignment?
Answer:
The new nations that got independence after the long period of colonial struggle found themselves in a very difficult situation with respect to economic development. So it was necessary to align with either of the blocs – United States of America (USA) or United Soviet Socialist Republic (USSR). Nehru, India’s first Prime Minister, was opposed to the rivalry of the two superpowers (America and Russia). So he chose the path of Non-Alignment.

Question 7.
In what ways are India’s global security concerns reflected?
Answer:
India’s global security concerns are reflected in its military modernisation, maritime security and nuclear policies.

Question 8.
List out the member countries of SAARC.
Answer:
The member countries are Afghanistan, Bangladesh, Bhutan, India, Nepal, Maldives, Pakistan and Sri Lanka.

Question 9.
Name the architects of the Non-Aligned movement.
Answer:
Jawaharlal Nehru of India, Tito of Yugoslavia, Nasser of Egypt, Sukarno of Indonesia, and KwameNkumarah of Ghana were the architects of Non Aligned Movement.

Question 10.
Mention the main tools of foreign policy.
Answer:
The main tools of foreign policy are treaties and executive agreements, appointing ambassadors, foreign aid, international trade and armed forces.

VI. Answer in detail.

Question 1.
Write a detailed note on Non-alignment.
Answer:

• The term ‘Non-Alignment’ was coined by V. Krishna Menon.
• Non-alignment has been regarded as the most important feature of India’s foreign policy.
• It aimed to maintain national independence in foreign affairs by not joining any military alliance.
• The Non-Aligned Movement (NAM) was formed with a membership of 120 countries and 17 states as observers and 10 international organisations.

The founding fathers of Non-Aligned Movement:
Jawaharlal Nehru of India, Tito of Yugoslavia, Nasser of Egypt, Sukarno of Indonesia, and Kwame Nkumarah of Ghana were the founding fathers of NAM.

• • Non-aligned countries have been successful in establishing a foundation of economic co-operation among underdeveloped countries.

Question 2.
Discuss the core determinants of India’s foreign policy?
Answer:

1. Geographical position and size of territory.
2. Nation’s history, traditions and philosophical basis.
3. Natural resources.
4. The compulsion of economic development.
5. Political stability and structure of Government.
6. The necessity of peace, disarmament and non – proliferation of nuclear weapons.
7. Military strength
8. International milieu.

Question 3.
Make a list on basic concepts followed by India to maintain friendly relations with its neighbours.
Answer:
(i) Indian foreign policy has always regarded the concept of neighbourhood as one of widening concentric circles, around the central axis of historical and cultural commonalties.

(ii) India gives political and diplomatic priority to her immediate neighbours and the Indian Ocean Island states such as Maldives.

(iii) India provides neighbours with support as needed in this form of resources equipment and training.

VII. Project and activity

Question 1.
Identify any two aspects of India’s foreign policy that you would like to retain and two that you would like to change if you were the decision maker.
Answer:
Do it yourself.

India’s Foreign Policy Additional Questions

I. Choose the correct answer.

Question 1.
India is a country with an unbounded faith in …………
(a) War
(b) Peace
(c) Love
Answer:
(b) Peace

Question 2.
Find out the main tools of the foreign policy of the following.
(a) Treaties
(b) International trade
(c) Foreign Aid
(d) All the above
Answer:
(d) All the above

Question 3.
Apartheid was abolished on …………
(a) 1990
(b) 1991
(c) 1890
Answer:
(a) 1990

Question 4.
In which place the Foreign Service Training Institute was established?
(a) New Delhi
(b) Mumbai
(c) Calcutta
(d) None of the above
Answer:
(a) New Delhi

Question 5.
Apartheid was abolished by …………..
(a) Jawaharlal Nehru
(b) Nelson Mandela
(c) Gandhi
Answer:
(b) Nelson Mandela

Question 6.
In which year Panchsheel was signed?
(a) 1951
(b) 1952
(c) 1954
(d) 1956
Answer:
(c) 1954

Question 7.
SAARC’s first meeting was held at ………….
(a) Colombo
(b) Cairo
(c) Dacca
Answer:
(c) Dacca

Question 8.
Who was opposed to the rivalry of the two super powers? (America and Russia)
(a) V. Krishna Menon
(b) Nasser
(c) Jawaharlal Nehru
(d) None
Answer:
(c) Jawaharlal Nehru

Question 9.
The first SAARC’s meeting was held at Dacca in the year …………..
(a) 1985
(b) 1965
(c) 1995
Answer:
(c) 1995

Question 10.
Which year did India conduct its first nuclear test at Pokhran?
(a) 1971
(b) 1972
(c) 1974
(d) 1976
Answer:
(c) 1974

II. Fill in the blanks :

1. India followed the policy of …………..
2. India has rendered whole-hearted support to the ………….. to bring World Peace.
3. ………….. is an economic and geopolitical organization of eight countries are particularly located in South Asia.
4. SAARC Disaster Management Centre was set up at ………….
5. …………policy is the nation’s plan for dealing issues within its own nation.
Answers:
1. Non-alignment
2. UNO
3. SAARC
4. New Delhi
5. Domestic

III. Match the following.

Answer:
1. (c)
2. (d)
3. (e)
4. (b)
5. (a)

IV. Answer briefly:

Question 1.
Mention any three objectives of our Foreign Policy.
Answer:

1. National security
2. National prosperity
3. Achieving world peace and enable every nation to peacefully co-exist.

Question 2.
Why is world peace an essential one?
Answer:
The economic development of the nations can be achieved only through world peace. So world peace is very essential not only for the economic development of India but also for all the developing countries of the world.

Question 3.
Explain the foreign policy stance of India?
Answer:
The foreign policy stance of India was:

1. Supporting the cause of decolonisation.
2. Staunch opponent of the apartheid regime in South Africa.
3. Accepted the importance of defence preparedness.

Question 4.
Name the areas identified by the SAARC countries for mutual co-operations.
Answer:
The SAARC countries identified mutual co-operation in the following areas. They are transportation, postal service, tourism, shipping meteorology, health, agriculture, rural construction and telecommunications.

Question 5.
What are the two themes of India’s nuclear doctrine?
Answer:
The two themes of India’s nuclear doctrine are:

1. No first use
2. Credible minimum deterrence.

V. Detail.

Question 1.
Write a paragraph about Panchsheel and the policy of Non-alignment.
Answer:
Panchsheel:
India is called by the name of “A Great Peace Maker”. It followed five principles which are popularly known as ‘Panchsheel’. Jawaharlal Nehru said, stress on these five principles. They are:

1. Each country should respect the territorial integrity and sovereignty of others.
2. No country should attack any other country.
3. No one should try to interference in the internal affairs of others.
4. All countries shall strive for equality and mutual benefit.
5. Every country should try to follow the policy of peaceful co-existence.

These Panchsheel greatly added to the international status of India.
Policy of Non-allignments:

1. After the Second World War the world was divided into two hostile blocs. The
   American Bloc and The Russian Bloc.
2. Both of them are trying to increase their influence at the cost of the other.
3. But India has not joined either of these two blocs.
4. Whenever any difference arises between there blocs, India tries to remove that difference. Thus India has contributed substantially towards world peace.

Question 2.
Write about policy of Disarmament.
Answer:

1. Our tradition and national ethos is to practice disarmament.
2. Since Independence, global non-proliferation has been a dominant theme of India’s nuclear policy.
3. So India supported UN disarmament progrmme.
4. Indian nuclear programme in 1974 and 1998 is only done for strategic purposes.

Question 3.
Point out the basic concepts of India’s foreign policy:
Answer:

• Preservation of national interest.
• Achievement of world peace.
• Disarmament
• Fostering cordial relationship with other countries.
• Solving conflicts by peaceful means.
• Independence of thought and action as per the principle of NAM.
• Equality in conducting international relations.
• Anti-Colonialism, anti-imperialism anti- racism.

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Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 3.1 ஒரு வேண்டுகோள்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
மயிலும் மானும் வனத்திற்கு ………………………… தருகின்ற ன.
அ) களைப்பு
ஆ) வனப்பு
இ) மலைப்பு
ஈ) உழைப்பு
Answer:
ஆ) வனப்பு

Question 2.
மிளகாய் வற்றலின் ……………… தும்மலை வரவழைக்கும்.
அ) நெடி
ஆ) காட்சி
இ) மணம்
ஈ) ஓசை
Answer:
அ) நெடி

Question 3.
அன்னை தான் பெற்ற ………….. ….. சிரிப்பில் மகிழ்ச்சி அடைகிறார்.
அ) தங்கையின்
ஆ) தம்பியின்
இ) மழலையின்
ஈ) கணவனின்
Answer:
இ) மழலையின்

Question 4.
வனப்பில்லை ‘ என்னும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது …………………
அ) வனம் + இல்லை
ஆ) வனப்பு + இல்லை
இ) வனப்பு + யில்லை
ஈ) வனப் + பில்லை
Answer:
ஆ) வனப்பு + இல்லை

Question 5.
வார்ப்பு + எனில்’ என்பதனைச் சேர்தெழுதக் கிடைக்கும் சொல்
அ) வார்ப்எனில்
ஆ) வார்ப்பினில்
இ) வார்ப்பெனில்
ஈ) வார்ப்பு எனில்
Answer:
இ) வார்ப்பெனில்

நயம் அறிக

ஒரே எழுத்திலோ ஓசையிலோ முடியும் இயைபுச் சொற்களைப் பாடலில் இருந்து எடுத்து எழுதுக.

குறுவினா

Question 1.
தாய்மையின் ஓவியத்தில் நிறைந்திருக்க வேண்டியவை யாவை?
Answer:
அன்பும் பாசமும் தாய்மையின் ஓவியத்தில் நிறைந்திருக்க வேண்டும்.

Question 2.
ஒரு கலை எப்பொழுது உயிர்ப்புடையதாக அமையும்?
Answer:
மானுடப் பண்பு நிறைந்திருந்தால் ஒரு கலை உயிர்ப்புடையதாக அமையும்.

சிறுவினா

சிற்பங்களும் ஓவியங்களும் எவ்வாறு அமைய வேண்டும் என்று கவிஞர் கூறுகிறார்?
Answer:
(i) சிற்பங்கள் : ஒரு சிற்பி, பாறை உடைப்பவரின் சிலையைச் செதுக்கினால் அதிக வியர்வை நாற்றம் வீச வேண்டும். உழவனின் உருவச் சிலையாக இருந்தால் ஈரமண் வாசம் வீச வேண்டும்.

(ii) ஓவியங்கள் : ஓர் ஓவியன், தாயின் உருவத்தைத் தீட்டினால் அன்பும் பண்பும் மேலோங்கிட வேண்டும். சிறு குழந்தையின் சித்திரமானால் உடலெங்கும் பால் மணம் கமழ வேண்டும். சிற்பங்களும் ஓவியங்களும் இவ்வாறு அமைவதே சிறப்பு என்று கவிஞர் கூறுகிறார்.

சிந்தனை வினா

நீங்கள் ஒரு கலைஞராக இருந்தால் எத்தகைய படைப்புகளை உருவாக்குவீர்கள்?
Answer:
நான் ஒரு கலைஞராக இருந்தால் பச்சைப்பசேல் என விளங்கும் மலைகள், அங்கு விழும் அருவிகள், பயமறியாமல் பறக்கும் பறவைகள், புலி, மான், சிங்கம் என அனைத்து வனவாழ் விலங்குகளும் அச்சமின்றி அருவியில் நீர் அருந்துதல், ஒன்றையொன்று நட்புடன் நோக்குதல் இவற்றை உருவாக்குவேன்.

கற்பவை கற்றபின்

Question 1.
உங்களுக்குப் பிடித்த ஏதேனும் ஒரு கலை பற்றிய தகவல்களைத் திரட்டுக.
Answer:
மயிலாட்டம்: மயில் வடிவுள்ள கூட்டுக்குள் ஒருவர், தன் உருவத்தை மறைத்துக் கொண்டு, நையாண்டி மேளத்திற்கேற்ப ஆடும் ஆட்டமே மயிலாட்டமாகும். நையாண்டி மேளம் இசைக்க, காலில் கட்டப்பட்டுள்ள சலங்கை ஒலிக்க மயிலின் அசைவுகளை ஆடிக்காட்டுவர்.

கரகாட்டத்தின் துணையாட்டமாகவும் மயிலாட்டம் ஆடப்படுகிறது. ஊர்ந்து ஆடுதல், மிதந்து ஆடுதல், சுற்றி ஆடுதல், இறகை விரித்தாடுதல், தலையைச் சாய்த்தாடுதல், தாவியாடுதல், இருபுறமும் சுற்றியாடுதல், அகவுதல், தண்ணீ ர் குடித்துக் கொண்டே ஆடுதல் ஆகிய அடவுகளைக் கலைஞர்கள் இவ்வாட்டத்தில் ஆடிக் காட்டுவர்.

Question 2.
உழைப்பாளர்களின் பெருமையைக் கூறும் கவிதைகளைத் தொகுத்து வந்து வகுப்பறையில் பகிர்க.
Answer:
உழைப்பு இல்லையேல்
உணவும், மகிழ்ச்சியும்
இல்லை இன்று
உழைத்தால் வயதான போது
உட்கார்ந்து உண்ணலாம்
உழைக்க வேண்டிய காலத்தில்
உழைக்கவில்லை என்றால்
ஓய்வு எடுக்க வேண்டிய
காலத்தில் உழைக்க வேண்டும்
உயிரினங்கள் கூட உழைக்கின்றன
கையிருந்தும் உழைக்காமல்
இருந்தால் வாழ்க்கை இல்லாமலாகும்.

கூடுதல் வினாக்கள்

சொல்லும் பொருளும் :

1. பிரும்மாக்கள் – படைப்பாளர்கள்
2. நெடி – நாற்றம்
3. மழலை – குழந்தை
4. வனப்பு – அழகு
5. பூரிப்பு – மகிழ்ச்சி
6. மேனி – உடல்

நிரப்புக.

Question 1.
பூரிப்பு என்பதன் பொருள்
Answer:
மகிழ்ச்சி

Question 2.
தேனரசன் …………….. பணியாற்றியவர்.
Answer:
தமிழாசிரியராகப்

Question 3.
ஒரு வேண்டுகோள் கவிதை ……………. என்னும் நூலிலிருந்து எடுத்துத் தரப்பட்டுள்ளது.
Answer:
பெய்து பழகிய மேகம்

விடையளி :

Question 1.
தேனரசன் எழுதிய நூல்கள் யாவை?
Answer:

• மண்வாசல்
• வெள்ளை ரோஜா
• பெய்து பழகிய மேகம்.

Question 2.
தேனரசன் எந்த இதழ்களில் கவிதைகள் எழுதினார்?
Answer:

• வானம்பாடி
• குயில்
• தென்றல்

பாடலின் பொருள்

கலையுலகப் படைப்பாளர்களே! மண்ணின் அழகுக்கு அழகு சேர்ப்பவர்களே! உங்களுக்கு ஒரு மனித சமுதாயத்தின் வேண்டுகோள்!

நீங்கள் பாறை உடைப்பவரின் சிலையைச் செதுக்கினால், அதில் வியர்வை நாற்றம் வீசவேண்டும். உழவரின் உருவ வார்ப்பாக இருந்தால், அதில் ஈரமண்ணின் மணம் வீச வேண்டும்.

தாயின் மகிழ்ச்சியான உருவத்தை ஓவியமாக வரைந்தால், அவரின் முகத்தில் அன்பும் பாசமும் நிறைந்திருக்க வேண்டும். சிறு குழந்தையின் சித்திரத்தைத் தீட்டினால் அதன் உடலில் பால் மணம் கமழ வேண்டும்.

ஆல்ப்ஸ் மலைச் சிகரங்கள், அட்லாண்டிக் பெருங்கடல் அலைகள், அமேசான் காடுகள், பனிபடர் பள்ளத்தாக்குகள், தொங்கும் தோட்டங்கள் என இயற்கையின் விந்தைத் தோற்றங்கள் எவையும் கலைவடிவம் பெறலாம். ஆனால் அதில் மானுடப் பண்பு கட்டாயமாக இருக்க வேண்டும். மானுடம் இல்லாத எந்த அழகும் அழகன்று. மனிதன் கலக்காத எதிலும் உயிர்ப்பில்லை

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

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Samacheer Kalvi 11th Chemistry Chapter 2 Quantum Mechanical Model of Atom Textual Evaluation Solved

I. Choose the correct answer
Quantum Mechanical Model Of Atom Class 11 Book Back Answers Question 1.
Electronic configuration of species M²⁺ is 1s² 2s² 2p⁶3s² 3p⁶ 3d⁶ and its atomic weight is 56. The number of neutrons in the nucleus of species M is ………..
(a) 26
(b) 22
(c) 30
(d) 24
Answer:
(c) 30
Solution:
M²⁺ : 1s² 2s² 2p⁶3s² 3p⁶ 3d⁶
M : 1s² 2s² 2p⁶3s² 3p⁶ 3d⁸
Atomic number = 26
Mass number = 56
No. of neutrons = 56 – 26 = 30.

11th Chemistry Quantum Mechanical Model Of Atom Question 2.
The energy of light of wavelength 45 nm is
(a) 6.67 x 10¹⁵ J
(b) 6.67 x 10¹¹ J
(c) 4.42 .x 10¹⁸ J
(d) 4.42 x 10^-15 J
Answer:
(c) 4.42 .x 10¹⁸ J
Solution:
E = hv = hc / λ
\(\frac{6.626 \times 10^{-34} \mathrm{J} \mathrm{s} \times 3 \times 10^{8} \mathrm{ms}^{-1}}{45 \times 10^{-9} \mathrm{m}}\) = 4.42 .x 10¹⁸ J.

11th Chemistry Chapter 2 Book Back Answers Question 3.
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. λ₁ and λ₂ will be …………
(a) \(\frac{\lambda_{1}}{\lambda_{2}}=1\)
(b) λ₁ = 2 λ₂
(c) λ₁ = \(\sqrt{25 \times 50} \lambda_{2}\)
(d) 2 λ₁ = λ₂
Answer:
(b) λ₁ = 2 λ₂
Solution:
\(\frac{E l}{E 2}\) = \(\frac{25eV}{50eV}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{hc}}{\lambda_{1}} \times \frac{\lambda_{2}}{\mathrm{hc}}\) = \(\frac{1}{2}\)
2λ₂ = λ_1.

Samacheer Kalvi Guru 11th Chemistry Question 4.
Splitting of spectral lines in an electric field is called …………….
(a) Zeeman effect
(b) Shielding effect
(c) Compton effect
(d) Stark effect
Answer:
(d) Stark effect
Solution:
Splitting of spectral lines in magnetic field is called Zeeman effect and splitting of spectral lines in electric field, is called Stark effect.

Samacheerkalvi.Guru 11th Chemistry Question 5.
Based on equation E = -2.178 x 10¹⁸ J \(\left(\frac{z^{2}}{n^{2}}\right)\) certain conclusions are written. Which of them is not correct ? (NEET)
(a) Equation can be used to calculate the change in energy when the electron changes orbit
(b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit
(c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
(d) Larger the value of n, the larger is the orbit radius.
Answer:
(b) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit
Solution:
Correct statement:
For n = 6, the electron has more negative energy than it does for n = 6 which means that the electron is strongly bound in the smallest allowed orbit.

11th Chemistry Samacheer Kalvi Question 6.
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
(a) n = 6 to n = 1
(b) n = 5 to n = 4
(c) n = 5 to n = 3
(d) n = 6 to n = 5
Answer:
(d) n = 6 to n = 5
Solution:
n = 6 to n = 5
E₆ = -13.6 / 6² ; E₅ = – 13.6 / 5²
E₆ – E₅ = (-13.6 / 6²) – (-13.6 / 5²)
= 0.166 eV atom^-1
E₅ – E₄ = (-13.6 / 5²) – (-13.6 / 4²)
= 0.306 eV atom^-1

Samacheer Kalvi.Guru 11th Chemistry Question 7.
Assertion : The spectrum of He⁺ is expected to be similar to that of hydrogen
Reason : He⁺ is also one electron system,
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false
(d) If both assertion and reason are false
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.

Samacheer Kalvi Guru 11 Chemistry Question 8.
Which of the following pairs of d-orbitals will have electron density along the axes ? (NEET Phase – II)
(a) d_z², d_xz
(b) d_xz, d_yz
(c) d_z², \(d_{x^{2}-y^{2}}\)
(d) d_xy, \(d_{x^{2}-y^{2}}\)
Answer:
(c) d_z², \(d_{x^{2}-y^{2}}\)

Samacheer Kalvi 11th Chemistry Question 9.
Two electrons occupying the same orbital are distinguished by …………
(a) azimuthal quantum number
(b) spin quantum number
(c) magnetic quantum number
(d) orbital quantum number
Answer:
(b) spin quantum number
Solution:
Spin quantum number For the first electron ms = +\(\frac {1}{2}\)
For the second electron ms = –\(\frac {1}{2}\).

Quantum Mechanical Model Of Atom Question 10.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are (NEET – Phase II)
(a) [Xe] 4f⁷ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷, 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ , 6s², [Xe] 4f⁸ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(d) [Xe] 4f⁸ 5d¹ 6s²[Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Answer:
(b) [Xe] 4f⁷, 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰,6s²

Chemistry Class 11 Samacheer Kalvi Question 11.
The maximum number of electrons in a sub shell is given by the expression …………..
(a) 2n²
(b) 21 + 1
(c) 41 + 2
(d) none of these
Answer:
(c) 41 + 2
Solution:
2 (21 + 1) = 41 + 2.

11th Chemistry 2nd Lesson Question 12.
For d-electron, the orbital angular momentum is ………….
(a) \(\frac{\sqrt{2} h}{2 \pi}\)
(b) \(\frac{\sqrt{2 \mathrm{h}}}{2 \pi}\)
(c) \(\sqrt{2 \times 4}\)
(d) \(\frac{\sqrt{6} h}{2 \pi}\)
Answer:
(d) \(\frac{\sqrt{6} h}{2 \pi}\)
Solution:
Orbital angular momentum
= \(\sqrt{(1(1+1)}) \mathrm{h} / 2 \pi\)
For d orbital = \(\sqrt{(2 × 3)} \mathrm{h} / 2 \pi\) = \(\sqrt{6} \mathrm{h} / 2 \pi\).

Quantum Mechanical Model Of Atom Class 11 Question 13.
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? n = 3,l = 1 and m = -1
(a) 4
(b) 6
(c) 2
(d) 10
Answer:
(c) 2
Solution:
n = 3; l = 1; m = -1 either 3p_x or 3p_y

Samacheer Kalvi Guru Chemistry Question 14.
Assertion: Number of radial and angular nodes for 3p orbital are l, l respectively. Reason: Number of radial and angular nodes depends only on principal quantum number.
(a) both assertion and reason are true and reason is the correct explanation of assertion.
(b) both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Solution:
No. of radial node = n- l – 1
No. of angular node = l for 3p orbital
No. of angular node = l =1
No. of radial node = n- l – 1 = 3 – 1 – 1 = 1.

Question 15.
The total number of orbitals associated with the principal quantum number n = 3 is ………..
(a) 9
(b) 8
(c) 5
(d) 7
Answer:
(a) 9
Solution:
n = 3; l = 0; m₁ = 0 – one s orbital n = 3; l = 1; m₁ = -1, 0, 1 – three p orbitals n = 3; l = 2; m₁ = -2, -1, 0, 1, 2 – five d orbitals, overall nine orbitals are possible.

Question 16.
If n = 6, the correct sequence for filling of electrons will be, …………
(a) ns → (n – 2) f → (n – 1)d → np
(b) ns → (n – 1) d → (n – 2) f → np
(c) ns → (n – 2) f → np → (n – 1) d
(d) none of these are correct
Answer:
(a) ns → (n – 2)f → (n – l)d → np
Solution:
n = 6 According Aufbau principle,
6s → 4f → 5d → 6p
ns → (n – 1)f → (n – 2)d → np.

Question 17.
Consider the following sets of quantum numbers:

Which of the following sets of quantum number is not possible ?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (z) and (iii)
(d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)
Solution:
(ii) l can have the values from 0 to n – 1 n = 2; possible l values are 0, 1 hence l = 2 is not possible.
(iv) for l = 0; m = -1 not possible
(v) for n = 3 l = 4 and m = 3 not possible.

Question 18.
How many electrons in an atom with atomic number 105 can have (n + 1) = 8 ?
(a) 30
(6) 17
(c) 15
(d) unpredictable
Answer:
(b) 17
Solution:
n + 1 = 8
Electronic configuration of atom with atomic number 105 is [Rn] 5f¹⁴ 6d³ 7s²

Question 19.
Electron density in the yz plane of 3 d_x²_-y² orbital is …………….
(a) zero
(b) 0.50
(c) 0.75
(d) 0.90
Answer:
(a) zero

Question 20.
If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is ……….
(a) \(\frac{1}{m} \sqrt{\frac{h}{\pi}}\)
(b) \(\sqrt{\frac{\mathrm{h}}{\pi}}\)
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
(d) \(\frac{\mathrm{h}}{4 \pi}\)
Answer:
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
Solution:

Question 21.
A macroscopic particle of mass 100 g and moving at a velocity of 100 cm s^-1d will have a de Broglie wavelength of ………….
(a) 6.6 x 10^-29 cm
(b) 6.6 x 10^-30 cm
(c) 6.6 x 10^-31 cm
(d) 6.6 x 10^-32 cm
Answer:
(c) 6.6 x 10^-31 cm
Solution:
m = 100 g = 100 x 10^-3 kg
v = 100 cm s^-1 = 100 x 10^-2 m s^-1
λ = \(\frac{h}{mv}\) =
= 6.626 x 10^-31 ms^-1
= 6.626 x 10^-31 cm s^-1

Question 22.
The ratio of de Broglie wavelengths of a deuterium atom to that of an a – particle, when the velocity of the former is five times greater than that of later, is ……………
(a) 4
(b) 0.2
(c) 2.5
(d) 0.4
Answer:
(d) 0.4

Question 23.
The energy of an electron in the 3rd orbit of hydrogen atom is -E. The energy of an electron in the first orbit will be ……………..
(a) – 3E
(b) – E /3
(c) – E / 9
(d) – 9E
Answer:
(c) – E / 9
Solution:
E_n  = \(\frac{-13.6}{n^{2}}\) eV atom^-1
E₁ = \(\frac{-13.6}{1^{2}}\)13.6 = \(\frac{-13.6}{9}\)
Given that,
E₃ = – E
\(\frac{-13.6}{9}\) = -E
13.6 = – 9E = E₁ = – 9E
E₁ = – 9E

Question 24.
Time independent Schnodinger wave equation is ………….

Answer:
(a) \(\widehat{\mathrm{H}} \psi=\mathrm{E} \psi\).

Question 25.
Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?
(a) ∆E.∆p ≥ h/4π
(b) ∆E.∆v ≥ h/4πm
(c) ∆E.∆t ≥ h/4π
(d) ∆E.∆x ≥ h/4π
Answer:
(d) ∆E.∆x ≥ h/4π.

II. Write brief answer to the following questions

Question 26.
Which quantum number reveal information about the shape, energy, orientation and size of orbitals?
Answer:
Magnetic quantum number reveal information about the shape, energy, orientation and size of orbitals.

Question 27.
How many orbitals are possible for n =4?
Answer:
If n = 4, the possible number of orbitals are calculated as follows –
n = 4, main shell = N
If n = 4, l values are 0, 1, 2, 3
If l = 0,  4s orbital = 1 orbital
If l = 1,  m = -1,0, +1 = 3 orbitals
If l = 2,  m = -2,-1,0, +1,+2 = 5 orbitals
If l = 3,  m = -3,-2,-1,0, +1,+2,+3 = 7 orbitals
∴ Total number of orbitals = 16 orbitals

Question 28.
How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
Answer:
Formula for total number of nodes = n – 1

1. For 2s orbital: Number of radial nodes =1.

2. For 4p orbital: Number of radial nodes = n – l – 1. = 4 – 1 – 1 = 2
Number of angular nodes = l
∴ Number of angular nodes = 1
So, 4p orbital has 2 radial nodes and 1 angular node.

3. For 5d orbital:
Total number of nodes = n – 1 = 5 – 1 = 4 nodes
Number of radial nodes = n – l – 1 = 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴ 5d orbital have 2 radial nodes and 2 angular nodes.

4. For 4f orbital:
Total number of nodes = n – 1 = 4 – 1 = 3 nodes
Number of radial nodes = n – 7 – 1 = 4 – 3 – 1 = 0 node.
Number of angular nodes = l = 3 nodes
∴ 4f orbital have 0 radial node and 3 angular nodes.

Question 29.
The stabilization of a half filled d-orbital is more pronounced than that of the p-orbital why?
Answer:
The exactly half filled orbitals have greater stability. The reason for their stability are –

1. symmetry
2. exchange energy.

(1) Symmetry:
The half filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability.

(2) Exchange energy:
The electrons with same spin in the different orbitals of the same sub shell can exchange their position. Each such exchange release energy and this is known as exchange energy. Greater the number of exchanges, greater the exchange energy and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, d – orbital with 5 unpaired electrons (10 exchanges)n i.e. half filled is more stable than p – orbital with 3 unpaired electrons (6 exchanges).

Question 30.
Consider the following electronic arrangements for the d5 configuration.

(1) Which of these represents the ground state
(2) Which configuration has the maximum exchange energy.
Answer:
(1)  – This represents the ground state.
(2)   – This represents the maximum exchange energy.

Question 31.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) 1s¹.
One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + \(\frac {1}{2}\). For helium Z = 2. He: 1s². In this one electron has the quantum number same as that of hydrogen, n = 1,l = 0, m = 0 and s = +½ For other electron, fourth quantum number is different, i.e. n = 1, l = 0, m = 0 and s = – ½.

Question 32.
Define orbital? What are the n and l values for 3p_x and 4 d_x²_-y² electron?
Answer:
(i) Orbital is a three dimensional space which the probability of finding the electron is maximum.

Question 33.
Explain briefly the time independent Schrodinger wave equation?
Answer:
The time independent Schrodinger equation can be expressed as
\(\widehat{\mathrm{H}}\) Ψ = EΨ ……………(1)
Where \(\widehat{\mathrm{H}}\) is called Hamiltonian operator.
Ψ is the wave function.
E is the energy of the system.

Since Ψ is a function of position coordinates of the particle and is denoted by Ψ (x, y, z)
∴ Equation (1) can be written as,

Multiply the equation (3) by \(\widehat{\mathrm{H}}\) and rearranging

The above equation (4) Schrodinger wave equation does not contain time as a variable and is referred to as time independent Schrodinger wave equation.

Question 34.
Calculate the uncertainty in position of an electron, if ∆v = 0.1% and n = 2.2 x 10⁶ ms^-1.
Answer:
Mass of an electron = m = 9.1 x 10^-31 kg.
∆v = Uncertainty in velocity = \(\frac {0.1}{100}\) x 2.2 x 10³ ms^-1 .
∆v = 0.22 x 10⁴ = 2.2 x 10³ ms^-1
∆x . ∆v . m = \(\frac {h}{4π}\)
∆x = \(\frac {h}{∆v . m x 4π}\)

= 0.02635 x 10^-6
∆x = 2.635 x 10^-8
Uncertainty in position = 2.635 x 10^-8.

Question 35.
Determine the values of all the four quantum numbers of the 8th electron in O – atom and 15^th electron in Cl atom and the last electron in chromium.
Answer:
(1) O (Z = 8) 1s² 2s² 2p_x² 2p_y¹ 2p_z¹
Four quantum numbers for 2p_x¹ electron in oxygen atom:
n = principal quantum number = 2
l = azimuthal quantum number =1
m = magnetic quantum number =+1
s = spin quantum number = +\(\frac {1}{2}\)

(2) Cl (Z = 17) 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹
Four quantum numbers for 15^th electron in chlorine atom:
n = 3, l = 1, m = 0, s = + ½

(3) Cr (Z = 24) 1s² 2s² 2p² 3s² 3p² 3d² 4s¹
n = 3, l = 2, m = +2, s = + ½

Question 36.
The quantum mechanical treatment of the hydrogen atom gives the energy value:
E_n = \(\frac{-13.6}{n^{2}}\) eV atom^-1

1. use this expression to find ∆E between n = 3 and n = 4
2. Calculate the wavelength corresponding to the above transition.

Answer:
(1) When n = 3
E₃ = \(\frac{-13.6}{3^{2}}\) = \(\frac {-13.6}{9}\) = – 1.511 eV atom^-1
When n = 4 E₄ = \(\frac{-13.6}{4^{2}}\) = – 0.85 eV atom^-1
∆E = E₄ – E₃ = – 0.85 – (-1.511) = + 0.661 eV atom
∆E = E₃ – E₄
= – 1.511 – (-0.85)
= – 0.661 eV atom^-1

(2) Wave length = λ
∆E = \(\frac {hc}{ λ}\)
λ = \(\frac {hc}{∆E}\)
h = Planck’s constant = 6.626 x 10^-34 Js^-1
c = 3 x 10⁸ m/s
λ = \(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.661}\)
= 10.02 x 10^-34 x 3 x 10⁸
= 30 x 10^-26
λ = 3 x 10^-25 m.

Question 37.
How fast must a 54 g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400 Å?
Answer:
m = mass of tennis ball = 54 g = 5.4 x 10^-2 kg.
λ = de Broglie wavelength = 5400 Å. = 5400 x 10^-10 m.
V = velocity of the ball = ?
λ = \(\frac {h}{mV}\)
V = \(\frac {h}{λ.m}\)

= 0.2238 x 10^-24
= 2.238 x 10^-25 m.

Question 38.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals.

1. n = 4, l = 2,
2. n = 5, l = 3
3. n = 7, l = 0

Answer:
1. n = 4, l = 2
If l = 2, ‘m’ values are -2, -1, 0, +1, +2
So, 5 orbitals such as d_xy,d_yz,d_xz,\(d_{x^{2}-y^{2}}\) and d^_z

2. n = 5 , l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as f z, fxz, fyz, fxyz, fz(x2 y2)’ ^x(x2-3y2)’ ^y(3×2 ??y

3. n = 7 , l = 0
If l = 0, ‘m’ values are 0. Only one value.
So, 1 orbital such as 7s orbital.

Question 39.
Give the electronic configuration of Mn²⁺ and Cr³⁺
Answer:
1. Mn (Z = 25)
Mn → Mn²⁺ + 2e^–
Mn²⁺ electronic configuration is 1s 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

2. Cr (Z = 24)
Cr → Cr³⁺ + 3e^–
Cr³⁺ electronic configuration is Is² 2s² 2p⁶ 3s²3p⁶ 3d³

Question 40.
Describe the Aufbau principle.
Answer:
In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals.
The order of filling of various orbitals as per Aufbau principle is –
1 s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d ………..
For e.g., K (Z =19)
The electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
After filling 4s orbital only we have to fill up 3d orbital.

Question 41.
A n atom of an element contains 35 electrons and 45 neutrons. Deduce

1. the number of protons
2. the electronic configuration for the element
3. All the four quantum numbers for the last electron

Answer:
An element X contains 35 electrons, 45 neutrons

1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m =+1, s = + \(\frac {1}{2}\)

Question 42.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving around the nucleus.
Answer:
In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron wave is out of phase.
mvr = nh / 2π, 2πr = nλ,
where mvr = angular momentum
where 2πr = circumference of the orbit

n = 3, n = 4

Question 43.
Calculate the energy required for the process.
He⁺_(g) → He²⁺_(g) + e^–
The ionization energy for the H atom in its ground state is – 13.6 eV atom^-1.
Answer:
The ionization energy for the H atom in its ground state =-13.6 eV atom^-1.
Ionization energy = \(\frac{13.6 z^{2}}{n^{2}}\) eV
Z = atomic number
n = principal quantum number or shell number
For He, n = 1, z = 2
IE = \(\frac{-13.6 \times 2^{2}}{1^{2}}\)eV.

Question 44.
An ion with mass number 37 possesses unit negative charge. If the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion.
Answer:
Let the number of electrons in an ion = x
number of neutrons = n = x + \(\frac{11.1}{100}\) eV = 1.111 x
(As the number of neutrons are 11.1% more than the number of electrons)
In the neutral of atom, number of electron.
e^– = x – 1 (as the ion carries -1 charge)
Similarly number of protons = P = x – 1 Number of protons + number of neutrons = mass number = 37
(x – 1) + 1.111 x = 37 .
2.111 x = 37 +1
2.111 x = 38
x = \(\frac{38}{2.111}\) = 18.009 = 18
∴ Number of protons = atomic number – 1 = 18-1 = 17
∴ The symbol of the ion = \(_{17}^{37} \mathrm{Cl}\).

Question 45.
The Li²⁺ ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4^th orbit.
Answer:
Li²⁺ hydrogen like ion.
Bohr radius of the third orbit = r₃ = ?
r₃ = \(\frac{(0.529) n^{2}}{Z}\) A
Where n = shell number, Z = atomic number.
r₃ = \(\frac{(0.529) 3^{2}}{3}\) A [∴for lithium Z = 3, n = 3]
= \(\frac{0.529 x 9}{3}\)
r₃ = l.587Å
E_n = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom^-1.
E₄ = Energy of the fourth orbit = ?
E₄ = \(\frac{(-13.6) \times 3^{2}}{4^{2}}\) = \(\frac{-13.6 \times 9}{16}\) = -7.65 eV atom^-1
E₄ = – 7.65 eV atom^-1

Question 46.
Protons can be accelerated in particle accelerators. Calculate the wavelength (in Å)of such accelerated proton moving at 2.85 × 108 ms^-1 (the mass of proton is 1.673 x 10^-27 Kg).
Answer:
m = mass of the proton = 1.673 x 10^-27 Kg
v = velocity of the proton = 2.85 x 10⁸ ms^-1
λ = \(\frac{h}{mv}\)
h = Planck’s constant = 6.626 x 10³⁴ Kg m² s^-1

Wavelength of proton = λ = 1.389 x 10^-15 m.

Question 47.
What is the de Broglie wavelength (in cm) of a 160g cricket ball travelling at 140 Km hr^-1.
Answer:
m = mass of the cricket ball = 160g = 0.16 kg.
v = velocity of the cricket ball =140 Km h^-1
= \(\frac {140 x 5}{18}\) = 38.88 ms^-1
de Broglie equation = λ = \(\frac{h}{mv}\)
h = Planck’s constant = 6.626 x 10^-34 kg m² s^-1

λ = 1.065 x 10^-34m
Wave length in cm = 1.065 x 10^-34 x 100
= 1.065 x 10^-32 cm.

Question 48.
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 A. What is the uncertainty in its momentum?.
Answer:
∆x = uncertainty in position of an electron = 0.6 Å = 0.6 x 10^-10 m.
∆p = uncertainty in momentum = ?
Heisenberg’s uncertainty principle states that,
∆x.∆p ≥ \(\frac{h}{4π}\)
∆p = \(\frac{h}{4π.∆x}\)
h = Planck’s constant = 6.626 x 10^-34 kg m² s^-1

Uncertainty in momentum = 0.8792 x 10^-24 kg ms^-1 (or) = 8.792 x 10^-25 kg ms^-1

Question 49.
Show that if the measurement of the uncertainty in .the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity is equal to its velocity /4π
Answer:
If, uncertainty in position = ∆x = λ , the value of uncertainty in velocity = \(\frac{v}{4π}\)
Heisenberg’s principle states that
∆x.∆v. m = \(\frac{h}{4π}\) …………(1)
de Broglie equation states that
λ = \(\frac{h}{mv}\) ………….(2)
∴ h = λ .m.v …………(3)
∆x = \(\frac{h}{∆v.4π}\) ………….(4)
Substituting the value of h in equation (4)
∆x = \(\frac{λ x m. v}{∆v.4π.m}\)
if ∆x = λ
∆v =  = \(\frac{v}{4π}\)

Question 50.
What is the de Broglie wave length of an electron, which is accelerated from the rest, through a potential difference of 100V?
Answer:
Potential difference = V = 100 V
Potential energy = eV = 1.609 x 10^-19c x 100V
\(\frac{v}{4π}\) m v² = 1.609 x 10^-19 x 100
\(\frac{v}{4π}\) m v² = 1.609 x 10^-19 J
v² = \(\frac{2 \times 1.609 \times 10^{-17}}{m}\)
m = mass of electron = 9.1 x 10^-31 Kg

v = 5.93 x 10⁶ m/s
λ = \(\frac{h}{mv}\) where h = 6.62 x 10^-34 JS
= \(\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^{6}}\)
= 1.2x 10^-10m
A= 1.2 Å.

Question 51.
Identify the missing quantum numbers and the sub energy level

Answer:

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom In-Text Questions – Evaluate Yourself

Question 1.
Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 k eV.
Answer:
λ = \(\frac{h}{mv}\)
Potential difference of an electron = V = 1 keV.
Potential energy = \(\frac{1}{2}\) mv² = eV
e = charge of an electron = 1.609 x 10^-19c
l k V = 1000 V
:. Potential energy = 1.609 x 10^-19 x 1000 = 1.609 x 10^-19
\(\frac{1}{2}\) mv² = 1.609 x 10^-16V
m = 9.1 x 10^-31 kg
λ = \(\frac{h}{mv}\)

= 1.2 x 10^-11 m
λ = 1.2 x 10^-11 m.

Question 2.
Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 x 10 s ms^-1.
Answer:
Uncertainty in velocity = Av = 5.7 x 10⁵ ms^-1
Mass of an electron = m = 9.1 x 10^-31 kg.
Uncertainty in position = ∆x = ?
∆x.m.∆v = \(\frac{h}{4π}\)

= 1 x 10^-10m.
Uncertainty in position = 1 x 10^-10 m.

Question 3.
How many orbitals are possible in the 4th energy level? (n = 4)
Answer:
n = 4
Number of orbitals in 4^th energy level = ?
When n = 4, l = 0,1,2,3
If l = 0 orbital = 4s =1
If l = 1 orbital = 4p_x, 4p_y, 4p_z = 2
If l = 2 orbital = \(4 \mathrm{d}_{\mathrm{xy}}, 4 \mathrm{d}_{\mathrm{yz}}, 4 \mathrm{d}_{\mathrm{zx}}, 4 \mathrm{d}_{\mathrm{x}} 2_{\mathrm{y}}, 2,4 \mathrm{d}_{\mathrm{z}^{2}}\) = 5
If l = 3 orbital = -3,-2, -1, 0, +1, +2, +3 = 7
Number of orbitals in 4th energy level = 16.

Question 4.
Calculate the total number of angular nodes and radial nodes present in 3d and 4f orbitals.
Answer:
Number of angular nodes in 3d orbital = ?
Number of radial nodes in 3d orbital = ?
Number of angular nodes = l
Number of radial nodes = n – l – 1

1. For 3d orbital:
Number of angular nodes = 2 because l = 2
Number of radial nodes = 3 – 2 -1 = 0
Total number of nodes in 3d orbital = 2

2. For 4f orbital:
Number of angular nodes = 3 because l = 3
Number of radial nodes = n – l – l =4 – 3 – 1 = 0
Total number of nodes in 4f orbital = 3.

Question 5.
Energy of an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the second excited state?
Answer:
Energy of an electron in ground state = -13.6 eV.
∴ Energy of an electron in the second excited state = E₂.
n = 2
E₂ = \(\frac{-13.6 \mathrm{eV}}{\mathrm{n}^{2}}\) = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 eV.

Question 6.
How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn2+ (z = 25) and argon (z=18)?
Answer:

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

Question 7.
Explain the meaning of the symbol 4f². Write all the four quantum numbers for these electrons.
Answer:
4f² : It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac {1}{2}\) – \(\frac {1}{2}\).

Question 8.
Which has the stable electronic configuration? Ni²⁺ or Fe³⁺
Answer:
Ni (Z = 28). 1s² 2s² 2p⁶ 3s² 3p⁶4s²3d⁸
Ni²⁺ electronic configuration = Is² 2s²2p⁶ 3s² 3p⁶ 3d⁸
Fe (Z = 26). 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Fe³⁺ Is² 2s² 2p⁶ 3s²3p⁶ 3d⁵
If d orbital is half filled, according to Aufbau principle, it is more stable. So Fe³⁺ is more stable than Ni²⁺.

Samacheer Kalvi 11th Chemistry Solutions Quantum Mechanical Model of Atom Additional Questions Solved

I. Choose the correct answer

Question 1.
Which of the following experiment proves the presence of an electron in an atom?
(a) Rutherford’s α-ray scattering experiment
(b) Davisson and Germer experiment
(c) J.J. Thomson cathode ray experiment
(d) G.R Thomson gold foil experiment
Answer:
(c) J.J. Thomson cathode ray experiment.

Question 2.
Consider the following statements regarding Rutherford’s α-ray scattering experiment.
i. Most of the α-particles were deflected through a small angle.
ii. Some of α-particles passed through the foil.
iii. Very few α-particles were reflected back by 180°.
Which of the above statements is/are not correct.
(a) i and ii
(b) ii and iii
(c) i and iii
(d) i ii and iii
Answer:
(a) i and ii.

Question 3.
Considering Bohr’s model which of the following statements is correct?
(a) The energies of electrons are continuously reduced in the form of radiation.
(b) The electron is revolving around the nucleus in a dynamic orbital.
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2 π.
(d) In an atom, electrons are embedded like seeds in watermelon.
Answer:
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π.

Question 4.
The energy of an electron of hydrogen atom in 2nd main shell is equal to
(a) – 13.6 eV atom^-1
(b) – 6.8 eV atom^-1
(c) – 0.34 eV atom^-1
(d) – 3.4 eV atom^-1
Answer:
(d) -3.4 eV atom^-1
Hints:
Energy of an electron in 2nd main shell = \(\frac{(-13.6) Z^{2}}{n^{2}}\); Z = 1, n = 2
E = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 atom^-1.

Question 5.
The energy of an electron of Li²⁺ in the 3^rd main shell is …………..
(a) – 1.51 eV atom^-1
(b) – 6.8 eV atom^-1
(c) + 1.51 eV atom^-1
(d) – 3.4 eV atom^-1
Answer:
(a) -1.51 eV atom^-1
Hints:
E = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom^-1
Li²⁺= H atom. So Z = 1, n = 3.
E = \(\frac{(-13.6) 1^{2}}{3^{2}}\) = \(\frac{-13.6}{9}\) = -1.51 eV atom^-1

Question 6.
The energy of an electron of hydrogen atom in main shell in terms of U mold is
(a) – 1312.8 k J mol^-1
(b) – 82.05 k J mol^-1
(c) – 328.2 kJ mol^-1
(d) – 656.4 k J mol^-1
Answer:
(b) – 82.05 k J mol^-1
Hints:
E = \(\frac{(-13.6) Z^{2}}{n^{2}}\) kJ mol^-1 , Z = 1, n = 4
∴ E = \(\frac{-1312.8}{16}\) = -82.50 kJ mol^-1

Question 7.
The Bohr’s radius of Li2 0f21d orbit is
(a) 0.529 Å
(b) 0.0753 Å
(c) 0.7053 Å
(d) 0.0529 Å
Answer:
(c) 0.7053 Å
Hints:
r_n = \(\frac{(0.529) Z^{2}}{n^{2}}\)Å, n = 2, Z = 3(for Li²⁺)
r = \(\frac{(0.529) 3^{2}}{2^{2}}\) = \(\frac{0.529 x 4}{3}\) = 0.7053 Å.

Question 8.
The formula used to calculate the Boh’s radius is ………..
(a) r_n = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom^-1
(b) r_n = \(\frac{(0.529) Z^{2}}{n^{2}}\) A
(c) r_n= \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol^-1
(d) r_n = \(\frac{(+1312.8) Z^{2}}{n^{2}}\) kJ mol^-1
Answer:
(b) r_n = \(\frac{(0.529) Z^{2}}{n^{2}}\) A.

Question 9.
Who proposed the dual nature of light to all forms of matter?
(a) John Dalton
(b) Neils Bohr
(c) Albert Einstein
(d) J.J. Thomson
Answer:
(c) Albert Einstein

Question 10.
dc Brogue equation is ………..
(a) E = h γ
(b) E = mc²
(c) γ = \(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
(d) λ = \(\frac{h}{mv}\)
Answer:
(d) λ = \(\frac{h}{mv}\).

Question 11.
The crystal used in Davison and Germer experiment is …………….
(a) nickel
(b) zinc suiphide
(c) gold foil
(d) NaCl
Answer:
(a) nickel.

Question 12.
Which one of the following is the time independent Schrodinger wave equation?

Answer:

Question 13.
Match the list-I and list-II correctly using the code given below the list.
List – I
A. Principal quantum number
B. Azimuthal quantum number
C. Magnetic quantum number
D. Spin quantum number

List – II
1. represents the directional orientation of orbital
2. represents the spin of the electron
3. represents the main shell
4. represents the sub shell

Answer:

Question 14.
The maximum number of electrons that can be accommodated in N shell is …………..
(a) 8
(b) 18
(c) 32
(d) 36
Answer:
(c) 32
Hints:
Number of electrons in the main shell = 2n² n = 4, for N shell.
∴ Maximum number of electrons in N shell = 2(4)² = 32.

Question 15.
The maximum number of electrons that can be accommodated in f orbital is ………….
(a) 10
(b) 14
(c) 16
(d) 6
Answer:
(6) 14
Hints:
forbital – l = 3.
Maximum number of electrons in sub shell = 2(2l + 1)
∴ For ‘f’ orbital, the maximum number of electrons = 2(2 x 3 + l) = 14.

Question 16.
When l = 0, the number of electrons that can be accommodated in the sub shell is ……………..
(a) 0
(b) 2
(c) 6
(d) 8
Answer:
(b) 2
Hints:
If l = 0, number of electrons = (2l + 1)
= 2 (2 x 0 + 1) = 2.

Question 17.
Which one of the quantum number is used to calculate the angular momentum of an atom?
(a) n
(b) m
(c) l
(d) s
Answer:
(c) l

Question 18.
What is the formula used to calculate the angular momentum?
(a) \(\sqrt{l^{(l+1)}} \frac{h}{2 \pi}\)
(b) \(\frac{\mathrm{mvr}}{2 \pi}\)
(c) \(\frac{mvr}{2}\)
(d) m . ∆v
Answer:
(a) \(\sqrt{l^{(l+1)}} \frac{h}{2 \pi}\).

Question 19.
Which of the following provides the experimental justification of magnetic quantum number?
(a) Zeeman effect
(b) Stark effect
(c) Uncertainty principle
(d) Quantum condition
Answer:
(a) Zeeman effect.

Question 20.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1, 0
(b) 3, 1,-l, +½
(c) 3, 2, +1, -½
(d) 3, 0, 0, +½]
Answer:
(b) 3, 1, -1, +½
Hint:
3p_x electron ; n = 3 (main shell)
for p_x orbitaI, l = 1, m = -1, s = \(\frac {1}{2}\).

Question 21.
Identify the quantum number for \(4 d_{x^{2}-y^{2}}\) electron.
(a) 4, 2, -2, +½
(b) 4, 0, 0, +½
(c) 4, 3, 2, +½
(d) 4, 3, 2, -½
Answer:
(a) 4, 2, -2, +½.

Question 22.
How many orbitals are possible in 3^rd energy level?
(a) 16
(6) 9
(c) 3
(d) 27
Answer:
(b) 9
Hints:
3^rd energy level Number of orbitals = ?
n = 3 main shell = m
l = 0, 1,2 m = 0, -1,0, +1
Total = 9 orbitals.

Question 23.
The region where the probability density function of electron reduces to zero is called
(a) orbit
(b) orbital
(c) nodal surface
(d) sub shell
Answer:
(c) nodal surface.

Question 24.
Consider the following statements.
(i) The region where the probability density of electron is zero, called nodal surface.
(ii) The probability of finding the electron is independent of the direction of the nucleus.
(iii) The number of radial nodes is equal to n + l + 1 Which of the above statements is/are correct?
(a) (i) and (iii)
(b) (i) and (ii)
(c) (iii) only
(d) (ii) and (iii)
Answer:
(b) (i) and (ii).

Question 25.
Match the list-I and list-II correctly using the code given below the list.
List-I
A. s – orbital
B. p – orbital
C. d – orbital
D. f – orbital

List-II
1. complex three-dimensional shape
2. symmetrical sphere
3. dumb-bell shape
4. clover leaf shape

Answer:

Question 26.
Which one of the following is the correct increasing order of effective nuclear charge felt by an electron?
(a) s>p>d>f
(b) s<p<d<f
(c) s>p>f>d
(d) f<p<d<s
Answer:
(a) s>p>d>f.

Question 27.
The value of n, l, m and s of 8th electron in an oxygen atom are respectively
(a) 1, 0, 0, + ½
(b) 2, 1, +1, – ½
(c) 2, 1, -1, – ½
(d) 2, 1, 0, +½
Answer:
(a) 2, 1, +1, – ½.

Question 28.
The number of impaired electrons in carbon atom in the gaseous state is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question 29.
Which of the following is not used in writing electronic configuration of an atom?
(a) Aufbau principle
(b) Hund’s rule
(c) Pauli’s exclusion principle
(d) Heisenberg’s uncertainty principle
Answer:
(d) Heisenberg’s uncertainty principle.

Question 30.
Which of the following is the expected configuration of Cr (Z = 24)?
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s³
Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²

Question 31.
Which of the following is the actual configuration of Cr (Z = 24)?
(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(d) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s³
Answer:
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Question 32.
Assertion (A) : Cr with electronic configuration [Ar] 3d⁵ 4s¹ is more stable than [Ar] 3d⁴ 4s¹.
Reason(R ): Half filled orbitals have been found to have extra stability than partially filled orbitals.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct the explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 33.
Assertion (A): Copper (Z = 29) with electronic configuration [Ar] 4s¹ 3d¹⁰ is more stable than [Ar] 4s¹ 3d¹⁰.
Reason(R): Copper with [Ar] 4s² 3d⁹ is more stable due to symmetrical distribution and exchange energies of d electrons.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 34.
In a sodium atom (atomic number = 11 and mass number = 23) and the number of neutrons is …………..
(a) equal to the number of protons
(b) less than the number of protons
(c) greater than the number of protons
(d) none of these
Answer:
(c) greater than the number of protons.

Question 35.
The idea of stationary orbits was first given by …………
(a) Rutherford
(b) J.J. Thomson
(c) Nails Bohr
(d) Max Planck
Answer:
(c) Niels Bohr.

Question 36.
de Broglie equation is ……………
(a) λ = \(\frac {h}{mv}\)
(b) λ = \(\frac {hv}{m}\)
(c) λ = \(\frac {mv}{h}\)
(d) λ = hmv
Answer:
(a) λ = \(\frac {h}{mv}\).

Question 37.
The orbital with n = 3 and l = 2 is …………..
(a) 3s
(b) 3p
(c) 3d
(d) 3J
Answer:
(c) 3d

Question 38.
The outermost electronic configuration of manganese (at. no. = 25) is …………
(a) 3d⁵ 4s²
(b) 3d⁶ 4s¹
(c) 3d⁷ 4s⁰
(d) 3d⁶ 4s²
Answer:
(a) 3d⁵ 4s²

Question 39.
The maximum number of electrons in a sub-shell is given by the equation
(a) n²
(b) 2 n²
(c) 2 l – l
(d) 2 l + 1
Answer:
(d) 2 l + 1

Question 40.
Which of the following statements is correct for an electron that has the quantum numbers n = 4 and m = -2.
(a) The electron may be in 2 p orbital
(b) The electron may be in 4 d orbital
(c) The electron is in the second main shell
(d) The electron must have spin quantum number as +\(\frac {1}{2}\).
Answer:
(b) The electron may be in 4d orbital.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 2 – Marks Questions

Question 1.
Write a note about J.J. Thomson’s atomic model.
Answer:

• J.J. Thomson’s cathode ray experiment revealed that atoms consist of negatively charged particles called electrons.
• He proposed that atom is a positively charged sphere in which the electrons are embedded like the seeds in the watermelon.

Question 2.
Explain about theory of electromagnetic radiation.
Answer:

• The theory of electromagnetic radiation states that a moving charged particle should continuously loose its energy in the form of radiation.
• O Therefore, the moving electron in an atom should continuously loose its energy and finally collide with nucleus resulting in the collapse of the atom.

Question 3.
Explain how matter has dual character?
Answer:

• Albert Einstein proposed that light has dual nature, i.e. like photons behave both like a particle and as a wave.
• Louis de Broglie extended this concept and proposed that all forms of matter showed dual character.
• He combined the following two equations of energy of which one represents wave character (hυ) and the other represents the particle nature (mc²).

Question 4.
Explain about the significance of de Broglie equation.
Answer:

• X = \(\frac {h}{mv}\). This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
• For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
• For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
• For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 5.
How many electrons can be accommodated in the main shell l, m and n?
Answer:

Question 6.
How many electrons that can be accommodated in the sub shell s, p, d, f ?
Answer:

Question 7.
What are quantum numbers?
Answer:

• The electron in an atom can be characterized by a set of four quantum numbers, namely principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s).
• When Schrodinger equation is solved for a wave function T, the solution contains the first three quantum numbers n, l
• and m.
• The fourth quantum number arises due to the spinning of the electron about its own axis.

Question 8.
How many orbitals are possible in the 3rd energy level?
Answer:
n = 3, main shell is m.
Total number of orbitals in 3rd energy level = ?

Total number of orbitals = 9.

Question 9.
What are Ψ and Ψ² ?
Answer:

• Ψ itself has no physical meaning but it represents an atomic orbital.
• Ψ² is related to the probability of finding the electrons within a given volume of space.

Question 10.
What is meant by nodal surface?
Answer:

• The. region where there is probability density function reduces to zero is called nodal surface or a radial node.
• For ns orbital, (n-1) nodes are found in it.

Question 11.
Mention the shape of s, p, d orbitals.
Answer:

• Shape of s – orbital – sphere
• Shape of p – orbital – dumb bell
• Shape of d – orbital – clover leaf

Question 12.
Calculate the total number of angular nodes and radial nodes present in 4p and 4d orbitals.
Answer:
1. For 4p orbital:
Number of angular nodes = l
For 4p orbital 7 = l
Number of angular nodes = l
Number of radial nodes = n – l – 1 = 4 -1 -1 = 2
Total number of nodes = n -1 = 4 – 1 = 3
1 angular node and 2 radial nodes.

2. For 4d orbital:
Number of angular nodes = l
For 4d orbital l = 2
Number of angular nodes = 2
Number of radial nodes = n – l – 1 = 4 – 2 – 1 = 1
Total number of nodes = n – l = 4 – l = 3
1 radial nodes and 2 angular node.

Question 13.
Write the equation to calculate the energy of nth orbit.
Answer:
E_n = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol^-1
Where Z = atomic number, n = principal quantum number.

Question 14.
what are degenerate orbitals?
Answer:

• Three different orientations in space that are possible for a p-orbital. All the three p-orbitals, namely p_x, p_y and p_z have same energies and are called degenerate orbitals.
• In the presence of magnetic or electric field, the degeneracy is lost.

Question 15.
Energy of an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the third excited state?
Answer:
E₁ = – 13.6 eV
E₃ = \(\frac{-13.6}{n^{2}}\) Where n = 3
E₃ = \(\frac{-13.6}{9}\) = 1.511 eV
Energy of the electron in the third excited state = 1.511 eV.

Question 16.
The energies of the same orbital decreases with an increase in the atomic number. Justify this statement.
Answer:
The energy of the 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on because H (Z =1), Li (Z = 3) and Na (Z = 11). When atomic number increases, the energies of the same orbital decreases. E_2s(H) > E_2s(Li) > E_2s(Na) > E_2s(K) ………….

Question 17.
State Hund’s rule of maximum multiplicity.
Answer:
It states that electron pairing in the degenerate orbitals does not take place until all the available orbitals contain one electron each.

Question 18.
How many unpaired electrons are present in the ground state of –
1. Cr³⁺ (Z = 24)
2. Ne (Z = 10)
Answer:
1. Cr³⁺ (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.

2. Ne (Z = 10) 1s²2s²2p⁶. No unpaired electrons in it.

Question 19.
What is meant by electronic configuration? Write the electronic configuration of N (Z = 7).
Answer:
The distribution of electrons into various orbitals of an atom is called its electronic configuration.

Question 20.
Which is the actual configuration of Cr (Z = 24) Why?
Answer:
Cr (Z = 24) 1s²2s²2p⁶.
The reason for this is, Cr with 3d⁵ configuration is half filled and it will be more stable. Chromium has [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s² due to the symmetrical distribution and exchange energies of d electrons.

Question 21.
What is the actual configuration of copper (Z = 29)? Explain about its stability.
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability. Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons. Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 3 – Mark Questions

Question 1.
What are the conclusions of Rutherford’s α – rays scattering experiment?
Answer:

• Rutherford bombarded a thin gold foil with a stream of fast moving α – particles.
• It was observed that most of the a-particles passed through the foil.
• Some of them were deflected through a small angle.
• Very few α- particles were reflected back by 180°.
• Based on these observations, he proposed that in an atom, there is a tiny positively charged nucleus and the electrons are moving around the nucleus with high speed.

Question 2.
What are the limitations of Bohr’s atom model?
Answer:

• The Bohr’s atom model is applicable only to species having one electron such as hydrogen, Li²⁺ etc and not applicable to multi – electron atoms.
• It was unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
• Bohr’s theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to nh / 2π.

Question 3.
Illustrate the significance of de Broglie equation with an iron ball and an electron.

1. 6.626 kg iron ball moving with 10 ms^-1.
2. An electron moving at 72.73 ms^-1.

Answer:
1. λ_{iron ball} = \(\frac{h}{mv}\)

= 1 x 10^-35m

2. λ_{iron ball} = \(\frac{h}{mv}\)

= \(\frac{6.626}{662.6}\) x 10^-3m = 1 x 10⁵m
For an electron, the de Broglie wavelength is significant and measurable while for iron ball it is too small to measure, hence it becomes insignificant.

Question 4.
Explain Davisson and Germer experiment.
Answer:

• The wave nature of electron was experimentally confirmed by Davisson and Germer.
• They allowed the accelerated beam of electrons to fall on a nickel crystal and recorded the diffraction pattern.
• The resultant diffraction pattern is similar to the X – ray diffraction pattern.
• The finding of wave nature of electron leads to the development of various experimental ’ techniques such as electron microscope, low energy electron diffraction etc.

Question 5.
Bohr radius of 1st orbit of hydrogen atom is 0.529 Å. Assuming that the position of an electron in this orbit is determined with the accuracy of 0.5% of the radius, calculate the uncertainty in the velocity of the electron in hydrogen atom.
Answer:
Uncertainty in position = ∆x
= \(\frac{0.5}{100}\) x 0.529 Å
= \(\frac{0.5}{100}\) x 10^-10 x 0.529 m
∆x = 2.645 x 10^-13 m
From Heisenberg’s uncertainty principle,
∆x.∆p ≥ \(\frac{h}{4π}\)
∆x.m.∆p ≥ \(\frac{h}{4π}\)
∆v ≥ \(\frac{h}{∆x.m.4π}\)
∆v =
∆v = 2.189 x 10⁸m.

Question 6.
Write a note about principal quantum number.
Answer:

• The principal quantum number represents the energy level in which electron revolves around the nucleus and is denoted by the symbol ‘n’.
• The ‘n’ can have the values 1, 2, 3,… n = 1 represents K shell; n=2 represents L shell and n = 3, 4, 5 represent the M, N, O shells, respectively.
• The maximum number of electrons that can be accommodated in a given shell is 2n².
• ‘n’ gives the energy of the electron,

E_n = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) KJ mol^-1 and the distance of the electron from the nucleus is given by r_n = \(\frac{(-0.529) n^{2}}{Z}\) A.

Question 7.
Explain about azimuthal quantum number.
Answer:

• It is represented by the letter 7′ and can take integral values from zero to n – 1, where n is the principal quantum number.
• Each l value represents a subshell (orbital). l = 0, 1, 2, 3 and 4 represents the s, p, d, f and g orbitals respectively.
• The maximum number of electrons that can be accommodated in a given subshell (orbital) is 2(2l + 1).
  It is used to calculate the orbital angular momentum using the expression Angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\).

Question 8.
Draw the shapes of 1s, 2s and 3s orbitals
Answer:

Question 9.
Explain how effective nuclear charge is related with stability of the orbital.
Answer:

• In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons.
• These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
• The net charge experienced by the electron is called effective nuclear charge.
• The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number l.
• The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
• Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Question 10.
Calculate the wavelength of an electron moving with a velocity of 2.05 x 10⁷ ms^-1.
Answer:
According to de Broglie’s equation, λ = \(\frac {h}{mv}\)
Mass of electron (m) = 9.1 x 10^-31 kg
Velocity of electron (υ) = 2.05 x 10⁷ ms^-1
Planck’s constant (h) = 6.626 x 10^-34 kg m² s^-1
λ =  =355 x 10^-4m.

Question 11.
The mass of an electron is 9.1 x 10^-31 kg. If its kinetic energy is 3.0 x 10^-25 J, calculate its wavelength.
Answer:
Step I.
Calculation of the velocity of electron
Kinetic energy = 1 / 2 mυ² = 3.0 x 10^-25 kg m² s^-2
υ² = 65.9 x 10⁴ m² s^-2
υ = (65.9 x 10⁴ m s^-2) = 8.12 x 10² ms^-1

Step II.
Calculation of wavelength of the electron
According to de Broglie’s equation,
λ = \(\frac {h}{mv}\) =
=  0.08967 x l0^-5 m = 8967 x 10^-10 m = 8967 Å (∴1Å = 10^-10m).

Question 12.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.

1. n = 0, l = 0, m_l = 0, m_s = + \(\frac {1}{2}\)
2. n = 1, l = 0, m_l = 0, m_s = – \(\frac {1}{2}\)
3. n = 1, l = 1, m_l = 0, m_s = + \(\frac {1}{2}\)
4. n = 1, l = 0, m_l = +1, m_s= +\(\frac {1}{2}\)
5. n = 3, l = 3, m_l = -3, m_s = + \(\frac {1}{2}\)
6. n = 3, l = 1, m_l = 0, m_s = +\(\frac {1}{2}\)

Answer:

1. The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
2. The set of quantum numbers is possible.
3. The set of quantum numbers is not possible because, for n = 1,1 cannot be equal to 1. It can have 0 value.
4. The set of quantum numbers is not possible because for l = 0, m_l; cannot be +1. It must be zero.
5. The set of quantum numbers is not possible because, for n = 3, l = 3.
6. The set of quantum numbers is possible.

Question 13.
How many electrons in an atom may have the following quantum numbers?
(a) n = 4; m_s = – ½
(b) n = 3, l = 0.
Answer:
(a) For n = 4
1 Total number of electrons = 2n² = 2 x 16 = 32
Half out of these will have ms = – \(\frac {1}{2}\)
Total electrons with m_s (-½) = 16.

(b) For n = 3
l = 0; m₁ = 0, m_s = + ½ – ½ (two e^–).

Question 14.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
According to Bohr’s theory,
mυr = \(\frac {nh}{2π}\) (n = 1,2,3, …… so on)
or 2πr = \(\frac {nh}{mυ}\) or mυ = \(\frac {nh}{2πr}\) ………..(i)
According to de Brogue equation,
λ = \(\frac {h}{mυ}\) or mυ = \(\frac {h}{λ}\) ……….(ii)
Comparing (i) and (ii),
\(\frac {nh}{2πr}\) = \(\frac {h}{λ}\) or 2πr = nλ
Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an into the de Broglie wave length.

Question 15.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac {30.4x }{100}\) = x + 0.304 x
Now, mass number of ion = Number of protons + Number of neutrons
= (x + 3) + (x + 0.304 x)
∴ 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 – 3 = 53
x = \(\frac {53}{2.304}\) = 23
Atomic number of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 16.
The uncertainty in the position of a moving bullet of mass 10 g is 10 s m. Calculate the uncertainty in its velocity?
Answer:
According to uncertainty principle,
∆x.m∆υ = \(\frac {h}{4π}\) or ∆υ = \(\frac {h}{4πm∆x}\);
h = 6.626 x 1o^-34 kg m₂ s^-1; m = 10 g = 10^-2 kg
∆x = 10^-5m; ∆v =  = 5.27 x 10^-28mv
= 1.6 x 10^-15 kg m² s^-15
Or
\(\frac {1}{2}\) mv² = 1.6 x 10^-15kg m²s^-2
v =  = 5.93 x 10⁷m^-1

Question 17.
The uncertainty in the position and velocity of a particle are 10^-10 m and 5.27 × 10^-24 ms^-1 respectively. Calculate the mass of the particle.
Answer:
According to uncertainty principle.
∆x. m∆υ = \(\frac {h}{4π}\)
or
m = \(\frac {h}{4π∆x.∆υ}\);
h = 6.626 x 10^-34 kg m² s^-1
∆x = 10^-10 m; ∆x = 5.27 x 10^-24ms^-1
m  = 0.1 kg.

Question 18.
With what velocity must an electron travel so that its momentum Is equal to that of a photon of wave length = 5200 A?
Answer:
According to de Brogue equation, λ = \(\frac {h}{mv}\)
Momentum of electron, mv = \(\frac {h}{λ}\)
= 1.274 x 10^-27 kg ms^-1 ………(i)
The momentum of electron can also be calculated as = mv = (9.1 x 10^-31kg) x v ………(ii)
Comparing (i) and (ii)
(9.1 X 10^-31kg) v = (1.274 x 10^-27 kg ms^-1)
v   = 1.4 x 10³ ms^-1

Question 19.
Using Aufbau principle, write the ground state electronic configuration of following atoms.

1. Boron (Z = 5)
2. Neon (Z = 10)
3. Aluminium (Z = 13)
4. Chlorine (Z = 17)
5. Calcium (Z = 20)
6. Rubidium (Z = 37)

Answer:

1. Boron (Z = 5) ; 1s² 2s² 2p¹
2. Neon (Z = 10) ; 1s² 2s² 2p⁶
3. Aluminium (Z = 13) ; 1s² 2s² 2p⁶ 3s² 3p¹
4. Chlorine(Z = 17) ; 1s² 2s² 2p⁶ 3s² 3p⁵
5. Calcium (Z = 20) ; 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
6. Rubidium (Z = 37) ; 1s² 2s²2p⁶ 3s² 3p⁶3d¹⁰ 4s² 4p⁶ 5s¹

Question 20.
Calculate the de Broglie wavelength of an electron moving with 1 % of the speed of light?
Answer:
According to de Brogue equation, A = \(\frac {h}{mv}\)
Mass of electron = 9.1 x 10^-31 kg; Planck’s constant 6.626 x 10^-34 kg m² s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 10⁶ ms^-1
Wavelength of electron (λ) = \(\frac {h}{mv}\)
= 2.43 x 10^-10m.

Question 21.
What is the wavelength for the electron accelerated by 1.0 X i0 volts?
Answer:
Step I.
Calculation of the velocity of electron
Energy (kinetic energy) of electron = 1.0 x 10⁴ volts.
= 1.0 x 10⁴ x 1.6 x 10^-19 J = 1.6 x 10^-15J.

Step II.
Calculation of wavelength of the electron
According to de Broglie equation,
λ = \(\frac {h}{mυ}\); λ 
= 1.22 x 10^-11m .

Question 22.
In a hydrogen atom, the energy of an electron in first Bohr’s orbit is 13.12 x 10⁵ J mol^-1. What is the energy required for its excitation to Bohr’s second orbit?
Answer:

The energy required for the excitation is:
∆E = E₂ – E₁ = (-3.28 x l0⁵) – (- 13.12 x 10⁵) = 9.84 x 10⁵ J mol^-1

Question 23.
Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ ms^-1, calculate de Broglie wavelength associated with this electron.
Answer:
λ = \(\frac {h}{mυ}\); λ =
= 0.455 x 10^{-34 + 25} m = 0.455 nm = 455 pm.

Question 24.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
∴ Number of neutrons = x + \(\frac {x × 31.7}{100}\) = (x + 0.317 x)
Now, Mass no. of element = No. of protons + No. of neutrons
81 = x + x + 0.317 x = 2.317 x
Or
x = \(\frac {81}{2.317}\) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 = 46
Atomic number of element (Z) = Number of protons = 35
The element with atomic number (Z) 35 is bromine ⁸¹₃₅Br.

Question 25.
The electron energy in hydrogen atom is given by E_n = (- 2.18 × 10^-18) / n² J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:
Step I.
Calculation of energy required
The energy required is the difference in the energy when the electron jumps from orbit with
n = ∞ to orbit with n = 2.
The energy required (∆E) = E_∞ – E₂
= 0 – \(\left(-\frac{2.18 \times 10^{-18}}{4} \mathrm{J}\right)\) = 5.45 x 10^-19 J.

Step II.
Calculation of the longest wavelength of light in cm used to cause the transition
∆E = hv = hc / λ.
λ = \(\frac {hc}{∆E}\) = n= 3.644 x 10^-7
m = 3.644 x 10^-7 x 10² = 3.645 x 10^-5 cm.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 5-Mark Questions

Question 1.
Describe about Bohr atom model.
Answer:
Assumptions of Bohr atom model.
1. The energies of electrons are quarantined

2. The electron is revolving around the nucleus in a certain fixed circular path called the stationary orbit.

3. Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π
mvr = \(\frac {nh}{2π}\) where n = 1,2,3,…etc.,

4. As long as an electron revolves in a fixed stationary orbit, it doesn’t lose its energy. But if an electron jumps from a higher energy state (E₂) to a lower energy state (E₁), the excess energy is emitted as radiation. The frequency of the emitted radiation is E₂ – E₁= hv.
∴ v = \(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit.

5. Bohr’s postulates are applied to a hydrogen like atom (H, He⁺ and Li²⁺ etc..) the radius of the nth orbit and the energy of the electron revolving in the th orbit were derived.
r_n = \(\frac{(0.529) n^{2}}{Z}\) A(0.529) n²
E_n = \(\frac{(-1 3.6) Z^{2}}{n}\) eV atom^-1
E_n = \(\frac{(1312.8) Z^{2}}{n}\) kJ mol^-1

Question 2.
Derive de Brogue equation and give its significance.
Answer:
1. Louis de Brogue extended the concept of dual nature of light to all forms of matter. To quantify this relation, he derived an equation for the wavelength of a matter-wave.

2. He combined the following two equations of the energy of which one represents wave character (hu) and the other represents the particle nature (mc²).
Planck’s quantum hypothesis:
E = hv ……….(1)
Einsteins mass-energy relationship:
E = mc² ………(2)
From (1) and (2)
hv = mc²
hc/λ = mc²
∴ λ = \(\frac{h}{mc}\) ………(3)
The equation (3) represents the wavelength of photons whose momentum is given by mc. (Photons have zero rest mass).

3. For a particle of matter with mass m and moving with a velocity y, the equation (3) can be written as λ = \(\frac{h}{mc}\) ………(4)

4. This is valid only when the particle travels at speed much less than the speed of Light.

5. This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle (i.e momentum).

6. Significance of de Brogue equation:
For a particle with high linear momentum, the wavelength will be too small and cannot be observed. For a microscopic particle such as an electron, the mass is 9.1 x 10^-31 kg. Hence the wavelength is much larger than the size of atom and it becomes significant.

Question 3.
What are the main features of the quantum mechanical model of an atom.
Answer:
1. The energy of electrons in an atom is quarantined.

2. The existence of quarantined electronic energy levels is a direct result of the wave-like properties of electrons. The solutions of the Schrodinger wave equation gives the allowed energy levels (orbits).

3. According to Heisenberg’s uncertainty principle, the exact position and momentum of an electron cannot be determined with absolute accuracy. As a consequence, quantum mechanics introduced the concept of orbital. Orbital is a three-dimensional space in which the probability of finding the electron is maximum.

4. The solution of the Schrodinger wave equation for the allowed energies of an atom gives the wave function Ψ, which represents an atomic orbital. The wave nature of the electron present in an orbital can be well defined by the wave function Ψ.

5. The wave function Ψ itself has no physical meaning. However, the probability of finding the electron in a small volume dx, dy, dz around a point (x,y,z) is proportional to |Ψ (x,y,z)|² dx dy dz |Ψ (x,y,z)|² is known as probability density and is always positive.

Question 4.
Explain about –
(1) Magnetic quantum number
(2) Spin quantum number
Answer:
(1) Magnetic quantum number:

• It is denoted by the letter m_l. It takes integral values ranging from – l to +l through 0.
  i.e. if l = 1; m = -1, 0 and +1.
• The Zeeman Effect (the splitting of spectral lines in a magnetic field) provides the experimental justification for this quantum number.
• The magnitude of the angular momentum is determined by the quantum number l while its direction is given by magnetic quantum number.

(2) Spin quantum number:

• The spin quantum number represents the spin of the electron and is denoted by the letter ‘m_s‘.
• The electron in an atom revolves not only around the nucleus but also spins. It is usual to write this as electron spins about its own axis either in a clockwise direction or in anti-clockwise direction.
• Corresponding to the clockwise and anti-clockwise spinning of the electron, maximum two values are possible for this quantum number.
• The values of ‘m_s‘ is equal to –\(\frac {1}{2}\) and +\(\frac {1}{2}\).

Question 5.
Explain about the shape of orbitals.
Answer:
Orbital: The solution to Schrodinger equation gives the permitted energy values called eigen values and the wave functions corresponding to the eigen values are called atomic orbitals.

Shape of orbital:
s – orbital:
For Is orbital, l = 0, m = 0, f(θ) = 1√2 and g(φ) = 1/√2π. Therefore, the angular distribution function is equal to 1/√2π. i.e. it is independent of the angle θ and φ. Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

p – orbital:
For p orbitals l = 1 and the corresponding m values are -1, 0 and +1. The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as p_x, p_y and p_z. The shape of p orbitals are dumb bell shape.

d – orbital:
For ‘d’ orbital 1 = 2 and the corresponding m values are -2, -1, 0, +l,+2. The shape of the d orbital looks like a clover leaf. The five m values give rise to five d orbitals namely d_xy, d_yz, d_zx, d_x²_-y² and d_z² The 3d orbitals contain two nodal planes.

f – orbital
For f orbital, 1 = 3 and the m values are -3, -2,-1, 0, +1, +2, +3 corresponding to seven f orbitals, \(\mathrm{f}_{\mathrm{z}^{3}}, \mathrm{f}_{\mathrm{xz}^{2}}, \mathrm{f}_{\mathrm{yz}^{2}}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}\). They contain 3 nodal planes.

Question 6.
What is exchange energy? How it is related with stability of atoms? Explain with suitable examples.
Answer:
1. If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy.

2. If more number of exchanges are possible, more exchange energy is released. More number of exchanges are possible only in the case of half filled and fully filled configurations.

3. For example, in chromium, the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges.

4. On the other hand only six exchanges are possible for [Ar] 3d⁴ 4s² configuration.

5. Hence, exchange energy for the half filled configuration is more This increases the stability of half filled 3d orbitals.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2

11th Maths Exercise 1.2 Solutions Question 1.
Discuss the following relations for reflexivity, symmetricity and transitivity:

(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.
Solution:
S = {set of all positive integers}

(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive

(b) mRn ⇒ m divides n but
nRm ⇒ n does not divide m
(i.e.,) mRn ≠ nRm
It is not symmetric

(c) mRn ⇒ nRr as n divides r
It is transitive

(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “lRm if l is perpendicular to m”.
Solution:
P = {set of all straight lines in a plane}
lRm ⇒ l is perpendicular to m

(a) lRl ⇒ l is not perpendicular to l
⇒ It is not reflexive

(b) lRm ⇒ l is perpendicular to m
mRl ⇒ m is perpendicular to l
It is symmetric

(c) l perpendicular to m ⇒ m perpendicular to n ⇒ l is parallel to n It is not transitive

(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if a is not a sister of b”.
Solution:
A = {set of all members of the family}
aRb is a is not a sister of b

(a) aRa ⇒ a is not a sister of a It is reflexive

(b) aRb ⇒ a is not a sister of b.
bRa ⇒ b is not a sister of a.
It is symmetric

(c) aRb ⇒ a is not a sister of b.
bRc ⇒ b is not a sister of c.
⇒ aRc ⇒ a can be a sister of c
It is not transitive.

(iv) Let A be the set consisting of all the female members of a family. The relation R
defined by “aRb if a is not a sister of b”.
Solution:
A = {set of all female members of a family}

(a) aRa ⇒ a is a sister of a
It is reflexive

(b) aRb ⇒ a is a sister of b
bRa ⇒ b is a sister of a
⇒ It is symmetric

(c) aRb ⇒ a is a sister of b bRc ⇒ b is a sister of c aRc ⇒ a can be sister of c It is not transitive.

(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.
Solution:
N= {1, 2, 3, 4, 5,….}
xRy if x + 2y = 1 R is an empty set

(a) xRx ⇒ x + 2x = 1 ⇒ x = \(\frac{1}{3}\) ∉ N. It is not reflexive
xRy = yRx ⇒ x + 2y = 1 It does not imply that y + 2x = 1 as y = \(\frac{1-x}{2}\) It is not symmetric.

(b) -x = y ⇒ (-1, 1) ∉ N
It is not transitive.

11th Maths Exercise 1.2 Answers Question 2.
Let X = {a, b, c, d} and R = {(a, a), {b, b), (a, c)}. Write down the minimum number
of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
X = {a, b, c, d}
R = {(a, a), (b, b), (a, c)}
(i) To make R reflexive we need to include (c, c) and (d, d)
(ii) To make R symmetric we need to include (c, a)
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c)
To make R symmetric we need to include (c, c) and (c, a) for transitive
∴ The relation now becomes
R = {(a, a), (b, b), (a, c), (c, c), (c, a)}
∴ R is equivalence relation.

Exercise 1.2 Class 11 Maths State Board Question 3.
Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
(i) (c, c)
(ii) (c, a)
(iii) nothing
(iv) (c, c) and (c, a)

11th Maths Exercise 1.2 Question 4.
Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is
similar to b. Prove that R is an equivalence relation.
Solution:
P = {set of all triangles in a plane}
aRb ⇒ a similar to b

(a) aRa ⇒ every triangle is similar to itself
∴ aRa is reflexive

(b) aRb ⇒ if a is similar to b ⇒ b is also similar to a.
⇒ It is symmetric

(c) aRb ⇒ bRc ⇒ aRc
a is similar to b and b is similar to c
⇒ a is similar to a
⇒ It is transitive
∴ R is an equivalence relation

11 Maths Exercise 1.2 Question 5.
On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
N = {set of natural numbers}
R ={(3, 8), (6, 6), (9, 4), (12, 2)}

(a) (3, 3) ∉ R ⇒ R is not reflexive
2a + 3b = 30
3b = 30 – 2a
b = \(\frac{30-2 a}{3}\)

(b) (3, 8) ∈ R(8, 3) ∉ R
⇒ R is not symmetric

(c) (a, b) (b, c) ∉ R ⇒ R is transitive
∴ It is not equivalence relation.

11th Maths Exercise 1.2 Answers In Tamil Question 6.
Prove that the relation “friendship” is not an equivalence relation on the set of all people in Chennai.
Solution:
(a) S = aRa (i.e. ) a person can be a friend to himself or herself.
So it is reflextive.

(b) aRb ⇒ bRa so it is symmetric

(c) aRb, bRc does not ⇒ aRc so it is not transitive
⇒ It is not an equivalence relation

11th Maths Chapter 1 Exercise 1.2 Question 7.
On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
Set of all natural numbers aRb if a + b ≤ 6
R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
(i) (5, 1) ∈ R but(5, 5) ∉ R
It is not reflexive

(ii) aRb ⇒ bRa ⇒ It is symmetric

(iii) (4, 2), (2, 3) ∈ R ⇒ (4, 3) ∉ R
∴ It is not transitive

(iv) ∴ It is not an equivalence relation

11th Std Maths Exercise 1.2 Answers Question 8.
Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?
Solution:
R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3

(i) R = {(a, a), {a, b), (a, c), (b, c), (b, b), {b, c), (c, a), (c, b), (c, c)}
n(R) = 9 is the largest cardinality of R to make it equivalence.

11 Maths Samacheer Kalvi Question 9.
In the set Z of integers, define mRn if m – n is divisible by 7. Prove that R is an equivalence relation.
Solution:
mRn if m – n is divisible by 7
(a) mRm = m – m = 0
0 is divisible by 7
∴ It is reflexive

(b) mRn = {m – n) is divisible by 7
nRm = (n – m) = – {m – n) is also divisible by 7
It is symmetric

It is transitive
mRn if m – n is divisible by 7
∴ R is an equivalence relation.

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2 Additional Questions

Samacheer Kalvi 11th Maths Guide Question 1.
Find the range of the function.
f = {(1, x), (1, y), (2, x), (2, y), (3, z)}
Solution:
The range of the function is {x, y, z}.

Samacheer Kalvi 11th Maths Example Sums Question 2.
For n, m ∈ N, nln means that tt is a factor of n&m. Then find whether the given relation is an equivalence relation.
Solution:
Since n is a factor of n. So the relation is reflexive.
When n is a factor of m (where m ≠ n) then m cannot be a factor of n.
So the relation is not symmetric when n is a factor of m and m is a factor of p then n will be a factor of p. So the given relation is transitive. So it is not an equivalence relation.

11th Maths Solution Samacheer Question 3.
Verify whether the relation “is greater than” is an equivalence relation.
Solution:
You can do it yourself.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6

9th Maths Exercise 3.6 Question 1.
Factorise the following:
(i) x² + 10x + 24
(ii) z² + 4z – 12
(iii) p² – 6p – 16
(iv) t² + 72 – 17t
(v) y² – 16y – 80
(vi) a² + 10a – 600
Solution:
(i) x² + 10x + 24
x² + 10x + 24 = x² + 6x + 4x + 24

= x(x + 6) + 4 (x + 6)
= (x + 6) (x + 4)

(ii) z² + 4z – 12
z² + 4z – 12 = z² + 6z – 2z- 12

= z (z + 6) – 2 (z + 6)
= (z + 6) (z – 2)

(iii) p² – 6p – 16
p² – 6p – 16 = p² – 8p + 2p – 16

= p(p – 8) + 2(p – 8)
= (p – 8)(p + 2)

(iv) t² + 72 – 17t
t² + 72 – 17t = t² – 17t + 72
= t² – 9t – 8t + 72

= t(t – 9) – 8 (t – 9)
= (t – 9) (t – 8)

(v) y² – 16y – 80
y² – 16y – 80 = y² – 20y + 4y – 80

= y(y – 20) + 4 (y – 20)
= (y – 20) (y + 4)

(vi) a² + 10a – 600
a² + 10a – 600 = a² + 30a – 20a – 600

= a(a + 30) -20 (a + 30)
= (a + 30) (a – 20)

Exercise 3.6 Class 9 Question 2.
Factorise the following
(i) 2a² + 9a + 10
(ii) 5x² – 29xy – 42y²
(iii) 9 – 18x + 18x²
(iv) 6x² + 16xy + 8y²
(v) 12x² + 36x²y + 27y²x²
(vi) (a + b)² + 9 (a + b) + 18
Solution:
(i) 2a² + 9a + 10
2a² + 9a + 10 = 2a² + 4a + 5a + 10

= 2a(a + 2) + 5 (a + 2)
= (a+ 2) (2a+ 5)

(ii) 5x² – 29xy – 42y²
5x² – 35xy + 6xy – 42y²

= 5x (x – 7) + 6y (x – 7)
= (x – 7) (5x + 6y)

(iii) 9 – 18x + 8x²
= 8x² – 18x + 9
= 8x² – 6x – 12x + 9

= 2x (4 x – 3) – 3 (4x – 3)
= (4x – 3) (2x – 3)

(iv) 6x² + 16xy + 8y²
= 2 (3x² + 8xy + 4y²)
= 2 (3x² + 8xy + 4y²)

= 2 (3x² + 6xy + 2xy + 4y²)
= 2 (3x (x + 2y) + 2y (x + 2y))
= 2 (x + 2y) (3x + 2y)

(v) 12x² + 36x²y + 27y²x²
= 27y²x² + 36x²y + 12x² = 3x²(9y² + 12y + 4)

= 3x² (9y² + 6y + 6y + 4) = 3x² (3y (3y + 2) + 2 (3y + 2))
= 3x² (3y + 2) (3y + 2) = 3x² (3y + 2) (3y + 2)

(vi) (a + b)² + 9 (a + 6) + 18
= (a + b)² + 6 (a + b) + 3 (a + b) + 18

= (a + b) ((a + b) + 6) + 3 ((a + b) + 6)
= ((a + 6) + 6) ((a + b) + 3) = (a + b + 6) (a + b + 3)

9th Maths 3.6 Question 3.
Factorise the following:
(i) (p – q)² – 6(p – q) – 16
(ii) m² + 2mn – 24n²
(iii) \(\sqrt{5} a^{2}\) + 2a – \(3 \sqrt{5}\)
(iv) a⁴ – 3a² + 2
(v) 8m³ – 2m²n – 15mn²
(vi) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{x y}\)
Solution:
(i) (p – q)² – 6 (p – q) – 16
= (p – q)² – 8(p – q) + 2(p – q) – 16

= (p – q)((p – q) – 8) + 2((p – q) – 8)
= (p – q – 8)(p – q + 2)

(ii) m² + 2mn – 24n²
= m² + 6mn – 4mn – 24n²

= m(m + 6n) – 4n(m + 6n)
= (m + 6n)(m – 4n)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.2

12th Maths Exercise 5.2 Samacheer Kalvi Question 1.
Find the equation of the parabola in each of the cases given below:
(i) focus (4, 0) and directrix x = – 4.
(ii) passes through (2, -3) and symmetric about y-axis.
(iii) vertex (1, – 2) and focus (4, – 2).
(iv) end points of latus rectum(4, – 8) and (4, 8) .
Solution:
(i) Focus = F = (4, 0)
⇒ a = 4
Equation of directrix x = – 4
⇒ The curve open to the right. So the equation will be of the form y² = 4 ax
Here a = 4
⇒ y² = 4 (4) x (i.e.,)y² = 16x

(ii) The parabola is symmetric about y axis. So the equation will be of the form
x² = 4 ay
It passes through (2, -3)
⇒ 2² = 4a(-3)
4 = -12a ⇒ a = \(-\frac{1}{3}\) ⇒ 4a = \(-\frac{4}{3}\)
∴ Equation of parabola is x² = \(-\frac{4}{3}\) y
3x²= – 4y.

(iii) The distance between vertex and focus = 3
(ie.,) a = 3
Parabola is open to the right.
So equation will be of the form y² = 4ax
Here a = 3 ⇒ y² = 12x
but the vertex is (1, -2)
So equation of the parabola is
(y + 2)² = 12(x – 1)

(iv) Focus = (4, 0)
Equation of the parabola will be of the form .
y² = 4ax
Here a = 4
⇒ y² = 16x

12th Maths Exercise 5.2 Question 2.
ind the equation of the ellipse in each of the cases given below:
(i) foci (± 3, 0), e = \(\frac{1}{2}\)
(ii) foci (0, ± 4) and end points of major axis are(0, ± 5).
(iii) length of latus rectum 8, eccentricity = \(\frac{3}{5}\) and major axis on x -axis.
(iv) length of latus rectum 4 , distance between foci \(4 \sqrt{2}\) and major axis as y -axis.
Solution:
(i) Given ae = 3 and e = \(\frac{1}{2}\)
⇒ a(\(\frac{1}{2}\)) = 3 ⇒ a = 6
So a² = 36
b² = a²(1 – e²) = 36 (1 – \(\frac{1}{4}\)) = 36 × \(\frac{3}{4}\) = 27
Since Foci = (± 3, 0), major axis is along x-axis
So equation of ellipse is \(\frac{x^{2}}{36}+\frac{y^{2}}{27}\) = 1

(ii) From the diagram we see that major axis is along y-axis.

Also a = 5 and ae = 4
⇒ 5e = 4 ⇒ e = \(\frac{4}{5}\)
Now a = 5 ⇒ a² = 25
ae = 4 ⇒ ae² = 16
We know b² = a² (1 – e²) = a² – a²e² = 25 – 16 = 9
Equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{25}\) = 1

(iv) Given \(\frac{2 b^{2}}{a}\) = 4 and 2ae = \(4 \sqrt{2}\)
Now \(\frac{2 b^{2}}{a}\) = 4 2b² = 4a
⇒ b² = 2a
2ae = \(4 \sqrt{2}\) ae = \(2 \sqrt{2}\)
So a²e² = 4(2) = 8
We know b² = a²(1 – e²) = a² – a²e²
⇒ 2a = a² – 8 ⇒ a² – 2a -8 = 0
⇒ (a – 4) (a +2) = 0 ⇒a = 4 or -2
As a cannot be negative
a = 4 So a² = 16 and b² = 2(4) = 8
Also major axis is along j-axis
So equation of ellipse is \(\frac{x^{2}}{8}+\frac{y^{2}}{16}\) = 1

12th Maths Exercise 5.2 8th Sum  Question 3.
Find the equation of the hyperbola in each of the cases given below:
(i) foci (± 2, 0), eccentricity = \(\frac{3}{2}\)
(ii) Centre (2,1), one of the foci (8,1) and corresponding directrix x = 4 .
(iii) passing through(5, -2) and length of the transverse axis along JC axis and of length 8 units.
Solution:
(i) Given
ae = 2 and e = \(\frac{3}{2}\)
a( \(\frac{3}{2}\)) = 2 ⇒ a = \(\frac{4}{3}\) So a² = \(\frac{16}{9}\)
b² = a²(e² – 1) = a² e² – a² = 4 – \(\frac{16}{9}\) = \(\frac{20}{9}\)
Since the foci are (± 2, 0), transverse axis is along x-axis
So equation of hyperbola is
\(\frac{x^{2}}{16 / 9}-\frac{y^{2}}{20 / 9}\) = 1 ⇒ \(\frac{9 x^{2}}{16}-\frac{9 y^{2}}{20}\) = 1

(ii) Given Centre = (2, 1)
ae = 6 (distance between (2, 1) and (8, 1)) ……………. (1)
Also \(\frac{a}{e}\) = 2 ⇒ a = 2e
Equation of directrix is x = 4 [(i.e.,) (x – 2 = 2) Since centre is (2, 1)]
⇒ \(\frac{a}{e}\) = 2
Given ae = 6 ⇒ a² e² = 36
(i.e.) (2e)² (e)² = 36
⇒ 4e⁴ = 36 ⇒ e⁴ = 9
⇒ e = \(\sqrt{3}\)
Now e = \(\sqrt{3}\) a = \(2\sqrt{3}\)
∴ a² = 4 × 3 = 12
b² = a² (e² – 1) = a² e² – a² = 36 – 12 = 24
So here Centre = (2, 1)
So equation of hyperbola is
\(\frac{(x-2)^{2}}{12}-\frac{(y-1)^{2}}{24}\) = 1

(iii) Length of the transverse axis = 8
2a = 8 ⇒ a = 4
Transverse axis is along x-axis
So of equation of hyperbola is will be

12th Maths 5th Chapter Samacheer Kalvi Question 4.
Find the vertex, focus, equation of directrix and length of the latus rectum of the following:
(i) y² = 16x
(ii) x² = 24y
(iii) y² = -8x
(iv) x² – 2x + 8y + 17 = 0
(v) y² – 4y – 8x + 12 = 0
Solution:
(i) y² = 16x
It is of the form y² = 4ax (type I)
Here 4a = 16 ⇒ a = 4
Vertex = (0, 0)
Focus = (a, 0) = (4, 0)
Equation of directrix x + 4 = 0 (or) x = – 4
Length of latus rectum = 4a = 16 .

(ii) x² = 24y
This is of the form x² = 4ay (type III)
4a = 24 ⇒ a = 6
Vertex = (0, 0)
Focus = (0, a) = (0, 6)
Equation of directrix is y + a= 0 (i.e.,) y + 6 = 0 (or) y = -6
Length of latus rectum = 4a = 24.

(iii) y² = -8x
This is of the form y² = – 4ax (type II)
Here 4a = 8 ⇒ a = 2
Vertex = (0, 0)
Focus = (- a, 0) = (-2, 0)
Equation of directrix is x – 2 = 0 (or) x = 2
Length of latus rectum = 4a = 8.

(iv) x² – 2x + 8y + 17 = 0
x² – 2x = -8y – 17
x² – 2x + 1 – 1 = – 8y — 17
(x – 1)² = – 8y – 17 + 1 = – 8y + 16
(x – 1)² = – 8 (y – 2)
Taking x – 1 = X and y – 2 = Y.
We get X² = – 8Y.
This is of the form x² = – 4ay (type IV)
Where 4a = 8 ⇒ a = 2

(v) y² – 4y = 8x – 12 = 0
y² – 4y + 4 = 8x – 12 + 4
(y – 2)² = 8x – 8 = 8 (x – 1)
Taking x – 1 = X and y – 2 = Y.
We get Y² = 8X.
This is of the form y² = 4ax (type IV)
Where 4a = 8 ⇒ a = 2

12th Maths 5.2 Exercise Question 5.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

Solution:
(i) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1
It is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, which is an ellipse
Here a² = 25, b² = 9
a = 5, b = 3
e² = \(\frac{a^{2}-b^{2}}{a^{2}}=\frac{25-9}{25}=\frac{16}{25}\) ⇒ e = \(\frac{4}{5}\)
Now e = \(\frac{4}{5}\) and a = 5 ⇒ ae = 4 and \(\frac{a}{e}=\frac{5}{4 / 5}=\frac{25}{4}\)
Here the major axis is along x axis
∴ Centre = (0, 0)
Foci = (± ae, 0) = (± 4, 0)
Vertices = (± a, 0) = (±5, 0)
Equation of directrix x = ± \(\frac{a}{e}\) (ie.,) x = ± \(\frac{25}{4}\)

(ii) \(\frac{x^{2}}{3}+\frac{y^{2}}{10}\) = 1
It is an ellipse and here (always a >b)

Here the major axis is alongy-axis
Centre = (0, 0)
Foci = (0, ± ae) = (0, ± \(\sqrt{7}\))
Vertices = (0, ± a) = (0, ± \(\sqrt{10}\) )
Equation of directrices y = ± \(\frac{10}{\sqrt{7}}\)

(iii) \(\frac{x^{2}}{25}-\frac{y^{2}}{144}\) = 1

Now, Here transverse axis is along x-axis
Centre = (0, 0)
Vertices = (± a, 0) = (± 5, 0)
Foci = (± ae, 0) = (± 13, 0)
Equation of directrices x = ± \(\frac{a}{e}\) (ie.,) x = ± \(\frac{25}{13}\)

(iv) \(\frac{y^{2}}{16}-\frac{x^{2}}{9}\) = 1
It is a hyperbola. Here transverse axis is along y-axis

Now Centre = (0, 0)
Vertices = (0, ± a) = (0, ± 4)
Foci = (0, ± ae) = (0, ± 5)
Equation of directrices y = ± \(\frac{16}{5}\)

Class 12 Maths Chapter 5 Exercise 5.2 Question 6.
Prove that the length of the latus rectum of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 is \(\frac{2 b^{2}}{a}\)
Solution:
The latus rectum LL’ of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 passes through the focus(ae, 0)

12th Maths 5.2 Question 7.
Show that the absolute value of the difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.
Solution:

∴ S’P – SP = (a + ex) – (ex – a)
a + ex – ex + a = 2a (transverse axis)

Question 8.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

Solution:

(v) 18x² + 12y² – 144x + 48y + 120 = 0
(18x² – 144x) + (12y² + 48y) = -120
18(x² – 8x) + 12 (y² + 4y) = -120
18(x² – 8x + 16 – 16) + 12(y² + 4y + 4 – 4) = -120
18(x – 4)² – 288 + 12(y + 2)² – 48 = – 120
18(x – 4)² + 12(y + 2)² = -120 + 288 + 48 = 216

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.2 Additional Problems

Question 1.
Find the equation of the ellipse if centre is (3, – 4), one of the foci is (3 + \(\sqrt{3}\), – 4) and e = \(\frac{\sqrt{3}}{2}\)
Solution:

Question 2.
Find the equation of the hyperbola if centre (1, -2); length of the transverse axis is 8; e = \(\frac{5}{4}\) and the transverse axis is parallel to X-axis.
Solution:

Here, centre = (1, -2) and transverse axis is parallel to X-axis.

12th Maths Chapter 5 Exercise 5.2 Question 3.
Find axis, vertex, focus and equation of directrix for y² + 8x – 6y + 1 = 0
Solution:
y² – 6y = – 8x – 1
y² – 6y + 9 = – 8x – 1 + 9
(y – 3)² = – 8x + 8 = – 8(x – 1)

Comparing this equation with Y² = – 4aX we get
4a = 8 or a = 2
Vertex is (0, 0)
x – 1 = 0 ⇒ x = 1, y – 3 = 0 ⇒ y = 3
Axis y – 3 = 0, Vertex = (1, 3)
Focus is (- a, 0) = (-2, 0) + (1, 3) = (-1, 3)
Equation of directrix is x – a = 0. i.e., X – 2 = 0
⇒ x – 1 – 2 = 0 ⇒ x – 3 = 0
Latus rectum x + a = 0 i.e., x – 1 + 2 = 0
x + 1 = 0
Length of latus rectum = 4a = 8

Question 4.
Find axis, Vertex focus and equation of directrix for x² – 6x – 12y – 3 = 0.
Solution:
x² – 6x = 12y + 3
x² – 6x + 9 = 12y + 3 + 9 = 12y + 12
(x – 3)² = 12(y + 1)

This equation is of the form X² = 4aY
4a = 12
⇒ a = 3
Vertex is x – 3 = 0 ; y + 1 = 0
⇒ x = 3 ; y = -1

Question 5.
Find the eccentricity, centre, foci and vertices of the following hyperbolas: x² – 4y² – 8x – 6y – 18 = 0
Solution:

Question 6.
Find the eccentricity, centre, foci, vertices of 9x² + 4y² = 36
Solution:

Ex 5.2 Class 12 Question 7.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
x² – 4y² + 6x + 16y – 11 = 0
Solution:

10th Maths Exercise 5.2 Samacheer Kalvi Question 8.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
x² – 3y² + 6x + 6y + 18 = 0
Solution:

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Tamilnadu Samacheer Kalvi 10th Social Science History Solutions Chapter 5 Social and Religious Reform Movements in the 19th Century

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Social and Religious Reform Movements in the 19th Century Textual Exercise

I. Choose the correct answer.

Social And Religious Reform Movements 10th Class Question 1.
In which year was Sati abolished?
(a) 1827
(b) 1829
(c) 1826
(d) 1927
Answer:
(b) 1829

Social And Religious Reform Movements In The 19th Century Book Back Answers Question 2.
What was the name of the Samaj founded by Dayanand Saraswati?
(a) Arya Samaj
(b) Brahmo Samaj
(c) Prarthana Samaj
(d) Adi Brahmo Samaj
Answer:
(a) Arya Samaj

10th Social History 5th Lesson Question 3.
Whose campaign and work led to the enactment of Widow Remarriage Reform Act of 1856?
(a) Iswarchandra Vidyasagar
(b) Raja Rammohan Roy
(c) Annie Besant
(d) Jyotiba Phule
Answer:
(a) Iswarchandra Vidyasagar

Social And Religious Reform Movements 10th Class Question Answer Question 4.
Whose voice was Rast Goftar?
(a) Parsi Movement
(b) Aligarh Movement
(c) Ramakrishna Mission
(d) Dravida Mahajana Sabha
Answer:
(a) Parsi Movement

Social And Religious Reform Movements 10th Class Notes Question 5.
Who was the founder of Namdhari Movement?
(a) Baba Dayal Das
(b) Baba Ramsingh
(c) Gurunanak
(d) Jyotiba Phule
Answer:
(b) Baba Ramsingh

Question 6.
Who was Swami 5hradhananda?
(a) a disciple of Swami Vivekananda
(b) one who caused a split in the Brahmo Samaj of India
(c) one who caused a split in the Arya Samaj
(d) founder of Samathuva Samajam.
Answer:
(c) one who caused a split in the Arya Samaj

Question 7.
Who was the founder of Widow Remarriage Association?
(a) M.G. Ranade
(b) Devendranath Tagore
(c) Jyotiba Phule
(d) Ayyankali
Answer:
(a) M.G. Ranade

Question 8.
Who was the author of the book Satyarthaprakash?
(a) Dayananda Saraswathi
(b) Vaikunda Swamy
(c) Annie Besant
(d) Swami Shradanatha
Answer:
(a) Dayananda Saraswathi

II. Fill in the blanks.

1. ……… founded the Samarasa Vedha Sanmarga Sangam.
2. The founder of Poona Sarvajanik Sabha was ………
3. Satyashodak Samaj was launched by ………
4. Gulumgir was written by ………
5. Satyarthaprakash enumerates the positive principles of ………
6. Ramakrishna Mission was established by ………
7. ……… was the forerunner of Akali Movement.
8. ……… brought tremendous changes in the caste structure in Kerala.
9. Oru paisa Tamilan was started by ………
Answers:
1. Ramalinga Swamigal
2. M.G. Ranade
3. Jyotiba Phule
4. Jyotiba Phule
5. conduct
6. Swami Vivekanand
7. Singh Sabha
8. Narayan Guru
9. Iyothee Thassar

III. Choose the correct statement.

Question 1.
(i) Raja Rammohan Roy preached monotheism
(ii) He encouraged idolatry
(iii) He published tracts condemning social evils
(iv) Raja Rammohan Roy was supported by Governor General William Bentinck
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (i) (ii) and (iii) are correct
(d) (i) (iii) and (iv) are correct
Answer:
(d) (i) (iii) and (iv) are correct

Question 2.
(i) Prarthana Samaj was founded by Dr. Atma Ram Pandurang.
(ii) Prarthana Samaj encouraged interdining and inter-caste marriage.
(iii) Jyotiba Phule worked for the upliftment of men.
(iv) Prarthana Samaj had it’s origin in the Punjab.
(a) (i) is correct
(b) (ii) is correct
(c) (i) and (ii) are correct
(d) (iii) and (iv) are correct
Answer:
(c) (i) and (ii) are correct

Question 3.
(i) Ramakrishna Mission was actively involved in social causes such as education, health care, relief in time of calamities.
(ii) Ramakrishna emphasised the spiritual union with god through ecstatic practices.
(iii) Ramakrishna established the Ramakrishna Mission
(iv) Ramakrishna opposed the Partition of Bengal
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (iii) is correct (d)
(iv) alone correct
Answer:
(b) (i) and (ii) are correct

Question 4.
Assertion (A): Jyotiba Phule opened orphanages and homes for widows
Reason (R): Jyotiba Phule opposed child marriage and supported widow remarriage
(a) Assertion is correct but reason is not apt to the assertion
(b) Assertion is correct and the reason is apt to the assertion
(c) Both are wrong
(d) Reason is correct but assertion is irrelevant
Answer:
(b) Assertion is correct and the reason is apt to the assertion

IV. Match the following.

Answers:
1. (d)
2. (e)
3. (b)
4. (a)
5. (c)

V. Answer briefly.

Question 1.
Mention the four articles of faith laid down by Maharishi Debendranath Tagore?
Answer:
He laid down four articles of faith.

1. In the beginning there was nothing. The one Supreme Being alone existed who created the universe.
2. He alone is the God of Truth, Infinite wisdom, Goodness and power, eternal, omnipresent.
3. Our Salvation depends on belief in Him and in His worship in this world.
4. Belief consists in loving Him and doing His will.

Question 2.
Discuss Mahadev Govind Ranade’s contribution to social reforms.
Answer:
Mahadev Govind Ranade was a great social reformer. He advocated for inter-caste dining, inter-caste marriage, widow remarriage and improvement of women and depressed classes. He founded the Widow Marriage Association, the Poona Sarvajanik Sabha and Decean Education society.

Question 3.
Assess the role of Ayyankali in fighting for the cause of “untouchables.”
Answer:
Ayyankali, as a child faced the problems that made him turn into a leader of an anti-caste movement. He fought for the basic rights including access to public spaces and entry to schools. He wore clothes associated with upper castes and rode an ox-cart which was a ban for untouchables on public roads.

Question 4.
Write a note on reforms of Ramalinga Adigal.
Answer:
Ramalinga swamigal emphasised the bonds of responsibility and compassion between living beings. He empressed the view that those who lack compassion for suffering beings are hard hearted, their wisdom clonded. He showed his compassion and mercy on all living beings including plants. He established the Samarasa Vedha Sanmarga Sangam in 1865 and it was renamed in 1872 as ‘Samarasa Suddha Sanmarga Sathya Sangam’ which means ‘Society for pure truth in universal self-hood’. Ramalinga also estabilished a free feeding house for everyone.

Question 5.
What was the impact of Swami Vivekananda’s activist ideology?
Answer:
Vivekananda’s activist ideology breed a sense of confidence among Indians. It kindled the desire for political change among many western-education young Bengal. He advocated that service to humanity is service to God. His ideology allowed the lower caste Hindus to be allowed in the Hindu rituals.

Question 6.
What are the differences between Reformist Movements and Revival Movements?
Answer:
While the reformist movements strived to change the fundamental system and structures of the society, through gradual changes within the existing institutions; revivalist movements tended to revive formers customs or practices and thus take the society back to the glorious past.

Question 7.
List the social evils eradicated by Brahmo Samaj.
Answer:

1. Abolition of Sati in 1829, Child marriage and Polygamy.
2. Advocated Education for women, western science in schools and colleges.

Question 8.
Highlight the work done by Jyotiba Phule for the welfare of the poor and the marginalized.
Answer:
Jyotiba Phule worked for the upliftment of depressed classes and the causes of women. He opened orphanges and homes for widows. He opened the first school for ‘untouchable’ in 1852 in Poona. He launched the Satyashodhak Samaj (Truth-Seekers Society) in 1870 to stir the non-Brahman masses to self-respect and ambition. He supported widow remarriage, which was prohibited particularly among high-caste Hindus. His work, Gulamgiri (slavery) is an important work that condemned the inequlities of caste.

Question 9.
What was the impact of lyotheeThassar’s visit to SriLanka.
Answer:
He converted himself to Buddhism. He founded the Sakya Buddhist society at Madras. He believed that the revival of Buddhism could liberate the people from the evil of caste that afflicted the Hindu society. He called the ‘Untouchables, Sathi Petham Atra Dravidar and urged them to register as castelss Dravidians.

VI. Answer all the questions given under each caption:

Question 1.
Aligarh Movement

(i) What is the main aim of this Movement?
Answer:
The main aim of the Aligarh Movement was to persuade the Muslims to acquire modem knowledge and English language.

(ii) Who is considered the soul of this Movement?
Answer:
Sir Sayed Ahmed Khan is considered the soul of the Aligarh Movement.

(iii)Why were English books translated into Urudu?
Answer:
Many English Books were translated into Urdu in order to enable the Muslims to accept the western science and take up government services.

(iv) Name the college which was later raised to the status of a University?
Answer:
Aligarh Muhammadan Anglo-Oriental College.

Question 2.
Ramalinga Adigal

(i) What is Jeevakarunya?
Answer:
Ramalinga Adigal showed his compassion and mercy on all living beings including plants . This he called as Jeevakarunya.

(ii) What are the Songs of Grace?
Answer:
His voluminous songs were compiled and published under the title Thiruvarutpa which are called Songs of Grace.

(iii) Point out the major contribution of Samarasa Vedha Sanmarga Sathya Sangam?
Answer:
The Major contribution of Samarasa Vedha Sanmarga Sathya Sangam were to establish free feeding house for everyone irrespective of caste at Vadalur in 1867.

(iv) Where did he establish his free feeding house?
Answer:
The free feeding house is at Vadalur, established in 1867.

Question 3.
Deoband Movement

(a) Who were the organizers of this Movement?
Answer:
The Deoband Movement was organised be the orthodox Muslim Ulemas.

(b) What were the two main objectives of the Movement?
Answer:
The two main objectives of this Movement were:

• to propagate the pure teachings of the Quran and the Hadith
• to encourage the spirit of Jihad against the foreign and un-Ismalic elements

(c) Who founded the school at Deoband?
Answer:
The Ulemas under the leadership of Muhammad Qusim Wanoutavi and Rashed Ahmad Gangohi founded the school at Deoband in the Saharanpur district of Uttar Pradesh in 1866.

(d) Against whom the fatwa was issued by Deoband Ulema?
Answer:
In 1888 Deoband Ulemas issued the fatwa (a religious decree) against Syed Ahmed Khan’s Orgnisation called ‘The United Patriotic Association’ and ‘The Muhammadan Anglo-Oriental Association.

VII. Answer in detail.

Question 1.
Compare and contrast the contributions of Revivalist Movements with that of Reform Movements.
Answer:

1. Reform Movements were the Brahmo Samaj, the Prarthana Samaj and the Aligarh movement.
2. Revival movements were the Arya Samaj, the Ramakrishna mission, Deoband Movement.
3. To compare, both the reform and revival movements were in some way or other attack the social evils that prevailed in the society like sati, female infanticide, child marriages and various superstitious beliefs and complex rituals.
4. The reform movements strived to change the fundamental system and structure of the society, while the reformist movements strived to revive the earlier customs and practices, protecting the cultural heritage of the country.
5. Revival movements proved that the rich cultural heritage of India was found superior to the Western culture. That’s why, Arya Samaj advocated ‘Go back to the Vedas’.
6. The Reform movement responded to the changes in the time period and scientific thinking of the modem era.
7. For example, Brahmo Samaj, considers other religions such as Christianity, Islam as equal to Hinduism.
8. Arya Samaj believes in the superiority of Hinduism.
9. The Reform movements were influenced by the Western culture, whereas the Revival movements were influenced by Indian culture only.

Question 2.
Discuss the circumstances that led to the Reform movements of 19th century.
Answer:

1. The nineteenth century India was plagued with a number of social evils such as sati, child marriage, female infanticide, polygamy and so on.
2. Women were subjugated by men and were not allowed to get education. They were restricted to home and hearth. They were considered inferior to men.
3. The condition of the depressed classes was miserable. They were subject to untouchable. Their entry to schools, temples and other public places, meant for upper castes, was banned. Hence, there was no education among the people belonging to lower castes.
4. There were other evil practices prevalent in the Indian society such as excessive superstitious religious beliefs, animal sacrifice, which needed attention of the social reformers for the benefit of the common people of the society.
5. There was total absence of reason in the society. The system of child marriage was prevalent which resulted to child widows. These widows were never accepted in the family and were destined to lead a pathetic life. These were the circumstances that led to the Reform movements in the nineteenth century.

Question 3.
Evaluate the contributions of Ramakrishna Paramahamsa and Swami
Answer:
Vivekananda to regenerate Indian society.

1. Ramakrishna Paramahamsa was a simple priest of Kolkata gained popularity by emphasising the spiritual union with God by singing bhajans.
2. He declared that the manifestations of the divine mother were infinite and said ‘Jiva is Siva’ which means all living beings are God.
3. He insisted on not mercy, rather service. Service for Man as regarded as Service to God.
4. After his death in 1886, his disciples, especially Vivekananda established the Ramakrishna Mission, which undertook not only religious activities but also social causes.
5. Education, Health Care, Relief in times of calamities were the main social works of the mission.
6. Vivekananda emphasised a cultural nationalism and made a call to Indian youth to regenerate Hindu society.
7. His ideas gave a sense of self-confidence among Indians who felt inferior in relation to the materialistic achievements of the west.
8. He suggested the orthodox Hindus that the lower castes should be allowed to engaged in the Hindu rituals from which they were excluded.
9. Vivekananda’s activist ideology rekindled the desire for political change among many Western education young Bengalis.
10. His ideas were inspiring to many of the militant nationalist before Indian independence and even for today’s people.

Question 4.
Write an essay on the role played by the 19th century reformers towards the cause of Women.
Answer:
Several social reformers emerged during the 19th century India who played a big role towards the cause of women:

(i) Raja Rammohun Roy was deeply concerned with the prevanting customs of sati, child marriage and polygamy. He advocated the rights of widows to remarry. He wanted – polygamy to end. He raised voice against sati system and forced the Governor-General William Bentinck to abolish this social evil in 1829. He condemned the subjugation of women and opposed the prevailing ideas that women were inferior to men. He strongly advocated education for women.

(ii) Ishwar Chandra Vidyasagar was lead against the burning of widows. He supported the idea of widow remarriage. He played a leading role in promoting education of girls and helped them in setting up a number of schools. He dedicated his entire life for the betterment of the child widows of the Hindu society. He led a movement that resulted in the widows’ Remarriage Reform Act of 1856.

(iii) Jyotiba Phule worked for the cause of women. He opposed child marriage and supported widow remarriage, which was prohibited particularly their lives for the uplift of the depressed classes and women. Jyotiba opened orphanages and homes for widows.

(iv) Swami Dayanand Saraswati said that the prohibition of widow remarriage had no scriptural sanction.

(v) Reformers like R.C. Bhandarkar and Justice Mahadev Govind Ranade devoted themselves to activities such as widow remarriage and improvement of women and depressed classes. M.G. Ranade founded the widow Remarriage Association in 1861.

Social and Religious Reform Movements in the 19th Century Additional Questions

I. Choose the correct answer:

Question 1.
The pioneer of the reform movements was
(a) Raja Rammohan Roy
(b) Keshab ChandraSen
(c) Devendranath Tagore
Answer:
(a) Raja Rammohan Roy

Question 2.
……………… dedicated his whole life for the betterment of the child widows of the Hindu Society.
(a) M.G. Ranade
(b) Iswar Chandra Vidyasagar
(c) Atma Ram Pandurang
(d) Ramakrishna
Answer:
(b) Iswar Chandra Vidyasagar

Question 3.
Swami Dayananda Saraswathi started the
(a) Brahmo Samaj
(b) Arya Samaj
(c) Prarthana Samaj
Answer:
(b) Arya Samaj

Question 4.
The first age of consent Act was passed in the year ………………
(a) 1830
(b) 1840
(c) 1850
(d) 1860
Answer:
(d) 1860

Question 5.
Vallalar’s devotional songs are compiled in a volume called
(a) Devaram
(b) Ettuthogai
(c) Thiru Arutpa
Answer:
(c) Thiru Arutpa

Question 6.
……………… wrote the book Satyarthaprakash.
(a) Shraddhananda
(b) Dayananda Saraswathi
(c) Jyotiba Phule
(d) Narayana Guru
Answer:
(b) Dayananda Saraswathi

Question 7.
Sir Syed Ahamed Khan started a school at
(a) Alipore
(b) Allepey
(c) Ghazipur
Answer:
(c) Ghazipur

Question 8.
Inspired by Sree Narayan Guru, ……………… founded the Sadhu Jana Paripalana Sangam.
(a) Deoband
(b) Ayyankali
(c) Sayyid Ahmed Khan
(d) Ramalinga Adigal
Answer:
(b) Ayyankali

Question 9.
“Go back to Vedas” was said by
(a) Swami Dayananda Saraswathi
(b) Annie Besant
(c) Swami Vivekananda
Answer:
(a) Swami Dayananda Saraswathi

Question 10.
Ramalinga Adigal was popularly known as ………………
(a) Nirankar
(b) Vallalar
(c) Vaikunda Swami
(d) Iyothee Thassar
Answer:
(b) Vallalar

II. Fill in the blanks:

1. ……… was the pioneer of reform movements.
2. Raja Rammohan Roy was called the ……… India.
3. Prarthana Samaj was founded in Bombay in 1867 by ………
4. Swami Dayananda Saraswathi founded the ……… in 1875.
5. ……… was founded by Madame Blavatsky and Henry S Olcott.
6. Annie Besant started the ……… movement.
7. ……… was founded by Swami Vivekananda.
8. Samarasa Sudha Sanmarka Sangam was started by ………
9. ……… was started by Sir Syed Ahamed Khan.
10. Jyotiba Phule formed the ………
11. The Hindus who had been converted to other religions were reconverted by ………
12. Theosophy means ………
13. Home Rule movement was started by ………
14. The childhood name of Swami Vivekananda was ………
15. Sathya Gnana Sabai was established by ………
16. “The Hindus and Muslims are the two eyes of the beautiful bird that was India” was said by ………
17. “Satya Shodhak Samaj” was formed by ………
Answers:
1. Raja Rammohan Roy
2. Herald of New Age
3. Dr. Atmaram Pandurang
4. Arya Saniaj
5. The Theosophical Society
6. Home Rule
7. The Ramakrishna Mission
8. Ramalinga Adigal
9. The Aligarh movement
10. Satya Shodhak Samaj
11. Suddhi movement
12. Wisdom of God
13. Annie Besant
14. Narendranath Dutta
15. ValIalar
16. Syed Ahamed Khan
17. Jyotiba Phule

III. Match the following:

Answers:
1. (e)
2. (a)
3. (c)
4. (b)
5. (d)

Answer:
1. (d)
2. (c)
3. (a)
4. (e)
5. (b)

Answers:
1. (d)
2. (c)
3. (a)
4. (e)
5. (b)

IV. Answer the following briefly:

Question 1.
Write a note on the Age of Consent Act.
Answer:
The Age of Consent Act was enacted by the efforts of the reformer Vidyasagar in 1860. Initially it was fixed for ten years. Later it was raised to twelve years in 1891 and thirteen years in 1925.

Question 2.
Name the services rendered by the Arya Samaj.
Answer:

1. Arya Samaj opposed Child marriage, Polygamy, Purdha System, Casteism and Sati.
2. It advocated women education, inter-caste marriage and inter-dining.
3. It insisted on education of the women and upliftment of the depressed classes.
4. The Samaj started a number of schools all over the country to impart English and Vedic education. These schools are called Dayananda Anglo-Vedic schools and colleges.

Question 3.
Who opposed Arya Samaj for its doctrinal purity?
Answer:
Swami Shraddhananda opposed Arya Samaj for its doctrinal purity. He was the opinion that the group running the DAV schools are too westernised and thereby ignore founder’s ideology. Therefore he started own network of schools, Gurukulas, emphasising the study of vedas.

Question 4.
What are the teachings of Vallalar?
Answer:

1. Vallalar condemned the inequalities based on birth.
2. He advocated that feeding the poor is the highest form of worship.
3. He insisted compassion and mercy on all living things.
4. According to Vallalar “Service to mankind is the path of Moksha”
5. God is the personification of mercy and knowledge.
6. The path of compassion and mercy are the only way to reach God.

Question 5.
What are the Islamic Reform movements originated in India and what do they mainly focus on?
Answer:
Scientific society founded of Sir Sayyid Ahmed Khan was his first initiative. Later he founded the Aligarh Movement emphasising education and started Aligarh University. Deoband movement organised by The orthodox Muslim Ulema to spread the spirit of Muslims against un-islamic elements.

Question 6.
Name the followers of Swami Dayananda Saraswathi.
Answer:

1. Lala Lajpat Rai
2. Lala Hansraj
3. Pandit Guru Dutt
4. Bala Gangathara Tilak
5. Gopala Krishna Gokhale

Question 7.
Who started the Sadharan Samaj? Why?
Answer:
In 1866, Keshab Chandra Sen left the Brahmo Samaj and founded a new organisation called as Debendranath’s organisation. Later it was called as Adi Brahmo Samaj. When Keshab’s 14 yrs old daughter married to an Indian prince, the opponents of child marriage left Brahmo Samaj and started the SadharanSamaj.

Question 8.
Write a note on Keshab Chandra Sen.
Answer:

1. He was one of the followers of Raja Rammohan Roy.
2. After the death of Raja Rammohan Roy, the works of the Brahmo Samaj were carried on by Keshab Chandra Sen.
3. Because of his efforts, an act was passed in 1872. It abolished polygamy and child marriage.

Question 9.
What are the Social reforms of Jyotiba Phule?
Answer:

1. He revolted against the domination of the Brahmins.
2. He worked for the rights of the peasants and other low caste people.
3. He formed the “Satya Shadhak Samaj” in 1873, to liberate the people of lower caste from the suppression by Brahmins.
4. He started Orphanage for the unfortunate children.
5. He decided to construct a common bathing tank outside his house for the people of lower caste people.

Question 10.
Write a short note about ‘Sree Narayana Guru’.
Answer:

1. He was a great social reformer from Kerala.
2. He started ‘Sree Narayana Guru Dharma Paripalana Yogam’ in 1903.
3. It worked for the social economic and educational development of the Ezhava community and other backward people.
4. He condemned animal sacrifices, casteism and other social evils.

Question 11.
What is called ‘Jeeva Karunya’?
Answer:

1. St. Ramalinga had the view that love is the “Master key to Spirituality”.
2. He showed compassion and mercy not only on human beings, but also on plants, insects, birds and animals. This is called ‘Jeeva Karunya’.

V. Answer all the questions given under each Captions:

Question 1.
Brahmo Samaj

(a) When was it founded? Who was its founder?
Answer:
Brahmo Samaj was founded in 1828 by Raja Ram Mohan Roy.

(b) What was the important Legislation passed due to the efforts of. its founder?
Answer:
Abolition of Sati in 1829 was the important Legislation passed due to the efforts of its founder.

(c) What does the Samaj forbade?
Answer:
The Samaj forbade idol-worship and condemned meaning less religious rituals an ceremonies.

(d) Whow took over the Samaj leadership after the death of its founder?
Answer:
Maharishi Debendranath Tagore, carried on the work of the founder after his death in 1833.

Question 2.
Arya Samaj

(a) What was the original name of Swamy Dayananda Saraswathi?
Answer:
Mul Shankar

(b) Who was his guru”?
Answer:
Swamy Viijanand was his guru.

(c) What was his motto?
Answer:
His motto was “Go back to Vedas”.

(d) What did the Samaj advocate?
Answer:
The samaj advocated women education, inter-caste marriage and inter-dining.

Question 3.
Social Reform Movements in Kerala and Tamil nadu

(a) Name few Social Reformers from Kerala and Tamil Nadu.
Answer:
Sri Narayana Guru and Ayyarkali from Kerala and Ramalinga Adigal, Vaikunta Swamigal from Tamil Nadu.

(b) How was Vaikundar called by his followers. What was his cult called as?
Answer:
His followers called him respectfully as Ayya (father) and his cult was known as Ayya vazhi (The path of Ayya).

(c) What happened to the voluminous songs composed by Ramalinga Adigal?
Answer:
His voluminous songs were compiled and published under the title Thiruvarutpa (songs of Grace).

(d) Name the Association founded by Ayyankali inspired by Sree Narayana Guru?
Answer:
Ayyankali founded the Sadhu Jana Paripalana Sangam (Association for the protection of the poor) in 1907, inspired by Sree Narayana Guru.

Question 4.
Aligarh Movement

(a) Name the first religious movement of the Muslims.
Answer:
Aligarh Movement was the first religious movement of the Muslims.

(b) What did Sir Syed Ahmed Khan strongly believe?
Answer:
He strongly believed in the Hindu-Muslim unity.

(c) What was his greatest achievement?
Answer:
His greatest achievement was the establishment of the Mohammaden Anglo Oriental College at Aligarh in 1875.

(d) Name the newspaper published by him.
Answer:
Tahzil-ud-Akhlaq

VI. Answer the following questions in detail:

Question 1.
Write about Brahmo Samaj.

(a) The Foundation of the Brahmo Samaj:
Answer:

1. Raja Rammohan Roy founded “Atmiya Saba” in 1815.
2. Later it developed into “Brahmo Samaj” in 1828.

(b) Aim of the Brahmo Samaj:
Answer:
To cure Hindu society and religion from all its evils and set it on right footing.

(c) Principle of the Brahmo Samaj:
Answer:
The Brahmo Samaj believed in a ‘Universal religion’ based on the principles of one Supreme God.

(d) Services of the Brahmo Samaj:
Answer:

1. The samaj condemned idol worship, costly rites and rituals, cartle distinctions, untouchability and the practice of Sati.
2. Because of Raja Rammohan Roy’s hard work, William Bentinck passed Sati prohibition Act in 1829.
3. The samaj also fought against polygamy and child marriage and supported inter-caste marriage and widow remarriage.
4. It tried to obtain a respectable position for women in the Indian Society.
5. It encouraged the study of English language and the western science in India. By western studies superstitions and blind faiths were removed from India. After the death of Raja Rammohan Roy, the works of the samaj was carried on by great men like Keshab Chandra Sen and Devendranath Tagore.

Question 2.
Write a note on Social Reformers of Tamil Nadu in the 19th century.
Answer:
Ramalinga Swamigal: Ramalinga Adigal was popularly known as Vallalar. He showed his compassion and mercy on all living beings including plants. This he called as Jeevakarunya. He established Samarasa Vedha Sanmarga Sangam in 1865. He also established a free feeding house for everyone irrespective of caste at Vadalur in 1867. He wrote many songs which were compiled in the name of Thiruvarutpa.

Vaikunda Swamigal: His original name was Mudichudum Perumal which changed as Muthukutty because of the objection by upper class. He calls himself Vaikuntar. He preached the idea of equality and condemned the ’ worship of idols. He was against animal sacrifice in the name of religion. He was against caste differences and wanted social integration of the society.

Iyothee Thassaar: Iyothee Thassar was scholar , writer ,siddha medicine practitioner and also journalist and political activist. He campaigned for the social justice and worked for building up a casteless society and removal of untouchability. He established several schools for untouchables in Tamil Nadu. He published a weekly ‘Oru Paisa Tamilan’ in 1907 and it lasted till 1914. He worked also for the revival of Buddhism.

Question 3.
“Ramalinga Adigal played a prominent role in the social and Religious Reform Movements” – Explain it.
Answer:

1. St. Ramalinga tried his best to find solution to the social evils by inculcating spiritual awareness.
2. St. Ramalinga led to unity and solidarity of the Tamils.
3. He favoured the creation of a casteless society to be guided by God.
4. He believed God to be in the form of “Arul Perum Jothi”.
5. Ramalinga Adigal believed that hunger and poverty are the evils of the society.
6. He advocated that feeding the poor is the highest form of worship.
7. Ramalinga Adigal condemned the inequalities based on birth and promoted universal love and brother-hood.
8. As a result, he founded Sathya Dharma Sala at Vadalur for feeding the poor. It provides food to everyone irrespective of caste and creed throughout the year.
9. He showed compassion and mercy not only on human beings, but also on plants, insects, birds and animals. This is called “Jeeva Karunya”.
10. He opposed superstitious beliefs and rituals.
11. He emphasized on being vegetarian.
12. He forbade the killing of animals for the sake of food.
13. In 1870, he established Sathya Gnana Sabai.
14. His devotional songs are compiled in a volume called “Thiru Arutpa”.

Important years and events:

We think the data given here clarify all your queries of Chapter 5 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science History Chapter 5 Social and Religious Reform Movements in the 19th Century Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 10 Types of Chemical Reactions

Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 10th Science Chapter 10 Types of Chemical Reactions Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 10 easily after studying the Tamilnadu State Board Class 10th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 10 Types of Chemical Reactions solutions of Class 10th by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 10th Science Types of Chemical Reactions Textual Problems Solved

Chemical Reactions Questions And Answers Pdf Question 1.
Calculate the pH of 0.01 M HNO₃?
Solution:

Samacheer Kalvi Guru 10th Science Question 2.
The hydroxyl ion concentration of a solution is 1 × 10^-9 M. What is the pOH of the solution?
Solution:

Samacheer Kalvi Question 3.
A solution has a pOH of 11.76. What is the pH of this solution?
Solution:
pH = 14 – pOH
pH = 14 – 11.76 = 2.24.

10th Science Samacheer Kalvi Question 4.
Calculate the pH of 0.001 molar solution of HCl.
Solution:
HCl is a strong acid and is completely dissociated in its solutions according to the process:
\(\mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
From this process it is clear that one mole of HCl would give one mole of H⁺ ions.
Therefore, the concentration of H⁺ ions would be equal to that of HCl, i.e., 0.001 molar or 1.0 × 10^-3 mol litre^-1.

Samacheer Kalvi 10th Science Question 5.
What would be the pH of an aqueous solution of sulphuric acid which is 5 × 10^-5 mol litre^-1 in concentration?
Solution:
Sulphuric acid dissociates in water as:
\(\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \rightarrow 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)
Each mole of sulphuric acid gives two moles of H⁺ ions in the solution. One litre of H₂SO₄ solution contains 5 × 10^-5 moles of H₂SO₄ which would give 2 × 5 × 10^-5 = 10 × 10^-5 or 1.0 × 10^-4 moles of H⁺ ion in one litre of the solution.
Therefore,

10th Samacheer Kalvi Science Question 6.
Calculate the pH of 1 × 10^-4 molar solution of NaOH.
Solution:
NaOH is a strong base and dissociates in its solution as:
\(\mathrm{NaOH}_{(\mathrm{aq})} \rightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
One mole of NaOH would give one mole of OH^– ions.
Therefore,

10th Science Solution Samacheer Kalvi Question 7.
Calculate the pH of a solution in which the concentration of the hydrogen ions is 1.0 × 10^-8 mol litre^-1.
Solution:
Here, although the solution is extremely dilute, the concentration given is not of an acid or a base but that of H⁺ ions. Hence, the pH can be calculated from the relation:

Samacheer Kalvi 10th Science Solutions Question 8.
If the pH of a solution is 4.5, what is its pOH?
Solution:
pH + pOH = 14
⇒ pOH = 14 – 4.5 = 9.5
⇒ pOH = 9.5.

Samacheer Kalvi 10th Science Types of Chemical Reactions Textual Evaluation Solved

I. Choose the correct answer.

Samacheer Kalvi 10th Science Solution Question 1.
\(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\) is a ______.
(a) Decomposition Reaction
(b) Combination Reaction
(c) Single Displacement Reaction
(d) Double Displacement Reaction.
Answer:
(b) Combination Reaction
Hint: It is a combination reaction between H₂ and Cl₂ to form HCl as product.

Science Solution Class 10 Samacheer Kalvi Question 2.
Photolysis is a decomposition reaction caused by:
(a) heat
(b) electricity
(c) light
(d) mechanical energy
Answer:
(c) light

Samacheer Kalvi 10 Science Question 3.
A reaction between carbon and oxygen is represented by \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}\) + Heat. In which of the type(s), the above reaction can be classified?
(i) Combination Reaction
(ii) Combustion Reaction
(iii) Decomposition Reaction
(iv) Irreversible Reaction
(a) i and ii
(b) i and iv
(c) i, ii and iii
(d) i, ii and iv.
Answer:
(d) i, ii and iv.
Hint:

• It is a combination reaction with one product.
• It involves the combustion of carbon to form CO₂.
• The reaction is irreversible and we cannot reverse the reaction.

Samacheer Kalvi Chemistry Question 4.
The chemical equation
Na₂SO_4(aq) + BaCI_2(aq) → BaSO_4(s)↓ + 2NaCl_(aq) represents which of the following types of reaction?
(a) Neutralisation
(b) Combustion
(c) Precipitation
(d) Single displacement
Answer:
(c) Precipitation

Samacheer Kalvi 10th Science Solution Book Question 5.
Which of the following statements are correct about a chemical equilibrium?
(i) It is dynamic in nature
(ii) The rate of the forward and backward reactions are equal at equilibrium
(iii) Irreversible reactions do not attain chemical equilibrium
(iv) The concentration of reactants and products may be different
(a) i, ii and iii
(b) i, ii and iv
(c) ii, iii and iv
(d) i, iii and iv.
Answer:
(a) i, ii and iii
Hint: Chemical equilibrium is dynamic in nature for a reversible reaction.
At equilibrium, Rate of forwarding reaction = Rate of backward reaction.

Samacheer Kalvi Guru 10th Science Book Back Answers Question 6.
A single displacement reaction is represented by
X_(s) + 2HCl_(aq) → XCl_2(aq) + H_2(g). Which of the following(s) could be X?
(i) Zn
(ii) Ag
(iii) Cu
(iv) Mg. Choose the best pair.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

Samacheer Kalvi 10th Standard Science Question 7.
Which of the following is not an “element + element → compound” type reaction?
(a) \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}\)
(b) \(2 \mathrm{K}_{(\mathrm{s})}+\mathrm{Br}_{2(l)} \rightarrow 2 \mathrm{KBr}_{(\mathrm{s})}\)
(c) \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})}\)
(d) \(4 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3(\mathrm{s})}\).
Answer:
(c) \(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})}\)
Hint: It involves one reactant compound i.e., carbon monoxide (CO).
\(2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2}\)

10th Standard Science Samacheer Kalvi Question 8.
Which of the following represents a precipitation reaction?
(a) \(\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{s})}\)
(b) \(\mathrm{A}_{(\mathrm{s})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{aq})}+\mathrm{D}_{(\mathrm{l})}\)
(c) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{aq})}\)
(d) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{aq})}+\mathrm{D}_{(\mathrm{l})}\).
Answer:
(c) \(\mathrm{A}_{(\mathrm{aq})}+\mathrm{B}_{(\mathrm{aq})} \rightarrow \mathrm{C}_{(\mathrm{s})}+\mathrm{D}_{(\mathrm{aq})}\)
Hint: It involves the formation of solid precipitation by mixing of two aqueous solutions.

10th Standard Samacheer Kalvi Science Question 9.
The pH of a solution is 3. Its [OH^–] concentration is:
(a) 1 × 10^-3 M
(b) 3 M
(c) 1 × 10^-11 M
(d) 11 M
Answer:
(c) 1 × 10^-11 M

Samacheer Kalvi 10th Science Book Question 10.
Powdered CaCO₃ reacts more rapidly than flaky CaCO₃ because of ______.
(a) large surface area
(b) high pressure
(c) high concentration
(d) high temperature.
Answer:
(a) large surface area
Hint: We know that greater the surface area, faster will be a chemical reaction. Hence powdered CaCO₃ reacts more rapidly.

II. Fill in the blanks.

Samacheer Kalvi Science 10th Question 1.
A reaction between an acid and a base is called ______.
Answer:
Neutralization.

Question 2.
When lithium metal is placed in hydrochloric acid, _______ gas is evolved.
Answer:
Hydrogen.

Question 3.
The equilibrium attained during the melting of ice is known as ______.
Answer:
Physical equilibrium.

Question 4.
The pH of a fruit juice is 5.6. If you add slaked lime to this juice, its pH ______ (increase/decrease)
Answer:
Increases.

Question 5.
The value of the ionic product of water at 25°C is ______.
Answer:
1 × 10^-14

Question 6.
The normal pH of human blood is ______.
Answer:
7.4.

Question 7.
Electrolysis is type of _______ reaction.
Answer:
Decomposition reaction.

Question 8.
The number of products formed in a synthesis reaction is ______.
Answer:
One.

Question 9.
Chemical volcano is an example for ______ type of reaction.
Answer:
Decomposition reaction.

Question 10.
The ion formed by dissolution of H⁺ in water is called ______.
Answer:
Hydronium ion.

III. Match the following

Question 1.
Identify the types of reaction.

Answer:
i – c, ii-a, iii – d, iv- b.

IV. True or False: (If false give the correct statement)

Question 1.
Silver metal can displace hydrogen gas from nitric acid.
Answer:
False.
Correct statement: In the activity series, any metals that are below hydrogen will not react with HNO₃.

Question 2.
The pH of rainwater containing dissolved gases like SO₃, CO₂, NO₂ will be less than 7.
Answer:
True.

Question 3.
At the equilibrium of a reversible reaction, the concentration of the reactants and the products will be equal.
Answer:
False.
Correct statement: At equilibrium rate of the forward reaction is equal to the rate of backward reaction.

Question 4.
Periodical removal of one of the products of a reversible reaction increases the yield.
Answer:
True.

Question 5.
On dipping a pH paper in a solution, it turns into yellow. Then the solution is basic.
Answer:
True.

V. Short Answer Questions

Question 1.
When an aqueous solution of potassium chloride is added to an aqueous solution of silver nitrate, a white precipitate is formed. Give the chemical equation of this reaction.
Answer:
\(\mathrm{KCl}_{(\mathrm{aq})}+\mathrm{AgNO}_{3(\mathrm{aq})} \rightarrow \mathrm{AgCl}_{(\mathrm{s})}+\mathrm{KNO}_{3(\mathrm{aq})}\)
Formation of white precipitate by the above reaction is due to formation of silver chloride (AgCl).

Question 2.
Why does the reaction rate of a reaction increase on raising the temperature?
Answer:
The rate of a reaction increases at higher temperature, because the heat to the reactants provide energy to breakup more bonds and speeds up the reaction.

Question 3.
Define a combination reaction. Give one example of an exothermic combination reaction.
Answer:
A chemical reaction in which 2 or more reactants combine to form a single product, the reaction is known as combination reaction. Most of the combination reaction are exothermic because they involve formation of new bonds. For example,
\(\mathrm{H}_{2(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2 \mathrm{HCl}_{(\mathrm{g})}\)
\(\mathrm{SiO}_{2(\mathrm{s})}+\mathrm{CaO}_{(\mathrm{s})} \rightarrow \mathrm{CaSiO}_{3(\mathrm{s})}\).

Question 4.
Differentiate reversible and irreversible reactions.
Answer:

VI. Answer in detail.

Question 1.
What is called thermolysis reactions?
Answer:
A chemical reaction is a process in which old bond breaks up and new chemical bond get formed. Thermolysis chemical reactions is a special type of chemical reaction in which the reactant get decomposed by heat. For example,
\(\mathrm{CaCO}_{3(\mathrm{s})} \stackrel{\text { Heat }}{\rightleftharpoons} \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\)
\(2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\mathrm{Heat}}{\longrightarrow} 2 \mathrm{Hg}_{(\mathrm{l})}+\mathrm{O}_{2(\mathrm{g})}\)
In these reactions heat is supplied to break the bonds, so generally they are endothermic in nature.

Question 2.
Explain the types of double displacement reactions with examples.
Answer:
When two compounds react, if their ions are interchanged, then the reaction is called double displacement reactions. There are two types of double displacement reactions. They are
(i) Precipitation reactions : When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction. Because the insoluble compound, formed as one of the products, is a precipitate and hence the reaction is so called.

(ii) When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, a double displacement reaction takes place between them.
Pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s)↓ + 2KNO_3(aq).

(iii) Potassium and lead displace or replace one other and form a yellow precipitate of lead (II) iodide.

(iv) Neutralization reactions: When an acid reacts with the base to form a salt and water. It is called ‘neutralization reaction’ as both acid and base neutralize each other.

(v) Reaction of sodium hydroxide with hydrochloric acid is a typical neutralization reaction. Here, sodium replaces hydrogen from hydrochloric acid forming sodium chloride, a neutral soluble salt.
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

Question 3.
Explain the factors influencing the rate of a reaction.
Answer:
The factors influencing the rate of a reaction are,
(i) Nature of the reactants: The reaction of sodium with hydrochloric acid is faster than that with acetic acid. Do you know why? Hydrochloric acid is a stronger acid than acetic acid and thus more reactive. So, the nature of the reactants influences the reaction rate.
\(\begin{aligned} 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} & \rightarrow 2 \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}(\mathrm{fast}) \\ 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} & \rightarrow 2 \mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}(\mathrm{slow}) \end{aligned}\).

(ii) The concentration of the reactants: Changing the number of reactants also increases the reaction rate. The amount of the substance present in a certain volume of the solution is called ‘concentration’. More the concentration, more particles per volume exist in it and hence faster the reaction. Granulated zinc reacts faster with 2M hydrochloric acid than 1M hydrochloric acid.

(iii) Temperature: Most of the reactions go faster at a higher temperature. Because adding heat to the reactants provides energy to break more bonds and thus speed up the reaction. Calcium carbonate reacts slowly with hydrochloric acid at room temperature. When the reaction mixture is heated the reaction rate increases.

(iv) Pressure: If the reactants are gases, increasing their pressure increases the reaction rate. This is because on increasing the pressure the reacting particles come closer and collide frequently.

(v) Catalyst: A catalyst is a substance which increases the reaction rate without being consumed in the reaction. In certain reactions, adding a substance as catalyst speeds up the reaction. For example, on heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.

(vi) The surface area of the reactants: When solid reactants are involved in a reaction, their powdered form reacts more readily. For example, powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Because powdering of the reactants increases the surface area and more energy is available on the collision of the reactant particles. Thus, the reaction rate is increased. You will study more about reaction rate in your higher classes.

Question 4.
How does pH play an important role in everyday life?
Answer:
Our body works within the pH range of 7.0 to 7.8. Living organisms can survive only in a narrow range of pH change. Different body fluids have different pH values.
Eg: pH of blood is ranging from 7.35 to 7.45. Any increase or decrease in this value leads to diseases. The ideal pH for blood is 7.4.

pH in our digestive system : It is very interesting to note that our stomach produces hydrochloric acid. It helps in the digestion of food without harming the stomach. During indigestion the stomach produces too much acid and this causes pain and irritation. pH of the stomach fluid is approximately 2.0.

pH changes as the cause of tooth decay : pH of the saliva normally ranges between 6.5 to 7.5. White enamel coating of our teeth is calcium phosphate, the hardest substance in our body. When the pH of the mouth saliva falls below 5.5, the enamel gets weathered. Toothpastes, which are generally basic ate used for cleaning the teeth that can neutralise the excess acid and prevent tooth decay.

pH of soil : In agriculture, the pH of the soil is very important. Citrus fruits require slightly alkaline soil, while rice requires acidic soil and sugarcane requires neutral soil.

pH of rain water : The pH of rain water is approximately 7, which means that it is neutral and also represents its high purity. If the atmospheric air is polluted with oxide gases of sulphur and nitrogen, they get dissolved in the rain water and make its pH less than 7. Thus, if the pH of rain water is less than 7, then it is called acid rain. When acid rain flows into the rivers it lowers the pH of the river water also. The survival of aquatic life in such rivers becomes difficult.

Question 5.
What is chemical equilibrium? What are its characteristics?
Answer:
Chemical equilibrium is the state for a reversible chemical reaction where the rate of forwarding direction is equally balanced by the rate of backward direction and the process seems like to be stopped.
Its characteristics are,

• The rate of the forward and backward reaction are equal in chemical equilibrium.
• The observable properties such as pressure, concentration, colour, density, viscosity, etc. of the system unchanged with time.
• In physical equilibrium, the volume of all phases remains constant.

VII. HOT Questions

Question 1.
A solid compound ‘A’ decomposes on heating into ‘B’ and a gas ‘C’ On passing the gas ‘C’ through water. It becomes acidic. Identify A, B and C.
\(\mathrm{CaCO}_{3(\mathrm{s})} \longrightarrow \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{g})}\)
Calcium carbonate decomposes into solid Calcium oxide and a gas CO₂. This CO₂ dissolves in water and forms carbonic acid.
\(\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}\) (Carbonic acid).

Question 2.
Can a nickel spatula be used to stir copper sulphate solution? Justify your answer.
Answer:
No, a nickel spatula cannot be used to stir CuSO₄ solution, because Nickel will displace copper from CuSO₃ solution and Cu gets deposited on the Ni spatula.

VIII. Solve the following problems.

Question 1.
Lemon juice has a pH = 2, what is the concentration of H⁺ ions?
Solution:

Question 2.
Calculate the pH of 1.0 × 10^-4 molar solution of HNO₃?
Solution:

Question 3.
What is the pH of 1.0 × 10^-5 molar solution of KOH?
Solution:

Question 4.
The hydroxide ion concentration of a solution is 1 × 10^-11. What is the pH of the solution?
Solution:

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Question Solved

I. Choose the best answer.

Question 1.
Methane + Oxygen → A + Water. Identify A ______.
(a) Carbon monoxide
(b) Carbon dioxide
(c) ethane
(d) LPG.
Answer:
(i) Carbon dioxide

Question 2.
The chemical reaction in which electricity is used to bring about the change is:
(a) Thermal decomposition
(b) Photo decomposition
(c) Single displacement reaction
(d) Electrolytic decomposition
Answer:
(d) Electrolytic decomposition

Question 3.
Decomposition of the molecule occurs on passing an electric current through its aqueous solution. This process is termed as ______.
(a) Thermolysis
(b) Photolysis
(c) Electrolysis
(d) None of these.
Answer:
(c) Electrolysis

Question 4.
The most reactive element in the activity series is:
(a) platinum
(b) potassium
(c) sodium
(d) gold
Answer:
(b) potassium

Question 5.
Which one of the following is a more reactive element?
(a) Cu
(b) Li
(c) Zn
(d) Pb.
Answer:
(b) Li

Question 6.
Which one of the following is the least reactive element?
(a) Au
(b) Fe
(c) Ca
(d) Na.
Answer:
(a) Au

Question 7.
Double displacement reaction is also called as ______.
(a) Metastasis
(b) Metathesis
(c) Methanolysis
(d) Metalysis.
Answer:
(b) Metathesis

Question 8.
The acidic solution among the following is:
(a) seawater
(b) coffee
(c) lime water
(d) antacid
Answer:
(b) coffee

Question 9.
The reaction of sodium hydroxide with hydrochloric acid is an example for ______.
(a) Combination reaction
(b) Thermolysis reaction
(c) Neutralization reaction
(d) Precipitation reaction.
Answer:
(c) Neutralization reaction

Question 10.
In agriculture, the nature of the soil for rice is:
(a) alkaline
(b) neutral
(c) acidic
(d) none of the above
Answer:
(c) acidic

Question 11.
The role of manganese dioxide in the decomposition of potassium chlorate is ______.
(a) act as a reactant
(b) act as a catalyst
(c) act as a reagent
(d) act as a reaction medium.
Answer:
(b) act as a catalyst

Question 12.
The ionic product of water K_w is ______.

Answer:
(c) \(\mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\)

Question 13.
Unit of ionic product of water is ______.
(a) mol^-2 dm^-6
(b) mol² dm^-6
(c) mol² dm⁶
(d) mol dm^-3.
Answer:
(b) mol² dm^-6

Question 14.
What is the value of ionic product of water?
(a) 1 × 10^-4
(b) 1 × 10^-1
(c) 1 × 10¹⁴
(d) 1 × 10^-14.
Answer:
(d) 1 × 10^-14.

Question 15.
pH of milk of magnesia is ______.
(a) 10
(b) 15
(c) 14
(d) 10.5.
Answer:
(a) 10

Question 16.
What is the relationship between pH and pOH?
(a) pH – pOH = 14
(b) pH + pOH = 14
(c) pH/pOH = 14
(d) pH + pOH = 1.4.
Answer:
(b) pH + pOH = 14

Question 17.
\(\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2} \uparrow\)
The above reaction is an example of ______.
(а) Combination reaction
(b) Double displacement reaction
(c) Displacement reaction
(d) Decomposition reaction.
Answer:
(c) Displacement reaction

Question 18.
A student tests the pH of pure water using a pH paper. It shows a green colour. If a pH paper is after adding lemon juice to water, what colour will he observe?
(a) Green
(b) Red
(c) Yellow
(d) Pink.
Answer:
(b) Red

Question 19.
When aqueous solution of silver nitrate and sodium chloride are mixed, ______ precipitate is immediately formed.
(a) white
(b) yellow
(c) red
(d) black.
Answer:
(a) white

Question 20.
pH = – log₁₀ [H⁺]. The pH of a solution containing hydrogen ion concentration of 0.001 M solution is ______.
(a) 3
(b) 11
(c) 14
(d) 15.
Answer:
(a) 3

Question 21.
Silver anklet has got tarnished due to the formation of ______.
(a) Ag₂O
(b) AgNO₃
(c) Ag₂S
(d) AgBr.
Answer:
(c) Ag₂S

Question 22.
The colour of the precipitate formed by the reaction of lead nitrate with potassium iodide is ______.
(a) white
(b) yellow
(c) black
(d) red.
Answer:
(b) yellow

Question 23.
The reaction of calcium oxide with water is a ______ reaction.
(a) exothermic
(b) endothermic
(c) isothermic
(d) adiabatic.
Answer:
(a) exothermic

Question 24.
The gas released when calcium carbonate reacts with dilute hydrochloric acid is ______.
(a) NO₂
(b) CO
(c) CO₂
(d) H₂.
Answer:
(c) CO₂

Question 25.
The chemical used in whitewashing on the walls is ______.
(a) CaO
(b) CaCl₂
(c) CaOCl₂
(d) Ca(OH)₂.
Answer:
(d) Ca(OH)₂.

Question 26.
The chemical formula of marble is ______.
(a) CaCO₃
(b) CaOCl₂
(c) CaSO₄. \(\frac { 1 }{ 2 }\) H₂O
(d) Ca(OH)₂.
Answer:
(a) CaCO₃

Question 27.
Combustion of coal is an example of _____ reaction.
(a) displacement
(b) reduction
(c) decomposition
(d) combination.
Answer:
(d) combination.

Question 28.
The colour change takes place when copper carbonate is strongly heated is ______.
(a) green to black
(b) green to blue
(c) blue to green
(d) blue to black.
Answer:
(a) green to black

Question 29.
The reaction of heat on copper carbonate into copper (II) oxide is _________ reaction.
(a) combination
(b) decomposition
(c) displacement
(d) redox.
Answer:
(b) decomposition

Question 30.
The dissolution of glucose in water is a _____ reaction.
(a) exothermic
(b) endothermic
(c) neutralisation
(d) combination.
Answer:
(b) endothermic

Question 31.
All combustion reactions are _______ reactions.
(a) combination
(b) exothermic
(c) endothermic
(d) neutralisation.
Answer:
(b) exothermic

Question 32.
The factor that affects the rate of the chemical reaction is ______.
(a) temperature
(b) concentration
(c) catalyst
(d) all the above.
Answer:
(d) all the above.

Question 33.
Magnesium ribbon reacts at a very faster rate with ______ acid.
(a) hydrochloric
(b) acetic
(c) formic
(d) oxalic.
Answer:
(a) hydrochloric

Question 34.
Our body metabolism is carried out by means of ______ secreted in our stomach.
(a) sulphuric acid
(b) hydrochloric acid
(c) nitric acid
(d) acetic acid.
Answer:
(b) hydrochloric acid

Question 35.
Which metals do not liberate gas on reaction with acids?
(a) Zn, Mg
(b) Ag, Cu
(c) Na, K
(d) Cr, Al.
Answer:
(b) Ag, Cu

Question 36.
When CO₂ is passed through lime water, it turns ______.
(a) milky
(b) black
(c) red
(d) blue.
Answer:
(a) milky

Question 37.
The physical form of calcium carbonate is ______.
(a) limestone
(b) chalk
(c) marble
(d) all the above.
Answer:
(d) all the above.

Question 38.
The colour change takes place when copper (II) oxide reacts with dilute hydrochloric acid is ______.
(a) blue to green
(b) black to green
(c) green to black
(d) green to blue.
Answer:
(b) black to green

Question 39.
Which of the following is a strong base?
(a) NH₄OH
(b) Ca(OH)₂
(c) Al(OH)₃
(d) NaOH.
Answer:
(d) NaOH.

Question 40.
Zinc reacts with sodium hydroxide to form ______.
(a) Zinc hydroxide + H₂O
(b) Sodium zincate + H₂
(c) Zinc oxide + H₂
(d) Zinc oxide + H₂O.
Answer:
(b) Sodium zincate + H₂

Question 41.
_____ reacts with sodium hydroxide.
(a) Cu
(b) Ag
(c) Cr
(d) Zn.
Answer:
(d) Zn.

Question 42.
________ is used as a medicine for stomach disorder.
(a) Sodium hydroxide
(b) Ammonium hydroxide
(c) Magnesium hydroxide
(d) Calcium hydroxide.
Answer:
(c) Magnesium hydroxide

Question 43.
The pH of stomach fluid is ______.
(a) 12
(b) 14
(c) 2
(d) 1.
Answer:
(c) 2

Question 42.
The hardest substance in the human body is ______.
(a) bone
(b) the enamel coating of teeth
(c) brain
(d) liver.
Answer:
(b) the enamel coating of teeth

Question 43.
Sugarcane requires _______ soil.
(a) acidic
(b) alkaline
(c) neutral
(d) amphoteric.
Answer:
(c) neutral

Question 44.
Rice requires _____ soil.
(a) acidic
(b) basic
(c) alkaline
(d) neutral.
Answer:
(a) acidic

Question 45.
______ is a double salt.
(a) Sodium chloride
(b) Washing soda
(c) Potash alum
(d) Bleaching powder.
Answer:
(c) Potash alum

II. Fill in the blanks.

Question 1.
During chemical changes _____ are formed and these changes are more _____ than physical changes.
Answer:
New products, permanent.

Question 2.
Calcium oxide reacts with water to produce ______ and the reaction is ______.
Answer:
Slaked lime, exothermic.

Question 3.
During whitewashing, ______ reacts slowly with carbon dioxide in the air to form a thin layer of _______ on the walls.
Answer:
Calcium hydroxide, calcium carbonate.

Question 4.
When copper carbonate is heated, the products formed are _____ , ______ and change of colour from ______ to ____ is observed.
Answer:
CuO, CO₂, green, black

Question 5.
When lead nitrate is heated, the gas liberated is ______ and its colour is ______.
Answer:
NO₂, Reddish – brown

Question 6.
Copper sulphate solution changes its blue colour into _____ colour when an iron nail is added to it and it acquires _____ colour.
Answer:
Green, brownish.

Question 7.
When Barium chloride reacts with sodium sulphate, the product formed is _____ and it is a ______ precipitate.
Answer:
Barium sulphate, white.

Question 8.
Powdered ____ reacts more quickly with hydrochloric acid than marble chips.
Answer:
CaCO₃

Question 9.
Calcium carbonate present in the marble reacts with hydrochloric acid at a faster rate at _______ temperature.
Answer:
Higher.

Question 10.
______ is a substance which furnishes H⁺ ions when dissolved in water and a ______ is a substance which furnishes OH^– ions in water.
Answer:
Acid, Base.

Question 11.
Acids present in plants and animals are ______ and the acids in rocks and minerals are ______.
Answer:
Organic acid, inorganic acid.

Question 12.
Metal displaces _______ gas from dilute acid and the flame goes off with a ______ sound.
Answer:
Hydrogen, popping.

Question 13.
_____ is used in the manufacturing of soap and ______ is used in whitewashing buildings.
Answer:
Sodium hydroxide, calcium hydroxide.

Question 14.
pH scale was introduced by ______ and the pH of the solution is 7, it is a ______ solution.
Answer:
S.P.L. Sorenson, neutral.

Question 15.
The pH of a normal healthy human skin is _____ and the pH of the stomach fluid is ______.
Answer:
7.4, 5.5.

Question 16.
The ideal pH for blood is _______ and pH of the mouth falls below ______.
Answer:
7.35 – 7.45, Sulphuric acid.

Question 17.
All photo decomposition reaction are _______ reactions.
Answer:
Endothermic.

Question 18.
Precipitation reactions give a _____ as the product.
Answer:
Insoluble salt.

Question 19.
Plants cannot grow in a ______ soil.
Answer:
Acidic.

Question 20.
Equilibrium is possible in a _____ system.
Answer:
Closed.

Question 21.
Pure water is a _______ electrolyte.
Answer:
Weak.

Question 22.
Most of the combination reactions are _______ in nature.
Answer:
Exothermic.

Question 23.
Silicon dioxide reacts with calcium oxide to form ______.
Answer:
Calcium silicate.

Question 24.
Our mobile phones get energy from its _____ battery by chemical reaction.
Answer:
Lithium.

Question 25.
A ______ is a substance which increases the reaction rate.
Answer:
Catalyst.

Question 26.
pH range of human saliva is ______.
Answer:
6 – 8.

Question 27.
pH range of fresh milk is ______.
Answer:
5.

Question 28.
The pH of a solution can be determined by using a _______ indicator.
Answer:
Universal.

Question 29.
pH is ______.
Answer:
-log₁₀ [H⁺].

Question 30.
The white enamel coating of our teeth is ______.
Answer:
Calcium phosphate.

III. Match the following.

Question 1.

Answer:
i – b, ii – a, iii – d, iv – c.

Question 2.

Answer:
i – c, ii – d, iii – b, iv – a.

Question 3.

Answer:
i – b, ii – a, iii – d, iv – c.

Question 4.

Answer:
i – c, ii – d, iii – a, iv – b.

Question 5.

Answer:
i – c, ii – d, iii – b, iv – a.

Question 6.

Answer:
i – d, ii-a, iii – b, iv – e.

IV. State true or false. If false, give the correct statement.

Question 1.
Chemical changes are reversible changes.
Answer:
False.
Correct statement: Chemical changes are irreversible changes.

Question 2.
The silver anklet has got tarnished when exposed to air due to the formation of silver oxide. (Ag₂O).
Answer:
False.
Correct statement: The silver anklet has got tarnished when exposed to air due to the formation of silver sulphide (Ag₂S).

Question 3.
Brisk effervescence takes place with the evolution of CO₂ when calcium carbonate reacts with dilute hydrochloric acid.
Answer:
True.

Question 4.
The chemical formula for marble is CaCl₂.
Answer:
False.
Correct statement: The chemical formula for marble is CaCO₃

Question 5.
A reaction in which a single product is formed from two or more reactants is known as displacement reaction.
Answer:
False.
Correct statement: A reaction in which a single product is formed from two or more reactants is known as a combination reaction.

Question 6.
Lead Nitrate on heating decomposes to give lead oxide with the evolution of reddish-brown gas NO₂ and O₂ gas.
Answer:
True.

Question 7.
Any reaction that produces a precipitate is called a redox reaction.
Answer:
False.
Correct statement: Any reaction that produces a precipitate is called a precipitation reaction.

Question 8.
The reaction in which a more reactive element displaces a less reactive element from its compound is called displacement reaction.
Answer:
True.

Question 9.
The chemical reactions which take place with the evolution of heat energy are called endothermic reaction.
Answer:
False.
Correct statement: The chemical reactions which take place with the evolution of heat energy are called exothermic reaction.

Question 10.
All combustion reactions are endothermic reactions.
Answer:
False.
Correct statement: All combustion reactions are exothermic reactions.

Question 11.
Our body metabolism is carried out by means of sulphuric acid secreted in our stomach.
Answer:
False.
Correct statement: Our body metabolism is carried out by means of hydrochloric acid secreted in our stomach.

Question 12.
Limestone, chalk and marble are different physical forms of calcium oxide.
Answer:
False.
Correct statement: Limestone, chalk and marble are different physical forms of calcium carbonate.

Question 13.
The atmosphere of earth is made up of thick white and yellowish clouds of sulphuric acid.
Answer:
False.
Correct statement: The atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid.

Question 14.
Bond breaking releaser energy whereas bond formation absorbs energy.
Answer:
False.
Correct statement: Bond breaking absorbs energy whereas bond formation releases energy.

Question 15.
More active elements readily displace less active elements from their aqueous solution.
Answer:
True.

Question 16.
The irreversible reaction is relatively slow.
Answer:
False.
Correct statement: Irreversible reaction is fast.

V. Assertion and Reason

Question 1.
Assertion (A): The lustrous white colour of the silver anklet slowly changes into slightly back colour.
Reason (R): silver anklet reacts with H₂S in the air to form Ag₂S (Silver Sulphide) which is black in colour.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) Both (A) and (R) are wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): The reaction of calcium oxide with water is an exothermic reaction.
Reason (R): The reaction is accompanied by a hissing sound and formation of bubbles leading to the absorption of a considerable amount of heat.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(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

Question 3.
Assertion (A): During the reaction of calcium carbonate with dilute hydrochloric acid, brisk effervescence takes place.
Reason (R): Brisk effervescence is due to the evolution of carbon dioxide gas.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 4.
Assertion: When copper carbonate is heated strongly, the green colour is changed into a black colour.
Reason (R): The colour change is due to the decomposition of copper carbonate into copper oxide.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 5.
Assertion (A): when lead nitrate is heated, the gas released has red – orange colour and it is lead oxide.
Reason (R): Lead nitrate on heating undergoes combination reaction.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are wrong

Question 6.
Assertion (A): iron js more reactive than copper.
Reason (R): iron js displaced from iron sulphate by copper.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(b) (A) is correct but (R) is wrong

Question 7.
Assertion (A): Copper displaces zinc (or) lead from the salt solution.
Reason (R): Copper is more reactive than zinc and lead.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(d) Both (A) and (R) are wrong.

Question 8.
Assertion (A): All combustion reactions are exothermic reactions.
Reason (R): During combustion reaction, heat energy is liberated.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(a) Both (A) and (R) are correct

Question 9.
Assertion (A): when glucose is kept on our tongue, a cooling effect is felt.
Reason (R): It is an endothermic reaction in which heat is absorbed.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): Powdered calcium carbonate reacts more quickly with hydrochloric acid than marble chips.
Reason (R): Powdered calcium carbonate offers a large surface area than marble chips. Because the greater the surface area, the greater is the rate of the reaction.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 11.
Assertion (A): Combination reaction is exothermic in nature.
Reason (R): During the formation of new bonds, releases a huge amount of energy in the form of heat.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) does not explain (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) (A) is correct and (R) explains (A)

Question 12.
Assertion (A): When glucose is kept on our tongue, a cooling effect is felt.
Reason (R): It is an endothermic reaction in which heat is absorbed.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) does not explain (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) (A) is correct and (R) explains (A)

VI. Short Answer Questions.

Question 1.
What type of chemical reaction takes place when
(i) limestone is heated
(ii) a magnesium ribbon is burnt in air?
Answer:
(i) When limestone is heated, the reaction takes place is a decomposition reaction.

(ii) When a magnesium ribbon is burnt in air, it is an oxidation reaction.

Question 2.
What is a combustion reaction?
Answer:
A combustion reaction is one in which the reactant rapidly combines with oxygen to form one or more oxides and energy (heat). So in combustion reactions, one of the reactants must be oxygen.

Question 3.
Which of the following is a combustion?

1. Digestion of food
2. Rusting of iron.

Answer:

1. Digestion of food : Not a combustion reaction, because it is a endothermic process and where energy is utilized.
2. Rusting of iron : Combustion reaction.

Question 4.
When the lead powder is added to copper chloride solution, a displacement reaction occurs and solid copper is formed.
(i) Write the equation for the reaction.
(ii) Why does the displacement reaction occur?
Answer:

(ii) Copper is less reactive than lead. So lead has displaced copper from copper chloride. It is a displacement reaction.

Question 5.
What happens when lead nitrate reacts with potassium iodide solution?
Answer:
Lead nitrate solution when reacts with a potassium iodide solution, a deep yellow precipitate of Pbl₂ is formed.

Question 6.
Differentiate Exothermic reaction and Endothermic reaction.
Answer:

Question 7.
Why the study of reaction rate is important?
Answer:
Faster the reaction, more will be the amount of the product in a specified time. So, the rate of a reaction is important for a chemist for designing a process to get a good yield of a product. Rate of reaction is also important for a food processor who hopes to slow down the reactions that cause food to spoil.

Question 8.
What is meant by decomposition reaction? Give an example.
Answer:
A single compound breaks down to produce two or more substances. Such type of reaction is called decomposition reaction.

Question 9.
Give reason.
(a) Granulated Zinc reacts faster with 2M HCl than 1M HCl.
Answer:
Reason: As the concentration of the reactant increases the rate of the reaction increases.

(b) Food kept at room temperature spoils faster than that kept in the refrigerator.
Answer:
Reason: In the refrigerator the temperature is lower than room temperature, so the reaction rate is less.

Question 10.
What is meant by double displacement reaction? Give an example.
Answer:
A double decomposition reaction is a reaction in which the exchange of ions between two reactants occur leading to the formation of two different products.

Question 11.
Why the toothpastes are generally basic in nature?
Answer:
The pH of the saliva is usually between 6.5 to 7.5, when the pH of the mouth saliva falls below 5.5 the enamel coating of our teeth calcium phosphate gets weathered. That is why the tooth pastes are generally basic in nature.

Question 12.
What happens to food containing fat and oil kept open for a long time?
Answer:
When food containing fat and oil is left as such for a long time, it becomes stale. The stale food develops bad taste and foul smell. This is very common in curd and cheese, particularly in summer. Oils and fats are slowly oxidised to certain foul – smelling compounds.

Question 13.
Define the rate of a chemical reaction.
Answer:
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or product per unit time.
\(\text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\).

Question 14.
Define catalyst.
Answer:
A substance which alters the rate of a reaction without undergoing any change in mass and composition is known as a catalyst.

Question 15.
What happens during a chemical reaction?
Answer:

• In a chemical reaction, the atoms of the reacting molecules or elements are rearranged to form new molecules.
• Old chemical bonds between atoms are broken and new chemical bonds are formed.
• Bond breaking absorbs energy whereas bond formation releases energy.

Question 16.
What; is a balanced chemical equation?
Answer:
A balanced chemical equation is the simplified representation of a chemical reaction which describes the chemical composition, the physical state of the reactants and the products, and the reaction conditions.

Question 17.
What are the main classes of decomposition reactions?
Answer:
There are three main classes of decomposition reactions. They are,

• Thermal decomposition reactions
• Electrolytic decomposition reactions
• Photo decomposition reactions

Question 18.
What are thermal decomposition reactions?
Answer:
In a thermal decomposition reaction, the reactant is decomposed by applying heat.
For example, on heating mercury (II) oxide is decomposed into mercury metal and oxygen gas.
\(2 \mathrm{HgO}_{(\mathrm{s})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Hg}_{(\mathrm{l})}+\mathrm{O}_{2(\mathrm{g})}\).

Question 19.
What are Electrolytic decomposition reactions?
Answer:
In this type of reaction, the reactant is decomposed by applying electricity.
For example, decomposition of sodium chloride occurs on passing electric current through its aqueous solution.
\(2 \mathrm{NaCl}_{(\mathrm{aq})} \stackrel{\text { Electricity }}{\longrightarrow} 2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{Cl}_{2(\mathrm{g})}\).

Question 20.
What is the photodecomposition reaction?
Answer:
In this type of reaction, the reactant is decomposed by applying light.
For example, when silver bromide is exposed to light, it breaks down into silver metal and bromine gas.
\(2 \mathrm{AgBr}_{(\mathrm{s})} \stackrel{\mathrm{Light}}{\longrightarrow} 2 \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Br}_{2(\mathrm{g})}\).

Question 21.
What is Metathesis reaction?
Answer:
The ion of one compound is replaced by the ion of another compound. Ions of identical charges are only interchanged, i.e. a cation can be replaced by another cation. This reaction is called metathesis reaction.

Question 22.
What is a precipitation reaction?
Answer:
When aqueous solution of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
e.g„ \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{aq})}+2 \mathrm{KI}_{(\mathrm{aq})} \longrightarrow \mathrm{PbI}_{2(\mathrm{s})} \downarrow+2 \mathrm{KNO}_{3(\mathrm{aq})}\).

Question 23.
How will you distinguish between combination and decomposition reactions?
Answer:

Question 24.
What is a neutralization reaction?
Answer:
It is another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called neutralization reaction as both acid and base neutralizes each other.

Question 25.
What is Combustion reaction?
Answer:
A combustion reaction is one in which the reactant rapidly combines with oxygen to form one or more oxides and energy (heat).

Question 26.
What are reversible and irreversible reactions?
Answer:
Reversible reaction : A reversible reaction is a reaction that can be reversed, i.e., the products can be converted back to the reactants.
e.g., \(\mathrm{PCl}_{5(\mathrm{g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})}\)
Irreversible reaction: The reaction that cannot be reversed is called irreversible reaction. The irreversible reactions are unidirectional.

Question 27.
Why is the reaction rate important?
Answer:
Faster the reaction, more will be the amount of the product in a specified time. So, the rate of a reaction is important for a chemist for designing a process to get a good yield of a product. Rate of reaction is also important for a food processor who hopes to slow down the reactions that cause food to spoil.

VII. Long Answer Questions.

Question 1.
Suggest a reason for each observation given below.
Answer:
(i) In fireworks, powdered Mg is used rather than Mg ribbon.
Powdered Mg will have larger surface area than Mg ribbon and the rate of the reaction increases.

(ii) Zn and dil. H₂SO₄ react much more quickly when a few drops of CuSO₄ solution are added.
When few drops of CuSO₄ are added to the solution containing Zn and dil. H₂SO₄, the rate of the reaction increases, because CuSO₄ acts as catalyst.

(iii) The reaction between MgCO₃ and dil. HCl speeds up when some con. HCl is added. As the concentration of the reactant increases, the rate of the reaction increases.

Question 2.
Observe the given chemical change and answer the following:

1. Identify ‘A’ and ‘B’.
2. Write the commercial name of calcium hydroxide.
3. Identify products ‘C’ and ‘D’ when HCl is allowed to oxide react with calcium oxide.
4. Say whether calcium oxide is acidic or basic.

Answer:

1. A is calcium carbonate. B is carbon-di-oxide.
2. Slaked lime is the commercial name of calcium hydroxide.
3. The products C and D are calcium chloride (CaCl₂) and water (H₂O).
4. Calcium oxide is basic in nature.

Question 3.
Take copper nitrate in a test tube and heat it over the flame.

1. What is the colour of cupric nitrate?
2. What do you observe?
3. Name the type of reaction that takes place.
4. Write the balanced equation.

Answer:

1. The colour of cupric nitrate is Blue.
2. When cupric nitrate is heated in a test tube, we can observe the evolution of reddish-brown gas (NO₂) Nitrogen dioxide.
3. The reaction takes place is a decomposition reaction.
4. 

Question 4.
Redox reactions are reactions during which electron transfer takes place. Here magnesium atom transfers two electrons one each to the two chlorine atoms.
(i) What are the products of this reaction?
(ii) Write the balanced equation for the complete reaction.
(iii) Which element is being oxidized?
(iv) Which element is being reduced?
(v) Write the reduction part of the reaction.
Answer:
(i) Magnesium atom is converted to Mg²⁺ ion. Two chlorine atoms are converted to 2Cl^– ions. So the products are Mg²⁺ ion and 2Cl^– ions.
(ii)

(iii) Mg atom is being oxidised by donating 2 electrons.
(iv) Cl₂ molecule is being reduced by accepting 2 electrons.
(v) Cl₂ + 2e^– → 2Cl^–. It is a reduction reaction in which gain of e^– take place.

Question 5.
Take Cu(NO₃)₂ in a test tube and heat it over the flame.
(i) What is the colour of Cu(NO₃)₂?
Answer:
Blue

(ii) What do you observe?
Answer:
Evolution of reddish-brown gas.

(iii) Name the type of reaction that takes place.
Answer:
Decomposition.

(iv) Write the balanced equation.
Answer:
Cu(NO₃)₂ → 2CuO + 4NO₂ + O₂↑

Question 6.
Sodium hydroxide and hydrochloric acid react as shown in this equation.
\(\mathbf{N a O H}_{(\mathrm{aq})}+\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathbf{H}_{2} \mathbf{O}_{(\mathrm{l})}\)
(i) Which type of chemical reaction is this?
(ii) The reaction is exothermic. Explain what that means.
(iii) Differentiate exothermic reaction and endothermic reaction.
(iv) What happens to the temperature of the solution as the chemicals react?
Answer:
(i) It is a neutralisation reaction.
(ii) An exothermic reaction is a chemical reaction in which the evolution of heat energy takes place.
(iii)

(iv) When NaOH reacts with HCl to give NaCl and water, heat is evolved. So the solution’s temperature increases.

VIII. HOT Questions.

Question 1.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. The amount and concentration taken for both the acids are same. In which test tube does the reaction occur more vigorously and why?
Answer:
In the test tube, the reaction occurs more vigorously.
Comparing hydrochloric acid and acetic acid, HCl (hydrochloric acid) is a strong acid and more reactive whereas acetic acid is a weak organic acid and less reactive.
\(\mathrm{Mg}+2 \mathrm{HCl} \stackrel{\text { Fast reaction }}{\longrightarrow} \mathrm{MgCl}_{2}+\mathrm{H}_{2} \uparrow\).

Question 2.
Classify the following reactions based on the rate of the reactions as very fast or instantaneous slow and moderate reactions.
(a) AgNO_3(aq) + NaCl_(aq) → AgCl ↓ + NaNO_3(aq)
Answer:
Very fast reaction (or) Instantaneous reaction.

(b) 2 Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g)
Answer:
Very fast reaction.

(c) Rusting of iron.
Answer:
Very slow reaction.

(d) Inversion of cane sugar into Glucose and fructose.
Answer:
Moderately slow reaction.

(e) Fermentation of sugar into alcohol.
Answer:
Very slow reaction.

Question 3.
What is the chemical reaction taken place in the tarnishing of silver anklet?
Answer:
The lustrous white colour of the silver anklet slowly changes into a slightly black colour. It is due to the formation of silver sulphide (Ag₂S) as a result of the reaction between silver and hydrogen sulphide in the air.

Question 4.
Why toothpaste are basic in nature?
Answer:
The white enamel coating of our teeth is calcium phosphate, the hardest substance in our body. It does not dissolve in water. If the pH of mouth falls below 5.5, the enamel gets corroded. Toothpaste is generally basic and used for cleaning the teeth can neutralize the excess acid and prevent tooth decay.

Question 5.
Why the solution of slaked lime is used for whitewashing?
Answer:
A solution of slaked lime is used for whitewashing walls. Calcium hydroxide reacts slowly with the carbon dioxide in the air to form a thin layer of calcium carbonate on the walls. Calcium carbonate is formed after two to three days of whitewashing and gives a shiny finish to the walls. It is interesting to note that the chemical formula for marble is also CaCO₃.

Question 6.
Complete the following reactions.

Answer:

Question 7.
Let us consider the following two reactions.

Which reaction will not occur. Why?
Answer:
The first reaction involves the displacement of chlorine from NaCl, by fluorine. In the second reaction, chlorine displaces fluorine from NaF. Out of these two, the second reaction will not occur. Because fluorine is more active than chlorine and occupies the upper position in the periodic table. So, in displacement reactions, the activity of the elements and their relative position in the periodic table are the key factors to determine the feasibility of the reactions. More active elements readily displace less active elements from their aqueous solutions.

Question 8.
Which of the metals displaces hydrogen gas from hydrochloric acid? Silver or zinc. Give the chemical equation of the reaction and justify your answer.
Answer:
Zinc displaces hydrogen gas from hydrochloric, acid.
Zinc is more reactive than silver.
Zn + 2HCl → ZnCl₂ + H₂ ↑.

Question 9.
Foods kept at room temperature spoils faster than that kept in the refrigerator. Why?
Answer:
Food kept at room temperature spoils faster than that kept in the refrigerator. In the refrigerator, the temperature is lower than the room temperature and hence the reaction rate is less.

Question 10.
How will you enhance the rate of decomposition of potassium chlorate?
Answer:
On heating potassium chlorate, it decomposes into potassium chloride and oxygen gas, but at a slower rate. If manganese dioxide is added, it increases the reaction rate.

Here MnO₂ is the catalyst. Therefore the addition of MnO₂ enhances the rate of decomposition of potassium chlorate.

Question 11.
Powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips. Why?
Answer:
Powdering of the reactants increases the surface area and more energy is available on the collision of the reactant particles. Thus, the reaction rate is increased. Hence powdered calcium carbonate reacts more readily with hydrochloric acid than marble chips.

Samacheer Kalvi 10th Science Types of Chemical Reactions Additional Problems Solved

Question 1.
The [OH^–] ion concentration of a solution is 1.0 × 10^-8 M. What is the pH of the solution?
Solution:
The concentration of hydroxide ion = [OH^–] = 1.0 × 10^-8 M.

Question 2.
The hydrogen ion concentration of a solution is 1.0 × 10^-9 m. What is the pH of the solution? Find out whether the given solution is acidic, basic or neutral.
Solution:

Question 3.
The hydroxide ion concentration of a solution is 0.001 m. What is the pH of the solution?
Solution:

pH = 14 – pOH
pH = 14 – 3 = 11
pH = 11.

Question 4.
The hydroxide ion concentration of a solution is 1.0 × 10^-9 m. What is the pH of the solution?
Solution:

Question 5.
The hydrogen ion concentration of a solution is 1 × 10^-4 m. Calculate the pH and pOH of that solution.
Solution:

Question 6.
Calculate the pH of sodium hydroxide solution having the concentration of OH^– 0.01 mL^-1.
Solution:

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions

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Samacheer Kalvi 10th Science Solutions Textual Problems Solved

I. Problems based on Solubility.

10th Science Samacheer Kalvi Question 1.
1.5 g of solute is dissolved in 15 g of water to form a saturated solution at 298K. Find out the solubility of the solute at the temperature.
Solution:
Mass of the solute = 1.5 g
Mass of the solvent = 15 g
Solubility of the solute = \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\)
= \(\frac{1.5}{15} \times 100=10 \mathrm{g}\).

Samacheer Kalvi 10th Science Question 2.
Find the mass of potassium chloride would be needed to form a saturated solution in 60 g of water at 303 K? Given that solubility of the KCl is 37 / 100 g at this temperature.
Solution:
Mass of potassium chloride in 100 g of water in saturated solution = 37 g
Mass of potassium chloride in 60 g of water in saturated solution = \(\frac{37}{100} \times 60\) = 22.2 g.

You can Download Samacheer Kalvi 10th Science Guide PDF help you to revise the complete Syllabus and score more marks in your examinations.

Samacheer Kalvi Class 10 Science Solutions Question 3.
What is the mass of sodium chloride that would be needed to form a saturated solution in 50 g of water at 30°C. Solubility of sodium chloride is 36 g at 30°C?
Solution:
At 30°C, 36 g of sodium chloride is dissolved in 100 g of water.
Mass of sodium chloride that would be need for 100 g of water = 36 g
Mass of sodium chloride dissolved in 50 g of water = \(\frac{36 \times 50}{100}\) = 18 g.

10th Samacheer Kalvi Science Question 4.
The Solubility of sodium nitrate at 50°C and 30°C is 114 g and 96 g respectively. Find the amount of salt that will be thrown out when a saturated solution of sodium nitrate-containing 50 g of water is cooled from 50°C to 30°C?
Answer:
Amount of sodium nitrate dissolved in 100 g of water at 50°C is 114 g
Amount of sodium nitrate dissolving in 50 g of water at 50°C is = \(\frac{114 \times 50}{100}\) = 57 g
Similarly amount of sodium nitrate dissolving in 50 g of water at 30°C is = \(\frac{96 \times 50}{100}\) = 48 g
Amount of sodium nitrate thrown when 50 g of water is cooled from 50°C to 30°C is 57 – 48 = 9 g.

II. Problems based on Mass percentage.

Samacheer Kalvi 10th Science Solutions Question 1.
A solution was prepared by dissolving 25 g of sugar in 100 g of water. Calculate the mass percentage of solute.
Solution:
Mass of the solute = 25 g
Mass of the solvent = 100 g

Class 10 Science Samacheer Kalvi Question 2.
16 grams of NaOH is dissolved in 100 grams of water at 25°C to form a saturated solution. Find the mass percentage of solute and solvent.
Solution:
Mass of the solute (NaOH) = 16 g
Mass of the solvent H₂O = 100 g
(i) Mass percentage of the solute
= \(\frac{\text { Mass of the solute }}{\text { Mass of the solute }+\text { Mass of the solvent }} \times 100\)
= \(\frac{16 \times 100}{16+100}=\frac{1600}{116}\)
Mass percentage of the solute = 13.79 %.

(ii) Mass percentage of solvent = 100 – (Mass percentage of the solute) = 100 – 13.79 = 86.21 %.

Science Solution Class 10 Samacheer Kalvi Question 3.
Find the amount of urea which is to be dissolved in water to get 500 g of 10 % w/w aqueous solution?
Solution:
Mass percentage = \(\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(10=\frac{\text { Mass of the urea }}{500} \times 100\)
Mass of urea = 50 g.

III. Problem based on Volume – Volume percentage.

Samacheer Kalvi 10th Science Book Solutions Question 1.
The solution is made from 35 ml of methanol and 65 ml of water. Calculate the volume percentage.
Solution:
The volume of the ethanol = 35 ml
The volume of the water = 65 ml

10th Science Solution Samacheer Kalvi Question 2.
Calculate the volume of ethanol in 200 ml solution of 20 % v/v aqueous solution of ethanol.
Solution:
Volume of aqueous solution = 200 ml
Volume percentage = 20 %

Activity

10th Std Science Solutions Samacheer Kalvi Activity 1.
Look at the following pictures. Label them as a dilute and concentrated solution and justify your answer.

Solution:
In the above picture, the first teacup is more concentrated and the first beaker is more concentrated CuSO₄ solution. The reason is the colour intensity is more in a concentrated solution.

Samacheer Kalvi 10th Science Solutions Textual Evaluation Solved

I. Choose the best answer.

Class 10 Science Solutions Samacheer Kalvi Question 1.
A solution is a ____ mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non – homogeneous.
Answer:
(a) homogeneous

Science 10th Samacheer Kalvi Question 2.
The number of components in a binary solution is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Samacheer Kalvi 10th Science Answers Question 3.
Which of the following is the universal solvent?
(a) Acetone
(b) Benzene
(c) Water
(d) Alcohol.
Answer:
(c) Water

Science Class 10 Samacheer Kalvi Question 4.
A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called:
(a) Saturated solution
(b) Unsaturated solution
(c) Supersaturated solution
(d) Dilute solution
Answer:
(a) Saturated solution

Samacheer Class 10 Science Solutions Question 5.
Identify the non – aqueous solution ______.
(a) sodium chloride in water
(b) glucose in water
(c) copper sulphate in water
(d) sulphur in carbon-di-sulphide.
Answer:
(d) sulphur in carbon-di-sulphide.

Samacheer Kalvi Guru 10th Science Question 6.
When pressure is increased at a constant temperature the solubility of gases in liquid:
(a) No change
(b) increases
(c) decreases
(d) no reaction
Answer:
(b) increases

Science Samacheer Kalvi Question 7.
The solubility of NaCl in 100 ml water is 36 g. If 25 g of salt is dissolved in 100 ml of water how much more salt is required for saturation ______.
(a) 12 g
(b) 11 g
(c) 16 g
(d) 20 g.
Answer:
(b) 11 g

Samacheer Kalvi 10 Science Question 8.
A 25% alcohol solution means:
(a) 25 ml alcohol in 100 ml of water
(b) 25 ml alcohol in 25 ml of water
(c) 25 ml alcohol in 75 ml of water
(d) 75 ml alcohol in 25 ml of water
Answer:
(c) 25 ml alcohol in 75 ml of water

Samacheer Kalvi Guru 10th Science Book Pdf Download Question 9.
Deliquescence is due to ______.
(a) Strong affinity to water
(b) Less affinity to water
(c) Strong hatred of water
(d) Inertness to water.
Answer:
(a) Strong affinity to water

Samacheer Kalvi.Guru 10th Science Question 10.
Which of the following is hygroscopic in nature?
(a) ferric chloride
(b) copper sulphate penta hydrate
(c) silica gel
(d) none of the above
Answer:
(c) silica gel

II. Fill in the blanks.

10th Standard Science Samacheer Kalvi Question 1.
The component present in a lesser amount, in a solution, is called ______.
Answer:
Solute.

Samacheer Kalvi 10th Standard Science Question 2.
Example for liquid in solid type solution is ______.
Answer:
Sodium chloride dissolved in water.

Question 3.
Solubility is the amount of solute dissolved in _______ g of solvent.
Answer:
100.

Question 4.
Polar compounds are soluble in ______ solvents.
Answer:
Polar.

Question 5.
Volume percentage decreases with increases in temperature because of ______.
Answer:
Expansion of liquid.

III. Match the following.

Question 1.

Answer:
1 – (c), 2 – (a), 3 – (d), 4 – (b).

IV. True or False: (If false give the correct statement)

Question 1.
Solutions which contain three components are called binary solution.
Answer:
False.
Correct Statement: Solutions which contain two components are called binary solution.

Question 2.
In a solution, the component which is present in a lesser amount is called solvent.
Answer:
False.
Correct Statement:

• In a solution, the component which is present in a larger amount is called a solvent.
• In a solution, the component which is present in a lesser amount is called solute.

Question 3.
Sodium chloride dissolved in water forms a non-aqueous solution.
Answer:
False.
Correct Statement: Sodium chloride dissolved in water forms an aqueous solution.

Question 4.
The molecular formula of green vitriol is MgSO₄.7H₂O
Answer:
False.
Correct Statement: The molecular formula of green vitriol is FeSO₄ .7H₂O
Question 5.
When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature.
Answer:
True.

V. Short Answer Questions.

Question 1.
Define the term Solution.
Answer:
A solution is a homogeneous mixture of two or more substances.

Question 2.
What is mean by binary solution?
Answer:
Solutions which are made of one solute and one solvent, then it is called binary solution.

Question 3.
Give an example each

1. gas in liquid
2. solid in liquid
3. solid in solid
4. gas in gas.

Answer:

1. Carbon – di – oxide dissolved in water (Soda water).
2. Sodium chloride dissolved in water.
3. Copper dissolved in gold (Alloy).
4. A mixture of Helium – Oxygen gases.

Question 4.
What is aqueous and non-aqueous solution? Give an example.
Answer:
(i) Aqueous solution : The solution in which water acts as a solvent.
(ii) Non-aqueous solution : The solution in which any liquid other than water acts as a solvent.
Eg: Alcohol, benzene, CS₂ acetone.

Question 5.
Define Volume percentage.
Answer:
Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of the solution.
Volume percentage = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\).

Question 6.
The aquatic animals live more in cold region Why?
Answer:
Aquatic animals live more in cold regions because the solubility of O₂ in water is more at low temperature and therefore the amount of dissolved O₂ is more in the water of cold regions.

Question 7.
Define Hydrated salt.
Answer:
The number of water molecules found in the crystalline substance or salts is called water of crystallization. Such salts are called hydrated salts.

Question 8.
A hot saturated solution of copper sulphate forms crystals as it cools. Why?
Answer:
A hot saturated solution of CuSO₄ forms crystal as it cools. Because on cooling the water molecules move closer together and there is less space for the solution to hold on to as much of the dissolved solid and so it forms crystals.

Question 9.
Classify the following substances into deliquescent, hygroscopic. Conc. Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride, and Gypsum salt.
Answer:

1. Deliquescent substances: Calcium chloride
2. Hygroscopic substances: Conc Sulphuric acid, Copper sulphate penta hydrate, Silica gel and Gypsum salt.

VI. Long Answer Questions.

Question 1.
Write notes on
(i) saturated solution
(ii) unsaturated solution
Answer:
(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
Eg: 36 g of sodium chloride in 100 g of water at 25°C forms saturated solution.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.
Eg: 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

Question 2.
Write notes on various factors affecting solubility.
Answer:
Factors affecting solubility: There are three main factors which govern the solubility of the solute. They are
(i) Nature of the solute and solvent
(ii) Temperature
(iii) Pressure

(i) Nature of the solute and solvent: The nature of the solute and solvent plays an important role insolubility. Although water dissolves an enormous variety of substances, both ionic and covalent, it does not dissolve everything. The phrase that scientists often use when predicting solubility is “like dissolves like.” This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example, Common salt is a polar compound and dissolves readily in polar solvent like water.

Non – polar compounds are soluble in non-polar solvents. For example, Fat dissolved in ether. But non-polar compounds, do not dissolve in polar solvents; polar compounds do not dissolve in non-polar solvents.

(ii) Effect of Temperature
The solubility of Solids in Liquid: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. For example, a greater amount of sugar will dissolve in warm water than in cold water.
In the endothermic process, solubility increases with increase in temperature. In the exothermic process, solubility decreases with increase in temperature.

The solubility of Gases in liquid: Solubility of gases in liquids decreases with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles. Aquatic animals live more in cold regions because more amount of dissolved oxygen is present in the water of cold regions. This shows that the solubility of oxygen in water is more at low temperatures.

(iii) Effect of Pressure: Effect of pressure is observed only in the case of solubility of a gas in a liquid. When the pressure is increased, the solubility of a gas in liquid increases.
The common examples for solubility of gases in liquids are carbonated beverages, i.e. soft drinks, household cleaners containing an aqueous solution of ammonia, formalin aqueous solution of formaldehyde, etc.

Question 3.
(a) What happens when MgSO₄.7H₂O is heated? Write the appropriate equation.
(b) Define solubility.
Answer:
(a) MgSO₄.7H₂O has a water of crystallization is 7. When magnesium sulphate heptahydrate crystals are gently heated, it loses seven water molecules and becomes anhydrous magnesium sulphate.

(b) Solubility: It is defined as the number of grams of a solute that can be dissolved in 100 g of a solvent to form its saturated solution at a given temperature and pressure.

Question 4.
In what way, hygroscopic substances differ from deliquescent substances.
Answer:
Difference between hygroscopic and deliquescent substances is in the extent to which each material can absorb moisture. This is because both of these terms are very much related to each other and they refer to the property of observing and the retention of moisture from the air. However, they differ in the extent of absorption of moisture where hygroscopic materials absorb moisture but not to the extent the original substance dissolves in it, which is the case with deliquescence. Therefore deliquescence can be regarded as an extreme condition of hygroscopic activity.

Difference between hygroscopic substances and deliquescence

Question 5.
A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.
Answer:
Mass of sugar (solute) = 45 g
Mass of water (solvent) = 180 g.

Question 6.
3.5 litres of ethanol is present in 15 litres of an aqueous solution of ethanol. Calculate volume per cent of the ethanol solution.
Answer:
Volume of ethanol (solute) = 3.5 litre
Volume of aqueous solution = 15 litre
Volume of the solution = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\)
= \(\frac{3.5}{15} \times 100\)
Volume percentage = 23.33 %.

VII. HOT Questions.

Question 1.
Vinu dissolves 50 g of sugar in 250 ml of hot water, Sarath dissolves 50 g of same sugar in 250 ml of cold water. Who will get faster dissolution of sugar? and Why?
Answer:
Vinu will get faster dissolution of sugar. Because Vinu dissolves 50 g of sugar in 250 ml of hot water, whereas Sarath dissolves 50 g of sugar in 250 ml of cold water. Solubility of a solid in liquid increases with increase in temperature.

Question 2.
‘A’ is a blue coloured crystalline salt. On heating, it loses a blue colour and to give ‘B’. When water is added, ‘B’ gives back to ‘A’. Identify A and B, write the equation.
Answer:

• Blue coloured crystalline salt is copper sulphate pentahydrate (A)
• On heating Copper sulphate pentahydrate it loses blue colour and to give anhydrous copper sulphate (B).
• When water is added to the anhydrous copper sulphate (B) gives back to copper sulphate pentahydrate (A).
• 

Question 3.
Will the cool drinks give more fizz at top of the hills or at the foot? Explain.
Answer:
Carbonated cool drinks give more fizz at the foot hill because solubility of gases in liquid decrease with increase in temperature. At higher altitudes, the temperature is low and the dissolved CO₂ will not escape as fizz, whereas at the foot hill it does.

Samacheer Kalvi 10th Science Solutions Additional Questions Solved

I. Choose the best answer.

Question 1.
Find out the homogeneous mixture.
(a) Salt + Water
(b) Sand + Water
(c) Clay + Water
(d) Gold + Water.
Answer:
(a) Salt + Water

Question 2.
The aqueous solution is:
(a) S in CS₂
(b) I₂ in CCl₄
(c) Salt in H₂O
(d) None of the above
Answer:
(c) Salt in H₂O

Question 3.
In a solution, the component which is present in a larger amount is called ______.
(a) solvent
(b) dissolution
(c) solute
(d) mole.
Answer:
(а) solvent

Question 4.
The solubility of ammonia gas at 25°C is:
(a) 36 g
(b) 48 g
(c) 80 g
(d) 184 g
Answer:
(b) 48 g

Question 5.
Example for liquid in liquid binary solution is ______.
(a) Copper dissolved in gold
(b) Water vapour in the air
(c) Ethyl alcohol dissolved in water
(d) NaCl dissolved in water.
Answer:
(c) Ethyl alcohol dissolved in water

Question 6.
Which one of the following is an aqueous solvent?
(a) Benzene
(b) Acetone
(c) Alcohol
(d) Water.
Answer:
(d) Water.

Question 7.
Mass percentage of a solution is independent of:
(a) temperature
(b) amount of solute
(c) amount of solvent
(d) chemical nature of the solute
Answer:
(a) temperature

Question 8.
Solubility is equal to ______.

Answer:
(c) \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\)

Question 9.
In which case solubility increases with increase in temperature?
(a) Endothermic process
(b) Exothermic process
(c) Both (a) and (b)
(b) None of these.
Answer:
(a) Endothermic process

Question 10.
The factor/factors which affect the solubility of a solute:
(a) Nature of the solute and solvent
(b) Temperature
(c) Pressure
(d) All of these
Answer:
(d) All of these

Question 11.
The effect of pressure on the solubility of a gas in a liquid is given by ______.
(a) Hess’s law
(b) Ohm’s law
(c) Henry’s law
(d) Gases law.
Answer:
(c) Henry’s law

Question 12.
5% sugar solution means ______.
(a) 5 g of sugar in 95 g of water
(b) 50 g of sugar in 50 g of water
(c) 20 g of sugar in 80 g of water
(d) 95 g of sugar in 5 g of water.
Answer:
(a) 5 g of sugar in 95 g of water

Question 13.
The water of crystallisation present in white vitriol is:
(a) 2
(b) 7
(c) 5
(d) 24
Answer:
(b) 7

Question 14.
Mass percentage is independent of ______.
(a) density
(b) volume
(c) pressure
(d) temperature.
Answer:
(d) temperature.

Question 15.
Volume percentage is expressed as ______.
(a) v/v
(b) w/w
(c) v/w
(d) w/v.
Answer:
(a) v/v

Question 16.
When the temperature increases, volume percentage ______.
(a) Increases
(b) Decreases
(c) no change
(d) increases then decrease.
Answer:
(b) Decreases

Question 17.
The mass of NaCl needed to form saturated solution in 50 g of water at 30°C. When the solubility of NaCl is 36 g at 30°C is:
(a) 9 g
(b) 18 g
(c) 57 g
(d) 19 g
Answer:
(b) 18 g

Question 18.
Zinc sulphate heptahydrate is also called as ______.
(a) Blue Vitriol
(b) Epsom salt
(c) Green Vitriol
(d) White Vitriol.
Answer:
(d) white Vitriol.
Question 19.
Water of crystallization of blue vitriol is ______.
(a) 1
(b) 5
(c) 4
(d) 2.
Answer:
(b) 5

Question 20.
On heating copper sulphate pentahydrate crystals ______.
(a) blue colour changes into green
(b) green colour changes into blue
(c) blue colour changes into colourless
(d) no colour changes.
Answer:
(c) blue colour changes into colourless

Question 21.
Water of crystallization of Epsom salt is ______.
(a) 1
(b) 5
(c) 4
(d) 1.
Answer:
(d) 7.

Question 22.
Hygroscopic substances are used as ______.
(a) drying agents
(b) hydrating agents
(c) freezing agents
(d) reducing agents.
Answer:
(a) drying agents

Question 23.
If two liquids are mutually soluble, they are called _____ liquids.
(a) miscible
(b) immiscible
(c) insoluble
(d) viscous.
Answer:
(a) miscible

Question 24.
Aquatic species are more comfortable in cold water because of ______.
(a) as the temperature decreases, the solubility of dissolved oxygen increases
(b) as the temperature increases, the solubility of dissolved oxygen increases
(c) as the temperature increases, the solubility of dissolved oxygen decreases
(d) None of these.
Answer:
(a) as the temperature decreases, the solubility of dissolved oxygen increases

Question 25.
The process of food assimilation by man is in the form of ______.
(a) solution
(b) solid
(c) solute
(d) solvent.
Answer:
(a) solution

Question 26.
Common salt in water is an example of ______.
(a) Colloidal solution
(b) Binary solution
(c) Suspension
(d) all the above.
Answer:
(b) Binary solution

Question 27.
36 g of NaCl in 100 g of water is a ____ solution.
(a) Non – aqueous
(b) Unsaturated
(c) Supersaturated
(d) Saturated.
Answer:
(d) Saturated.
Question 28.
Nitrogen in soil is an example of _____ solution in nature.
(a) saturated
(b) unsaturated
(c) supersaturated
(d) aqueous.
Answer:
(a) saturated

Question 29.
The solubility of CuSO₄ in H₂O is ______ at 20°C.
(a) 36 g
(b) 20.7 g
(c) 10 g
(d) 95 g.
Answer:
(b) 20.7 g

Question 30.
The solubility of NaNO₃ in 100 g water at 25°C is ______.
(a) 36 g
(b) 95 g
(c) 184 g
(d) 92 g.
Answer:
(d) 92 g.

Question 31.
Which of the following affect solubility?
(a) Temperature
(b) Nature of solute and solvent
(c) Pressure
(d) all the above.
Answer:
(d) all the above.

Question 32.
In _____ process, solubility increases with increase in temperature.
(a) exothermic
(b) endothermic
(c) isothermic
(d) adiabatic.
Answer:
(b) endothermic

Question 33.
In ______ process, solubility decreases with increase in temperature.
(a) exothermic
(b) adiabatic
(c) endothermic
(d) isothermic.
Answer:
(a) exothermic

Question 34.
Increase in ______ increases the solubility of gases.
(a) temperature
(b) pressure
(c) no. of moles
(d) concentration.
Answer:
(b) pressure

Question 35.
The solubility of NaBr in H₂O is ______.
(a) 36 g
(b) 95 g
(c) 184 g
(d) 92 g.
Answer:
(b) 95 g

Question 36.
The gas – filled in soft drinks in ______.
(a) O₂
(b) N₂
(c) CO₂
(d) H₂
Answer:
(c) CO₂

Question 37.
The solubility of KNO₃ in water is ______ process.
(a) Exothermic
(b) endothermic
(c) Isothermic
(d) Adiabatic.
Answer:
(b) endothermic

Question 38.
The solubility of CaO in water is ______ process.
(a) Exothermic
(b) endothermic
(c) Adiabatic
(d) Isothermic.
Answer:
(a) Exothermic

Question 39.
The solubility of NaI in water is ______.
(a) 184 g
(b) 92 g
(c) 95 g
(d) 36 g.
Answer:
(a) 184 g

II. Fill in the blanks.

Question 1.
A solution is a _____ mixture of two or more substances.
Answer:
Homogeneous.

Question 2.
Common salt dissolved in water is an example for _____ solution.
Answer:
Binary.

Question 3.
In a solution, the component present in a lesser amount by weight is called ____ and in a large amount by weight is called ______.
Answer:
Solute, solvent.

Question 4.
The solution in which water acts as a solvent is called ______ and the solution in which Benzene acts as a solvent is called ______.
Answer:

1. The aqueous solution,
2. Non – aqueous solution.

Question 5.
20 g of NaCl in 100 g of water is _____ solution.
Answer:
Unsaturated.

Question 6.
36 g of NaCl in 100 g of water at room temperature forms a ____ solution.
Answer:
Saturated.

Question 7.
Nitrogen in soil is an example for _____ in nature.
Answer:
Saturated solution.

Question 8.
A solution which has more of solute than the saturated solution at a given temperature is called ______ solution.
Answer:
Supersaturated.

Question 9.
In _____ process, solubility increases with ______ in temperature.
Answer:
Endothermic, increase.

Question 10.
In an ______ process, solubility decreases with ______ in temperature.
Answer:
Exothermic, increase.

Question 11.
The solubility of KNO₃ increases with the _____ in temperature.
Answer:
Increase.

Question 12.
The solubility of CaO decreases with the ______ in temperature.
Answer:
Increase.

Question 13.
The solubility of oxygen is _____ in cold water.
Answer:
More.

Question 14.
A polar compound is insoluble in _____ solvent.
Answer:
Non – polar.

Question 15.
An increase in ______ increases the solubility of a gas in a liquid.
Answer:
Pressure.

Question 16.
______ gas is filled in soft drinks using the effect of pressure.
Answer:
CO₂

Question 17.
Hygroscopic substances are used as ______.
Answer:
Drying agents.

Question 18.
Anhydrous calcium chloride is an example of ______.
Answer:
hygroscopic substance.

Question 19.
Hygroscopic substances absorb moisture without changing their ______.
Answer:
Physical state.

Question 20.
Deliquescent substances lose their ______.
Answer:
Crystalline shape.

Question 21.
Ferric chloride is an example of ______.
Answer:
Deliquescent substances.

Question 22.
On heating, hydrated crystalline salts lose their ______.
Answer:
Water of crystallization.

Question 23.
The number of water molecules in blue vitriol is ______.
Answer:
Five.

Question 24.
The number of water molecules in magnesium sulphate heptahydrate is ______.
Answer:
Seven.

Question 25.
Mass percentage is _______ to temperature.
Answer:
Independent.

Question 26.
To qualify the solute in a solution, we can use the term ______.
Answer:
Concentration.

Question 27.
Based on the amount of solute, the solution is classified into ________ types.
Answer:
Three.

Question 28.
Zinc dissolved in copper is an example of ______ solution.
Answer:
Solid in solid.

Question 29.
______ is an example of gas in the liquid.
Answer:
Carbon dioxide dissolved in water.

Question 30.
______ and _______ are in the form of solution to decide the physiological activity of human beings.
Answer:
Blood, Lymph.

III. Match the following.

Question 1.

Answer:
i – b, ii – d, iii – a, iv – c.

Question 2.

Answer:
i – d, ii – c, iii – b, iv – a.

Question 3.

Answer:
i – c, ii – a, iii – d, iv – b.

Question 4.

Answer:
i – b, ii – c, iii – d, iv – a.

Question 5.

Answer:
i – c, ii – d, iii – a, iv – b.

Question 6.

Answer:
i – b, ii – d, iii – a, iv – c.

IV. State whether true or false. If false, give the correct statement.

Question 1.
A solution is a heterogeneous mixture of two or more substances.
Answer:
False.
Correct statement: A solution is a homogeneous mixture of two or more substances.

Question 2.
A solution contains four components, it is called as a binary solution.
Answer:
False.
Correct statement: A solution contains two components, it is called as a binary solution.

Question 3.
The solution in which water acts as a solvent, it is called a non-aqueous solution.
Answer:
False.
Correct statement: The solution in which water acts as a solvent, it is called an aqueous solution.

Question 4.
The solution of sulphur in CS₂ is a suitable example of a non-aqueous solution.
Answer:
True.

Question 5.
Nitrogen in soil is an example of a supersaturated solution.
Answer:
False.
Correct statement: Nitrogen in the soil is an example of a saturated solution.

Question 6.
The solubility of CuSO₄ in H₂O is 36 g at 25°C.
Answer:
False.
Correct statement: Solubility of CuSO₄ in H₂O is 20.7 g at 20°C.

Question 7.
100 ml of water can dissolve 36 g of NaCl at 25°C to attain saturation.
Answer:
True.

Question 8.
In an endothermic process, solubility decreases with increase in temperature.
Answer:
False.
Correct statement: In an endothermic process, solubility increases with increase in temperature.

Question 9.
In an exothermic process, solubility decreases with increase in temperature.
Answer:
True.

Question 10.
The solubility of CaO increases with the increase in temperature.
Answer:
False.
Correct statement: The solubility of CaO decreases with the increase in temperature.

Question 11.
The solubility of oxygen is more in cold water.
Answer:
True.

Question 12.
SO₂ gas is filled in soft drinks using the effect of pressure.
Answer:
False.
Correct statement: CO₂ gas is filled in soft drinks using the effect of pressure.

Question 13.
Ether act as a universal solvent.
False.
Correct statement: Water acts universal solvent.

Question 14.
Covalent compounds are soluble in water.
Answer:
False.
Correct statement: Covalent compounds are insoluble in water.

Question 15.
36 g of sodium chloride in 100 g of water at 25°C forms a saturated solution.
Answer:
True.

Question 16.
40 g of sodium chloride in 100 g of water at 25°C forms an unsaturated solution.
Answer:
False.
Correct statement: 40 g of sodium chloride in 100 g of water at 25°C forms a supersaturated solution.

Question 17.
Aquatic animals live more in hot regions.
Answer:
False.
Correct statement: Aquatic animals live more in cold regions.

Question 18.
Mass percentage is dependent of temperature.
Answer:
False.
Correct statement: Mass percentage is independent of temperature.

Question 19.
Hygroscopic substances do not change their physical state on exposure to air.
Answer:
True.

Question 20.
Deliquescent substances do not change their physical state on exposure to air.
Answer:
False.
Correct statement: Deliquescent substances change its physical state on exposure to air.

V. Assertion and Reason

Question 1.
Assertion (A): Salt solution-common salt dissolved in water is an example of a binary solution.
Reason (R): A solution with two components is called a binary solution.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): Sugar in water is a true solution.
Reason (R): True solution is a homogeneous mixture that contains small solute particles that are dissolved throughout the solvent.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is wrong but (R) is correct
(d) (A) is correct but (R) is wrong.
Answer:
(b) Both (A) and (R) are correct

Question 3.
Assertion (A): Solution of sulphur in CS₂ is an example of a non-aqueous solution.
Reason (R): The solution in which any liquid other than water acts as a solvent is called the non – aqueous solution.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 4.
Assertion (A): Solubility of KNO₃ increases with increase in temperature.
Reason (R): Solubility of KNO₃ is an endothermic process.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 5.
Assertion (A): Aquatic species are more comfortable in cold water.
Reason (R): Solubility of oxygen is more in cold water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 6.
Assertion (A): CO₂ gas is filled in soft drinks using the effect of pressure.
Reason (R): A decrease in pressure increases the solubility of a gas in a liquid.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(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

Question 7.
Assertion (A): Aqueous copper sulphate is a binary solution.
Reason (R): Copper sulphate solution contains two components i.e., one solute-copper sulphate and one solvent-water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 8.
Assertion (A): Water act as a universal solvent.
Reason (R): Most of the substances are soluble in water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 9.
Assertion (A): When Blue vitriols are gently heated it turns colourless.
Reason (R): It loses five water molecules and becomes anhydrous compounds.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): Anhydrous calcium chloride is an example of a hygroscopic substance.
Reason (R): Anhydrous calcium chloride changes its physical state on exposure to air.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is wrong but (R) is correct
(d) (A) is correct but (R) is wrong
Answer:
(d) (A) is correct but (R) is wrong.

VI. Short Answer Questions.

Question 1.
State Henry’s Law.
Answer:
The solubility of a gas in liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

Question 2.
Define solute and solvent.
Answer:
In a solution, the component present in a lesser amount by weight is called solute and the component present in a large amount by weight is called solvent.

Question 3.
What is a
(i) saturated and
(ii) supersaturated solution?
Answer:
(i) Saturated : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature.
(ii) Super saturated solution : A solution which contains more solute than the saturated solution at a given temperature.

Question 4.
Why water is known as a universal solvent?
Answer:
Most of the solutes (substances) are soluble in water and so it is called universal solvent.

Question 5.
Common salt dissolves in water easily. Give reason.
Answer:
Common salt is an electrolyte. It is easily dissociated into its ions like Na⁺ and Cl^– in polar solvent water. So NaCl (common salt) an inorganic compound readily dissolves in water.

Question 6.
What is meant by the ternary solution?
Answer:
Solutions which contain three components are called ternary solutions.
E.g. If salt and sugar are added to water, both dissolves in water forming a solution. Here two solutes are dissolved in one solvent. Therefore it is a ternary solution.

Question 7.
Name the type of solution formed in the following cases;

1. 20 g of NaCl in 100 g of water at 25°C
2. 36 g of NaCl in 100 g of water at 25°C
3. Nitrogen in soil
4. Sulphur in CS₂.

Answer:

1. Unsaturated
2. Saturated
3. Saturated
4. Non-aqueous solution

Question 8.
What are concentrated and dilute solutions?
Answer:
Two solutions having the same solute and solvent, the one which contains a higher amount of solute per the given amount of solvent is said to be concentrated solution and the other is said to be dilute solution.

Question 9.
Define Henry’s law.
Answer:
To quantify the solute in a solution, we can use the term concentration. The concentration of a solution may be defined as the amount of solute present in a given amount of solution or solvent.
(or)
The solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

Question 10.
Define Mass percentage.
Answer:
Mass percentage of a solution is defined as the percentage by mass of the solute present in the solution.
Mass percentage,
\(=\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\).

Question 11.
What is water crystallization?
Answer:
The number of water molecules found in the crystalline substance is called water of crystallization.

Question 12.
What are hygroscopic substances?
Answer:
Certain substances, when exposed to the atmospheric air at ordinary temperature, absorb moisture without changing their physical state. Such substances are called hygroscopic substances and this property is called hygroscopy.

Question 13.
List out the examples for hygroscopic substances.
Answer:

• Conc.Sulphuric acid (H₂SO₄).
• Phosphorus Pentoxide (P₂O₅).
• Quick lime (CaO).
• Silica gel (SiO₂).
• Anhydrous calcium chloride (CaCl₂).

Question 14.
What are the Deliquescent substances?
Answer:
Certain substances which are so hygroscopic, when exposed to the atmospheric air at ordinary temperatures, absorb enough water and get completely dissolved. Such substances are called deliquescent substances and this property is called deliquescence.

Question 15.
When is deliquescence is maximum?
Answer:

• The temperature is low
• The atmosphere is humid.

Question 16.
Give examples for deliquescent substances.
Answer:

• Calcium chloride (CaCl₂)
• Caustic soda (NaOH)
• Caustic potash (KOH)
• Ferric chloride (FeCl₃).

VII. Long Answer Questions.

Question 1.
From the table given below, furnish your points of inference.

Answer:
Inferences:

• At 25°C, 36g NaCl is dissolved in 100 g water to give a saturated solution.
• At 25°C, 95 g NaBr is dissolved in 100 g water to get a saturated solution.
• At 25°C, 184g Nal is dissolved in 100 g water to get a saturated solution.
• The solubility of a solute at a given solvent at a particular temperature is defined as the number of grams of solute necessary to saturate 100 g of the solvent at that temperature.
• In the above tabular column, we infer that the solubility of NaI is the highest and the solubility of NaCl is the lowest.
• All the NaCl, NaBr, Nal solution in water are called aqueous solutions.
• Depending upon the amount of the solute, the solutions are classified as a saturated and unsaturated solution.
• The above solutions are saturated solutions.

Question 2.
Distinguish between the saturated and unsaturated solution at a temperature of 25°C using the data given below (Note: Solubility of NaCl is 36 g).

1. 16 g NaCl in 100 g water
2. 36 g NaCl in 100 g water.

Answer:
The solubility of NaCl is 36g.

1. 16 g NaCl in 100 g water is an unsaturated solution.
2. 36 g NaCl in 100 g water is a saturated solution.

Question 3.
Write a note on the type of solution based on the amount of solute present in a solution.
Answer:
Based on the amount of solute, in the given amount of solvent, solutions are classified into the following types.
(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution. Eg: 36 g of sodium chloride in 100 g of water at 25°C forms saturated solution.Further addition of sodium chloride, leave it undissolved.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature. Eg: 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

(iii) Super saturated solution : Supersaturated solution is one that contains more solute than the saturated solution at a given temperature. Eg: 40 g of sodium chloride in 100 g of water at 25°C forms super saturated solution. This state can be achieved by altering any other conditions liken temperature, pressure. Super saturated solutions are unstable, and the solute is reappearing as crystals when the solution is disturbed.

Question 4.
Find the concentration of a solution in terms of weight per cent if 20 g of common salt is dissolved in 50 g of water.
Answer:

VIII. HOT Questions.

Question 1.
Observe the diagram.

1. Which is a concentrated solution and why?
2. Which is dilute solution and why?
   

Answer:

1. Flask (b) is a concentrated solution. Because (b) flask contains a large number of solute particles than (a) flask.
2. Flask (a) is a dilute solution. Because (a) flask contains a lesser number of solute particles than (b) flask.

Question 2.
Where we use the phrase – “Like dissolves Like” and explain the meaning of the phrase?
Answer:
The phrase “like dissolves like” is often used for predicting solubility. This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example, Common salt is a polar compound and dissolves readily in polar solvent like water.

Question 3.
Why bubbling occurs when water is boiled?
Answer:
The solubility of a gas in liquid decrease with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles.

Question 4.
What happens when blue vitriol is heated? Explain.
Answer:
When blue vitriol (Copper sulphate pentahydrate) crystals are gently heated, it loses it five water molecules and becomes colourless anhydrous copper sulphate.

Samacheer Kalvi 10th Science Solutions Additional Problems Solved

Question 1.
Take 10 g of common salt and dissolve it in 40 g of water. Find the concentration of a solution in terms of weight per cent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{10}{10+40} \times 100=20 \%\).

Question 2.
2 g of potassium sulphate was dissolved in 12.5 ml of water. On cooling, the first crystals appeared at 60°C. What is the solubility of potassium sulphate in water at 60°C? Solution. 12.5 ml of water weighs 12.5 g.
Solution:
In 12.5 g of water, the amount of potassium sulphate dissolved is 2 g.
In 1 g of water, the amount of potassium sulphate dissolved is 2 / 12.5 g.
Hence in 100 g of water, the amount of potassium sulphate dissolved is (2 × 100) / 12.5 = 16 g.
The solubility of potassium sulphate in the water at 60°C is 16 g.