Prasanna

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science Civics Solutions Chapter 2 Election, Political Parties and Pressure Groups

Election, Political Parties and Pressure Groups Textual Exercise

I. Choose the correct answer.

Election Political Parties And Pressure Groups Question 1.
India has adapted the electoral system followed in the
(a) USA
(b) United Kingdom
(c) Canada
(d) Russia
Answer:
(b) United Kingdom

Election Political Parties And Pressure Groups Questions And Answers Question 2.
The Election Commission of India is a/an
(a) Independent body
(b) Statutory body
(c) Private body
(d) Public corporation
Answer:
(a) Independent body

Class 9 Civics Chapter 2 Question 3.
Which Article of the Constitution provides for an Election Commission?
(a) Article 280
(b) Article 315
(c) Article 324
(d) Article 325
Answer:
(c) Article 324

Political Parties And Pressure Groups Question 4.
Which part of the constitution of India says about the election commission?
(a) Part III
(b) Part XV
(c) Part XX
(d) Part XXII
Answer:
(b) Part XV

Chapter 2 Civics Class 9 Question 5.
Who accords recognition to various political parties as national or regional parties?
(a) The President
(b) The Election Commission
(c) The Parliament
(d) The President in consultation with the Election Commission
Answer:
(b) The Election Commission

Question 6.
Assertion (A): Indian Constitution provides for an independent Election Commission
Reason (R): To ensure free and fair elections in the country.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 7.
NOT A was introduced in the year ………..
(a) 2012
(b) 2013
(c) 2014
(d) 2015
Answer:
(c) 2014

Question 8.
The term pressure groups originated in …….
(a) USA
(b) UK
(c) USSR
(d) India
Answer:
(a) USA

Question 9.
Assertion (A): A large number of pressure groups exist in India.
Reason (R): Pressure Groups are not developed in India to the same extent as in the USA.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

II. Fill in the blanks.

1. The Election Commission of India is a body of ……… members.
2. National Voters day has been celebrated on ………
3. In India ……… party system is followed.
4. In 2017, there were ……. recognised national parties.
5. Narmada Bachao Andolan is a …………
Answers:
1. 3
2. 25^th January
3. Multi
4. Seven
5. Pressure group
III. Match the following.

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

IV. Give short answers.

Question 1.
Explain the electoral system in India.
Answer:

1. The electoral system in India has been adapted from the system followed in the United Kingdom.
2. India is a socialist, secular, democratic, republic and the largest democracy in the world.
3. The modem Indian nation state came into existence on 15th August 1947.

Question 2.
Give the meaning of a political party.
Answer:
A political party is an organisation formed by a group of people with a certain ideology and agenda to contest elections and hold power in the government.

A political party has three components: a leader, active members and the followers.

Question 3.
Distinguish between two-party system and the multi-party system.
Answer:

Question 4.
What is a pressure group?
Answer:

1. A pressure group is a group of people who are organised actively for promoting and defending their common interest. It is so called as it attempts to bring a change in the public policy by exerting pressure on the government.
2. The pressure groups are also called ‘interest groups’ or vested groups.
3. They are different from the political parties in that they neither contest elections nor try to capture political power.

V. Answer in detail.

Question 1.
Discuss merits and demerits of direct elections?
Answer:
Merits:

1. As the voters elect their representatives directly, direct elections are considered to be a more democratic method of election.
2. It educates people regarding the government activities and helps in choosing the appropriate candidates. Also, it encourages people to play an active role in politics.
3. It empowers people and makes the rulers accountable for their actions.

Demerits:

1. Direct elections are very expensive.
2. Illiterate voters sometimes get misguided by false propaganda and sometimes campaigning based on caste, religious and various other sectarian consideration spose serious challenges.
3. Since conducting direct elections is a massive exercise, ensuring free and fair elections at every polling station is a major challenge to the Election Commission.
4. There are instances of some political candidates influencing the voters through payments in the form of cash, goods or services.
5. Election campaigns sometimes results in violence, tension, law and order problems and
   affects the day-to-day life of people.

Question 2.
What are the functions of political parties?
Answer:

1. Parties contest elections. In most democracies, elections are fought mainly among the candidates put up by political parties.
2. Parties put forward their policies and programmes before the electorate to consider and choose.
3. Parties play a decisive role in making laws for a country. Formally, laws are debated and passed in the legislature.
4. Parties form and run the governments.
5. Those parties that lose in the elections play the role of the Opposition to the party or a group of coalition parties in power, by voicing different views and criticising the government for its failures or wrong policies.
6. Parties shape public opinion. They raise and highlight issues of importance.
7. Parties function as the useful link between people and the government machinery.

Question 3.
What are the functions of Pressure groups in India?
Answer:
Pressure groups are the interest groups that work to secure certain interest by influencing the public policy. They are non-aligned with any political party and work as an indirect yet powerful group to influence the policy decisions. Pressure groups carry out a range of functions including representation, political participation, education, policy formulation and policy implementation.

Political Participation: Pressure groups can be called the informal face of politics. They exert influence precisely by mobilising popular support through activities such as petitions, marches, demonstrations and other forms of political protest. Such forms of political participation have been particularly attractive to young people.

Education: Many pressure groups devote significant resources by carrying out research, maintaining websites, commenting on government policy and using high-profile academics, scientists and even celebrities to get their views across, with an emphasis to cultivate expert authority.

Policy Formulation: Though the pressure groups themselves are not policy-makers, yet it does not prevent many of them from participating in the policy-making process. Many pressure groups are vital sources of information and render advice to the government and therefore they are regularly consulted in the process of policy formulation.

VI. Project and Activity

Question 1.
Compare the policies, programmes and achievements of a national party and a state party.
Answer:

1. Refer the National policies, programmes and achievements from the Internet and library books.
2. The students are instructed to compare the policies programmes and achievements.
3. This is a group Activity.

VII. HOTS

Question 1.
“Elections are considered essential for any representative democracy”. Why?
Answer:
“A democracy requires a mechanism by which people can choose their representatives at regular intervals and change them if they wish to do so. Therefore, elections are considered essential for any representative democracy. In an election the voters make many choices.

1. This helps the public in choosing the development course which they want and want to adopt.
2. This directly provide the public the opportunity to select their representatives and these representatives decision can be legitimacy.
3. This also ensures the transparency as well as the accountability as the public representatives are chosen directly.

Question 2.
What is the principle of universal adult franchise? What is its importance?
Answer:
Principle: Universal Adult Franchise means that the right to vote should be given to all adult citizens without the discrimination of caste, class, colour, religion (or) gender. It is based on equality, which is a basic principle of democracy.

Importance: Under this system a government is elected that is accountable to the people it governs. Because every vote counts, issues in a society receive their appropriate weight in terms of importance and urgency.

Question 3.
Discuss merits and demerits of democracy.
Answer:

Question 4.
Discuss the multi-party system.
Answer:

1. A multi-party system is a system where multiple political parties that have ideas participate in the national elections.
2. A lot of countries that use this system have a coalition government, meaning many parties are in control, and they all work together to make laws.
3. Countries with a multi-party political system tend to have greater voter participation.
4. No democracy can survive without multi-party system.

VIII. Life Skill

Question 1.
Conduct a mock poll in your classroom.
Answer:

1. Help the students to understand the process of electing officials and the power of vote by holding a mock election.
2. These are great activities to enjoy during the presidential election.
3. Students explain the steps taken from party formation to National election.
4. The students will act out the campaigning and voting process by stimulating a real election in their own class room.

Steps set for a mock-poll in the class room:

1. Setting up the political parties.
2. Preparing a manifesto.
3. Running a campaign.
4. Holding the class room election.

Election, Political Parties and Pressure Groups Additional Questions

I. Choose the correct answer.

Question 1.
India is the democracy in the world.
(a) largest
(b) smallest
(c) strongest
(d) None of the above
Answer:
(a) largest

Question 2.
Kudavolai was the system of voting followed during the ……. period in Tamil Nadu.
(a) Chera
(b) Chola
(c) Pandya
(d) Pallava
Answer:
(b) Chola

Question 3.
Which country has single party system?
(a) USA
(b) UK
(c) Cuba
(d) France
Answer:
(c) Cuba

Question 4.
Assertion (A): Parties shape public opinion.
Reason (R): They raise and highlight issues of importance.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A).
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 5.
India is the ………… th country in the world to introduce NOTA.
(a) 10
(b) 12
(c) 14
(d) 16
Answer:
(c) 14

Question 6.
The ……. is elected by members of the Lok Sabha.
(a) Prime Minister
(b) President
(c) Governor
(d) Cabinet Minister
Answer:
(a) Prime Minister

II. Fill in the blanks.
1. ……. in elections are the best way to make your ‘voice’ heard.
2. Indirect elections are less ……..
3. ……… parties are an essential part of Democracy.
4. …… treats all the parties equally.
5. The pressure groups are also called ……. groups.
6. A political party has three components: a ………. and the ……….
Answers:
1. Voting
2. expensive
3. Political
4. Election Commission
5. Interest
6. a leader, acting members, followers

III. Match the following.

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

IV. Give short answers.

Question 1.
What do you know about Voters Verified Paper Audit Trail?
Answer:

1. Voters Verified Paper Audit Trail (WPAT) is the way forward to enhance credibility and transparency of the election process.
2. This system was first introduced in 2014 General election.

Question 2.
Mention the merits and demerits of Indirect elections.
Answer:
Merits:

1. Indirect elections are less expensive.
2. It is more suited to elections in large countries.

Demerits:

1. If the number of voters is very small, there exists the possibility of corruption, bribery, horse trading and other unfair activities.
2. It is less democratic because people do not have a direct opportunity to elect, but they instead do it through their representatives. So, this may not reflect the true will of the people.

Question 3.
Write a short note on “State Parties”.
Answer:
Other than the seven national parties, most of the major parties of the country are classified by the Election Commission as ‘state parties’. These are commonly referred to as regional parties. A party is recognised as a state party by the Election Commission of India based on certain percentage of votes secured or a certain number of seats won in the Assembly or Lok Sabha elections.

Question 4.
Classify pressure groups in India.
Answer:
The pressure groups in India can be broadly classified into the following categories:

1. Business groups
2. Trade unions
3. Agrarian groups
4. Professional associations
5. Student organisations
6. Religious organisations
7. Tribal organisations
8. Linguistic groups
9. Ideology-based groups
10. Environmental protection groups.

V. Answer in detail.

Question 1.
Give an account of Mobilisation and Democratic Participation.
Answer:
Mobilising people towards socially productive activities that lead to the overall betterment of people’s lives is essential. Sometimes earthquakes, tsunamis, floods and other such natural disasters on a massive scale occur and people’s immediate mobilisation for evacuation and emergency relief becomes most essential.

Democratic Participation: Democracy can-succeed only when smaller local groups and, in fact, every citizen can take action that supports the tax and revenue collection systems, observance of national norms in environmental protection, cleanliness, health and hygiene, sanitary drives and immunisation programmes like pulse polio.

However, we must keep in mind that there is no better form of government than Democratic government. To create a better society and nation, the people of India along with the union and state governments should come together to fight against the miseries of human life.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

11th Maths Exercise 1.1 Answers Question 1.
Write the following in roster form.
(i) {x ∈ N : x² < 121 and x is a prime}.
(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x² – 1) = 0.
(iii) {x ∈ N : 4x + 9 < 52}.
(iv) {x : \(\frac{x-4}{x+2}\) = 3, x ∈ R – {-2}}
Solution:
(i) A = {2, 3, 5, 7}
(ii) B = {1}
(iii) 4x + 9 < 52
4x + 9 – 9 < 52 – 9
4x < 43
x < \(\frac{43}{4}\) (i.e.) x < 10.75 4
But x ∈ N
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(i.e.) x – 4 = 3(x + 2)
x – 4 = 3x + 6
– 4 – 6 = 3x – x
2x = -10 ⇒ x = -5
A = {-5}

11th Std Maths Exercise 1.1 Answers Question 2.
Write the set {-1,1} in set builder form.
Solution:
A = {x: x² = 1}

11th Maths Book Exercise 1.1 Answers Question 3.
State whether the following sets are finite or infinite.
(i) {x ∈ N : x is an even prime number}
(ii) {x ∈ N : x is an odd prime number}
(iii) {x ∈ Z : x is even and less than 10}
(iv) {x ∈ R : x is a rational number}
(v) {x ∈ N : x is a rational number}
Solution:
(i) Finite set
(ii) Infinite set
(iii) Infinite
(iv) and
(v) infinite

Exercise 1.1 Class 11 Maths State Board Question 4.
By taking suitable sets A, B, C, verify the following results:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(if) A × (B ∪ C) = (A × B) ∪ (A × C).
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B).
(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A).
Solution:
To prove the following results let us take U = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
B = {2, 7, 8, 9}
C = {1, 5, 8, 7}
(i) To prove: A × (B ∩ C) = (A × B) ∩ (A × C)
B ∩ C = {8}; A = {1, 2, 5, 7}
So A × (B ∩ C) = {1, 2, 5, 7} × {8}
= {(1, 8), (2. 8), (5, 8), (7, 8)}
Now A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} …. ( 1)
A × C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∩ (A × C) = {(1, 8), (2, 8), (5, 8), (7, 8)} ……… (2)
(1) = (2)
⇒ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove A × (B ∪ C) = (A × B) (A × C)
B = {2, 7, 8, 9}, C = {1, 5, 8, 10)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) …. (1)
A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),
(7, 2), (7, 7), (7, 8), (7, 9)}
A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} …… (2)
(1) = (2) ⇒ A × (B ∪ C) = (A × B) ∪ (A × C)

(iii) A × B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)}
B × A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)}
L.H.S. (A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …. (1)
R.H.S. A ∩ B = {2, 7}
B ∩ A = {2, 7}
(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7}
= {(2, 2), (2, 7), (7, 2), (7, 7)} ……… (2)
(1) = (2) ⇒ LHS = RHS

(iv) To prove C – (B – A) = (C ∩ A) ∪ (C ∩ B)
B – A = {8, 9}
C = {1, 5, 8, 10}
∴ LHS = C – (B – A) = {1, 5, 10} …… (1)
C ∩ A = {1}
U = {1, 2, 5, 7, 8, 9, 10}
B = {2, 7, 8, 9} ∴ B’ = {1, 5, 10}
C ∩ B = {1, 5, 10}
R.H.S. (C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10}
= {1, 5, 10} ……. (2)
(1) = (2) ⇒ LHS = RHS

(v) To prove (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)
A= {1, 2, 5, 7}, B = {2, 7, 8, 9}, C = {1, 5, 8, 10}
Now B – A = {8, 9}
(B – A) ∩ C = {8} ……. (1)
B ∩ C = {8}
A = {1, 2, 5, 7}
So (B ∩ C) – A = {8} …… (2)
C – A = {8, 10}
B = {2, 7, 8, 9}
B ∩ (C – A) = {8} …. (3)
(1) = (2) = (3)

(vi) To prove (B – A) ∪ C ={1, 5, 8, 9, 10}
B – A = {8, 9},
C = {1, 5, 8, 10}
(B – A) ∪ C = {1, 5, 8, 9, 10} ……. (1)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A – C = {2, 7}
(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)
(1) = (2)
⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

11th Maths Exercise 1.1 Question 5.
Justify the trueness of the statement.
“An element of a set can never be a subset of itself.”
Solution:
A set itself can be a subset of itself (i.e.) A ⊆ A. But it cannot be a proper subset.

11th Maths Chapter 1 Exercise 1.1 Question 6.
If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).
Solution:
n(P( A)) = 1024 = 2¹⁰ ⇒ n( A) = 10
n(A ∪ B) = 15
n(P(B)) = 32 = 2⁵ ⇒ n(B) = 5
We know n(A ∪ B) = n{A) + n(B) – n(A ∩ B)
(i.e.) 15 = 10 + 5 – n(A ∩ B)
⇒ n(A ∩ B) = 15 – 15 = 0

11 Maths Exercise 1.1 Question 7.
If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A(A ∆ B)).
Solution:
n(A ∪ B) = 10; n(A ∩ B) = 3
n(A ∆ B) = 10 – 3 = 7
and n(P(A ∆ B)) = 27 = 128

11th Maths 1.1 Exercise Question 8.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.
Solution:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

11th Maths Exercise 1.1 Answers In Tamil Question 9.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
Solution:
n(A) = 3 ⇒ set A contains 3 elements
n(B) = 2 ⇒ set B contains 2 elements –
we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}

11th Maths Exercise 1.1 Answers State Board Question 10.
If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.
Solution:
n(A × A) = 16 ⇒ n( A) = 4
S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 Additional Questions

Samacheer Kalvi 11th Maths Example Sums Question 1.
Write the following sets in roster form
(a) {x ∈ N; x³ < 1000}
(b) {The set of positive roots of the equation (x² – 4) (x³ – 27) = 0}
Solution:
(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) B = {2, 3}

11th Maths Exercise 1.1 4th Sum Question 2.
By taking suitable sets A, B, C verify the following results
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) (B – A) ∪ C = (B ∪ C) – (A – C)
Solution:
Prove by yourself

11th Maths 1st Chapter Exercise 1.1 Question 3.
Given n(A) = 7; n(B) = 8 and n(A ∪ B) = 10 find n[P(A ∩ B)].
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i.e.,) 10 = 7 + 8 – n(A ∩ B)
⇒ n(A ∩ B) = 7 + 8 – 10 = 5
So n[P(A ∩ B)] = 25 = 32

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 8 Periodic Classification of Elements

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Samacheer Kalvi 10th Science Periodic Classification of Elements Textual Evaluation Solved

I. Choose the best answer.

Ex 3.14 Class 10 Samacheer Question 1.
The number of periods and groups in the periodic table are ____.
(a) 6, 16
(b) 7, 17
(c) 8, 18
(d) 7, 18.
Answer:
(d) 7, 48.

10th Maths Exercise 3.14 Samacheer Kalvi Question 2.
The basis of modern periodic law is:
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons
Answer:
(a) atomic number

Exercise 3.14 Class 10 Samacheer Question 3.
_____ group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th.
Answer:
(a) 17th

10th Maths Exercise 3.14 Question 4.
……….is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity
Answer:
(d) electronegativity

10th Maths Exercise 3.14 Solutions Question 5.
Chemical formula of rust is ____.
(a) FeO.xH₂O
(b) FeO₄.xH₂O
(c) Fe₂O₃.xH₂O
(d) FeO.
Answer:
(c) Fe₂O₃.xH₂O

10th Maths Exercise 3.14 Solution Question 6.
In the aluminothermic process the role of Al is ____.
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent.
Answer:
(b) reducing agent

Maths Book For Class 10 Samacheer Kalvi Question 7.
The process of coating the surface of metal with a thin layer of zinc is called:
(a) painting
(b) thinning
(c) galvanization
(d) electroplating
Answer:
(c) galvanization

10th Samacheer Maths Question 8.
Which of the following have inert gases 2 electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr.
Answer:
(a) He

Samacheer Kalvi 10th Guide Maths Question 9.
Neon shows zero electron affinity due to:
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density
Answer:
(b) stable configuration of electrons

10th Maths Solution Samacheer Kalvi Question 10.
____ is an important metal to form an amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al.
Answer:
(b) Hg

II. Fill in the blanks.

10th Standard Maths Algebra Question 1.
If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ____.
Answer:
Ionic bond.

10th Maths Samacheer Question 2.
____ is the longest period in the periodical table.
Answer:
Sixth period.

Maths Book Class 10 Samacheer Kalvi Question 3.
______ forms the basis of modem periodic table.
Answer:
Atomic number.

Question 4.
If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of the Cl atom is ____.
Answer:
0.99 Å.

Question 5.
Among the given species A^–, A⁺, and A, the smallest one in size is ______.
Answer:
A^+.

Question 6.
The scientist who propounded the modem periodic law is ______.
Answer:
Henry Moseley.

Question 7.
Across the period, ionic radii ______ (increases,decreases)
Answer:
Decreases.

Question 8.
_______ and ______ are called inner transition elements.
Answer:
Lanthanides and Actinides.

Question 9.
The chief ore of Aluminium is ______.
Answer:
Bauxite (Al₂O₃.2H₂O).

Question 10.
The chemical name of rust is ______.
Answer:
Hydrated ferric oxide.

III. Match the following.

Question 1.

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

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

Question 1.
Moseley’s periodic table is based on atomic mass.
Answer:
False.
Correct Statement: Moseley’s periodic table is based on atomic number.

Question 2.
Ionic radius increases across the period from left to right.
Answer:
False.
Correct Statement: Ionic radius decreases across the period from left to right.
Question 3.
All ores are minerals, but all minerals cannot be called as ores.
Answer:
True.

Question 4.
Al wires are used as electric cables due to their silvery – white colour.
Answer:
False.
Correct Statement: Al wires are used as electric cables due to their good conductor of electricity.

Question 5.
An alloy is a heterogeneous mixture of metals.
Answer:
False.
Correct Statement: An alloy is a homogenous mixture of metals.

V. Assertion and Reason:

Answer the following questions using the data given below.
(i) A and R are correct, R explains the A.
(ii) A is correct, R is wrong.
(iii) A is wrong, R is correct.
(iv) A and R are correct, R doesn’t explain A.

Question 1.
Assertion: The nature of bond in HF molecule is ionic
Reason: The electronegativity difference between H and F is 1.9.
Answer:
(i) A and R are correct, R explains the A.

Question 2.
Assertion: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
Answer:
(i) A and R are correct, R explains the A.

Question 3.
Assertion: An uncleaned copper vessel is covered with a greenish layer.
Reason: copper is not attacked by alkali
Answer:
(iv) A and R are correct, R doesn’t explain A.

VI. Short Answer Questions.

Question 1.
A is a reddish-brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Answer:
(i) Reddish brown metal (A) is copper.

(ii) (A) reacts with O₂ at bleow 1370 K gives Copper (II) oxide (B), which is black in colour.
2Cu + O₂ \(\xrightarrow [ 1370k ]{ below }\) 2CuO (Copper (II) oxide) (B).

(iii) (A) reacts with O₂ at above 1370 K gives Copper (I) oxide (C), which is red in colour.

Question 2.
A is a silvery – white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer:
(i) Silver – white metal (A) is Aluminium.

(ii) (A) combines with O₂ to form aluminium oxide at 800°C.

(iii) Duralumin is the alloy of Al, which is used to make aircraft

Question 3.
What is rust? Give the equation for formation of rust.
Answer:
Rust is hydrated ferric oxide, Fe₂O₃.xH₂O. It is formed when iron is exposed to moist air.
4 Fe + 3O₂ + xH₂O > 2Fe₂O₃.xH₂O

Question 4.
State two conditions necessary for rusting of iron.
Answer:
Conditions for rusting of iron:

• The presence of water and oxygen is essential for the rusting of iron.
• Impurities in the iron, the presence of water vapour, acids, salts and carbon dioxide hasten to rust.
• Pure iron does not rust in dry and CO₂ free air. It also does not rust in pure water, free from dissolved salts.

VII. Long Answer Questions

Question 1.
(a) State the reason for the addition of caustic alkali to bauxite ore during purification of bauxite.
(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Answer:
(a) Naturally, Bauxite is not soluble in normal solvents. Therefore the addition of caustic alkali to bauxite plays an important role while extraction of aluminium. Caustic alkali dissolves bauxite forming soluble sodium meta aluminate while impurities remain insoluble and precipitate as red mud.

(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture is Fluorspar. Adding of fluorspar lowers the fusion temperature of the electrolyte.

Question 2.
The electronic configuration of metal A is 2, 8, 18, 1. The metal A when exposed to air and moisture forms B a green layered compound. A with conc. H₂SO₄ forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Answer:
(i) The electronic configuration of metal (A) is 2, 8, 18, 1. A is copper (Z = 29)

(ii) (A) Copper exposed to air and moisture forms green layered compound (B) that is a copper carbonate.

(iii) Copper (A) reacts with conc.H₂SO₄ to give copper sulphate (C) and Sulphur dioxide (D).

Question 3.
Explain the smelting process.
Answer:
Smelting (in a Blast Furnace): The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top. There are three important regions in the furnace.

(a) The Lower Region (Combustion Zone): The temperature is at 1500°C. In this region, coke bums with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

(b) The Middle Region (Fusion Zone): The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

(c) The Upper Region (Reduction Zone): The temperature prevails at 400°C . In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.
\(\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Fe}+3 \mathrm{CO}_{2}\)
The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

VIII. HOT Questions.

Question 1.
Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions.
Answer:
(i) Metal (A) belongs to period 3 and group 13, is Aluminium (Al).

(ii) (A) Al in red hot condition reacts with steam to form Aluminium oxide (B).

(iii) Aluminium (A) reacts with strong alkali forms of sodium meta aluminate (C).

Question 2.
Name the acid that renders aluminium passive. Why?
Answer:
Dil. or cone. HNO₃ does not attack aluminium, but renders Al passive due to the formation of oxide film on its surface.

Question 3.
(a) Identify the bond between H and F in the HF molecule.
(b) What property forms the basis of identification?
(c) How does the property7 vary in periods and in groups?
Answer:
(a) The nature of the bond in the HF molecule is ionic.
(b) Electronegativity.
(c) Along the period, from left to right in the periodic table the electronegativity increases because of the increase in the nuclear charge which in turn attracts the electrons more strongly. On moving down a group, the electronegativity of the elements decreases because of the increased number of energy levels.

Samacheer Kalvi 10th Science Periodic Classification of Elements Additional Questions Solved

I. Choose the best answer.

Question 1.
Which period contains only two elements?
(a) Second
(b) First
(c) Third
(d) Fifth.
Answer:
(b) First

Question 2.
The sixth period contains inner transition elements.
(a) 18
(b) 14
(c) 10
(d) 8
Answer:
(b) 14

Question 3.
Lanthanides and Actinides are called as _____.
(a) Alkali metals
(b) Inner transition elements
(c) Transition elements
(d) Representative elements.
Answer:
(b) Inner transition elements

Question 4.
The number of valence electrons present in Halogens is:
(a) One
(b) Seven
(c) Zero
(d) Two
Answer:
(b) Seven

Question 5.
The distance between the two hydrogen nuclei of the molecule is 0.74 Å. So its covalent radius is _____.
(a) 0.74 Å
(b) 0.99 Å
(c) 0.37 Å
(d) 7.4 Å.
Answer:
(c) 0.37 Å

Question 6.
Along the groups, atomic radius _____.
(a) decreases
(b) increases
(c) decreases then increase
(d) no change.
Answer:
(b) increases.

Question 7.
Which one of the following elements will have the highest electronegativity?
(a) chlorine
(b) nitrogen
(c) caesium
(d) fluorine
Answer:
(d) fluorine

Question 8.
Electron affinity is measured in _____.
(a) kJ^-1
(b) mol^-1
(c) kJ/mol
(d) kJ/mol².
Answer:
(c) kJ/mol

Question 9.
Noble gases have ______ electron affinity.
(a) positive
(b) negative
(c) zero
(d) high.
Answer:
(c) zero

Question 10.
The element with positive electron gain enthalpy is:
(a) Hydrogen
(b) Sodium
(c) Argon
(d) Fluorine
Answer:
(b) Sodium

Question 11.
The mineral from which a metal can be readily and economically extracted on a large scale is said to be a / an _____.
(a) Ore
(b) Flux
(c) Slag
(d) Gangue.
Answer:
(a) Ore

Question 12.
Flux + Gangue → _____?
(a) Mineral
(b) Matrix
(c) Slag
(d) Smog.
Answer:
(c) Slag

Question 13.
………… is used for anodizing process.
(a) Zinc
(b) iron
(c) Aluminium
(d) Tin
Answer:
(c) Aluminium

Question 14.
The ore which can be purified by gravity separation method is _____.
(a) Haematite
(b) oxide ores
(c) sulphide ores
(d) both (a) and (b).
Answer:
(d) both (a) and (b).

Question 15.
The calcium silicate slag is formed in ………. zone.
(a) combustion zone
(b) fusion zone
(c) reduction
(d) molten
Answer:
(b) fusion zone

Question 16.
Zinc blende is purified by _____.
(a) Hydraulic method
(b) Magnetic separation method
(c) Froth floatation method
(d) Chemical method.
Answer:
(c) Froth floatation method

Question 17.
Bauxite ore is purified by _____.
(a) Leaching process
(b) Hydraulic method
(c) Froth floatation method
(d) Magnetic separation method.
Answer:
(a) Leaching process

Question 18.
When copper reacts with dil.HNO₃ it liberates …………. gas.
(a) SO₂
(b) NO₂
(C) CO₂
(d) NO
Answer:
(d) NO

Question 19.
Which metal process low melting point?
(a) Gallium
(b) Caesium
(c) Aluminium
(d) Copper.
Answer:
(a) Gallium

Question 20.
Which one of the following is not an ore of aluminium?
(a) Bauxite
(b) Haematite
(c) Cryolite
(d) Corundum.
Answer:
(b) Haematite

Question 21.
……….. is stored in Kerosene.
(a) Iron
(b) Silver
(c) Sodium
(d) Aluminium
Answer:
(c) Sodium

Question 22.
Electrolytic reduction of alumina into aluminium is ______.
(a) Hall’s process
(b) Alumino thermic process
(c) Baeyer’s process
(d) Bessemerisation process.
Answer:
(a) Hall’s process

Question 23.
In Hall’s process, cathode used is ______.
(a) Iron tank
(b) Graphite
(c) Pure alumina
(d) Iron tank linked with graphite.
Answer:
(d) Iron tank linked with graphite.

Question 24.
The metal oxides are usually……….. in nature.
(a) basic
(b) acidic
(c) amphoteric
(d) neither acidic nor basic
Answer:
(a) basic

Question 25.
A silvery-white metal is ______.
(a) Aluminium
(b) Copper
(c) Iron
(d) Zinc.
Answer:
(a) Aluminium

Question 26.
Aluminium reacts with NaOH to give ______.
(a) Al₂O₃
(b) AlCl₃
(c) NaAlO₂
(d) Al(OH)₃
Answer:
(c) NaAlO₂

Question 27.
Atoms of the elements belonging to the same group of periodic table will have:
(a) same number of protons
(b) same number of electrons in the valence shell
(c) same number of neutrons
(d) same number of electrons
Answer:
(b) same number of electrons in the valence shell

Question 28.
Chief ore of copper is ______.
(a) CuFeS₂
(b) Cu₂O
(c) Cu₂S
(d) CuSO₄
Answer:
(a) CuFeS₂

Question 29.
Molecular formula for copper pyrites ______.
(a) Cu₂O
(b) Cu₂S
(c) CuCO₃
(d) CuFeS₂
Answer:
(d) CuFeS₂

Question 30.
The electronegativity of fluorine is:
(a) 4.0
(b) 3
(c) 2.8
(d) 2.1
Answer:
(a) 4.0

Question 31.
The second most abundant metal available next to aluminium is ______.
(a) Cu
(b) Ag
(c) Au
(d) Fe.
Answer:
(d) Fe.

Question 32.
Most important ore of iron is ______.
(a) Haematite
(b) Magnetite
(c) Iron pyrite
(d) Cryolite.
Answer:
(a) Haematite

Question 33.
The volatile impurities present in haematite are:
(a) Carbon and Nitrogen
(b) Oxygen and Nitrogen
(c) Helium and Oxygen
(d) Sulphur and Phosphorus
Answer:
(d) Sulphur and Phosphorus

Question 34.
The solute present in brass is:
(a) copper
(b) zinc
(c) tin
(d) magnesium
Answer:
(b) zinc

Question 35.
Which one of the following is used for making pressure cookers?
(a) Brass
(b) Magnalium
(c) Duralumin
(d) Nickel steel
Answer:
(c) Duralumin

Question 36.
Which is used as propeller?
(a) Stainless steel
(b) Nickel steel
(c) Brass
(d) Magnalium.
Answer:
(b) Nickel steel

Question 37.
Gold does not occur in the combined form. It does not react with air or water. It is in the ______ state.
(a) native
(b) combined
(c) complex
(d) molten.
Answer:
(a) native

Question 38.
Which of the following metal is not found in a free state?
(a) Ag
(b) Au
(c) Pt
(d) Al.
Answer:
(d) Al.

Question 39.
Which one of the following does not react with copper?
(a) Oxygen
(b) Conc.H₂SO₄
(c) NaOH
(d) Conc.HNO₃
Answer:
(c) NaOH

Question 40.
An element which is an essential constituent of all organic compounds belongs to ______ group.
(a) 14th
(b) 15th
(c) 16th
(d) 17th.
Answer:
(a) 14th

Question 41.
The highest ionization energy is exhibited by ______.
(a) Halogens
(b) Alkaline earth metals
(c) Transition metals
(d) Nobel gases.
Answer:
(d) Nobel gases.

Question 42.
Which two elements of the following belongs to the same period? (Al, Si, Ba, O).
(a) Si, Ba
(b) Al, Ba
(c) Al, Si
(d) Al, O.
Answer:
(c) Al, Si

Question 43.
98% pure copper and 2% impurities is called ______.
(a) Matte
(b) Copper pyrites
(c) blister copper
(d) cuprite.
Answer:
(c) blister copper

Question 44.
_____ is used in making anchors and electromagnets.
(a) Steel
(b) Pig iron
(c) Cast iron
(d) Wrought iron.
Answer:
(d) Wrought iron.

Question 45.
Which reagent does not react with iron?
(a) Conc. HNO₃
(b) Conc.H₂SO₄
(c) Steam
(d) Dil. HNO₃
Answer:
(a) Conc. HNO₃

II. Fill in the blanks.

Question 1.
Matte is a mixture of ______.
Answer:
Cu₂S + FeS.

Question 2.
Second group elements are called ______.
Answer:
Alkaline earth metals.

Question 3.
Ionisation energy is measured in _____ unit.
Answer:
kJ/mol

Question 4.
The ionisation energy ______ along the period.
Answer:
Increases.

Question 5.
______ property, which predicts the nature of bonding between the atoms in a molecule.
Answer:
Electronegativity.

Question 6.
If the difference in electronegativity between two elements is 1.7, the bond has ____ and _____.
Answer:
50% ionic character, 50% covalent character.

Question 7.
If the difference in electronegativity between two elements is less than 1.7, the bond is considered to be ______.
Answer:
Covalent.

Question 8.
If the difference in electronegativity between two elements is greater than 1.7, the bond is considered to be ______.
Answer:
Ionic.

Question 9.
The process of extracting the ores from the earth’s crust is called ______.
Answer:
Mining.

Question 10.
______ is the main principle behind in Hydraulic method.
Answer:
Specific gravity.

Question 11.
Froth floatation process is preferable for ______ ores.
Answer:
Lighter
(or)
Sulphide.

Question 12.
On heating in air, iron forms ______.
Answer:
Fe₃O₄

Question 13.
Iron reacts with Chlorine to form _____ compound.
Answer:
Ferric chloride.

Question 14.
The corrosive action in the absence of moisture is called ______.
Answer:
Dry corrosion.

Question 15.
______ technique used to renovate the Pamban bridge.
Answer:
Protective coating.

Question 16.
The atomic number is the number of ____ in the nucleus or number of ______ revolving around the nucleus in an atom.
Answer:
Protons, electrons.

Question 17.
The long form of periodic table is based upon the _____ of elements.
Answer:
Electronic configuration.

Question 18.
In the periodic table, the horizontal rows are called _____ and vertical columns are called ______.
Answer:
Periods, groups.

Question 19.
The modem periodic table has been divided into ______ blocks known as _______ blocks.
Answer:
Four, s, p, d, f

Question 20.
The ______ of the elements in a period decreases from left to right and the atomic radii of the elements present in a group downwards.
Answer:
Atomic size, increases.

Question 21.
_______ period is the longest period and it contains ______ elements.
Answer:
Seventh, 32.

Question 22.
In the periodic table, there are ______ groups and _____ periods.
Answer:
18, 7.

Question 23.
Metals like Ti, Cr, Mn, Zr find their application in the manufacturing of defence equipment called ______.
Answer:
Strategic metals.

Question 24.
The metal ______ plays a vital role in nuclear reactions releasing nuclear energy and used in nuclear weapons.
Answer:
Uranium.

Question 25.
Copper, silver and gold are called _______ as they are used in making ____ and ______.
Answer:
Coinage metals, coins and jewellery.

Question 26.
Purity of gold is expressed in ______ and ______ is pure gold.
Answer:
Carats, 24 – carat gold.

Question 27.
_______ is an ore of aluminium and _____ is its mineral.
Answer:
Bauxite, clay.

Question 28.
All ______ cannot be called as ores but all ______ are minerals.
Answer:
Minerals, ores.

Question 29.
The process of extracting the ores from the earth’s crust is called ______.
Answer:
Mining.

Question 30.
The rocky impurity associated with the ore is called ____ or ______.
Answer:
Gangue, matrix.

Question 31.
_____ is a substance added to the ore to reduce the fusion temperature and to remove impurities.
Answer:
Flux.

Question 32.
____ is the process of reducing the roasted metallic oxide to metal.
Answer:
Smelting.

Question 33.
Slag is the fusible product formed when ____ reacts with ____ during the extraction of metals.
Answer:
Flux, gangue

Question 34.
The temperature applied in Hall’s process is _____ and the voltage used in ______.
Answer:
900 – 950°C, 5 – 6V

Question 35.
_____ is used in making manhole covers and drain pipes and _______ is used in making transmission cables and T.V. towers.
Answer:
Pig iron, steel.

Question 36.
______ is defined as the slow and steady destruction of a metal by the environment.
Answer:
Corrosion.

Question 37.
_______ is a process of coating zinc on iron sheets by using electric current.
Answer:
Galvanization.

III. Match the following.

Question 1.

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

Question 2.

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

Question 3.

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

Question 4.

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

Question 5.

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

Question 6.

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

Question 7.

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

Question 8.

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

Question 9.

Answer:

Question 10.

Answer:

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

Question 1.
First period contains only one element.
Answer:
False.
Correct Statement: First period contains two elements.(Hydrogen & Ftelium)

Question 2.
The valency of all alkali metals is one.
Answer:
True.
Question 3.
Noble gases are more reactive.
Answer:
False.
Correct Statement: Noble gases are less reactive.

Question 4.
The atomic radius decreases from Li to B?
Answer:
True.

Question 5.
As the positive charge increases, the size of the cation also increases.
Answer:
False.
Correct Statement: As the positive charge increases, the size of the cation also decreases.

Question 6.
Copper pyrite ore is concentrated by gravity separation method.
False
Correct Statement: Copper pyrite ore is concentrated by froth floatation process.

Question 7.
Aluminium alloyed with gold and silver for making coins and jewels.
Answer:
False.
Correct Statement: Copper alloyed with gold and silver for making coins and jewels.

Question 8.
The corrosive action in the presence of moisture is called wet corrosion.
Answer:
True

Question 9.
The physical and chemical properties of elements are the periodic function of their atomic numbers – modern periodic law.
Answer:
True.

Question 10.
The long form of periodic table consists of horizontal rows called groups and vertical columns called periods.
Answer:
False.
Correct statement: The long form of periodic table consists of horizontal rows called periods and vertical columns called groups.

Question 11.
The first period in the periodic table is the shortest period and contains 8 elements from Lithium to Neon.
Answer:
False.
Correct statement: The first period in the periodic table is the shortest period and contains 2 elements Hydrogen and Helium.
(OR)
The second period in the periodic table is the short period and contains 8 elements from Lithium to Neon.

Question 12.
The sixth period in the periodic table is the longest period and contains 32 elements.
Answer:
True.

Question 13.
Group 1, 2 and 13 – 18 are called normal elements.
(or)
Main group elements.
(or)
Representative elements.
Answer:
True.

Question 14.
The atomic size of the elements in a period increases from left to right.
Answer:
False.
Correct statement: The atomic size of the elements in a period decreases from left to right.

Question 15.
In a period, the metallic character of the element increases while their non-metallic character decreases.
Answer:
False.
Correct statement: In a period, the metallic character of the element decreases while their non-metallic character increases.

Question 16.
The last element authenticated by IUPAC is Cn 112 [Copemicium].
Answer:
True.

Question 17.
Silver was the first metal to be used in making utensils and weapons.
Answer:
False.
Correct statement: Copper was the first metal to be used in making utensils and weapons.

Question 18.
The strategic metals such as copper, silver and gold are used in the manufacturing of defence equipment.
Answer:
False.
Correct statement: The strategic metals such as titanium, chromium manganese, zirconium are used in the manufacturing of defence equipment.

Question 19.
Copper, silver and gold are called coinage metals.
Answer:
True.

Question 20.
For making ornaments, 24 – carat gold is used which is pure gold.
Answer:
False.
Correct statement: For making ornaments, 22 – carat gold is used which contains 22 parts of gold by weight and 2 parts of copper by weight.

Question 21.
The mineral from which a metal can be readily and economically extracted on a large scale is said to be ore.
Answer:
True.

Question 22.
The rocky impurity associated with the ore is called flux.
Answer:
False.
Correct statement: The rocky impurity associated with the ore is called gangue or matrix.

Question 23.
Slag is the fusible product formed when flux reacts with gangue during the extraction of metals.
Answer:
True.

Question 24.
Metals which have high chemical reactivity are found in a free state or in the native state.
Answer:
False.
Correct statement: Metals which have low chemical reactivity are found in a free state or in the native state.

Question 25.
Aluminium is the metal found most abundantly in the earth’s crust.
Answer:
True

Question 26.
Aluminium is a reddish-brown metal and it is a bad conductor of heat and electricity.
Answer:
False.
Correct statement: Aluminium is a silvery – white metal and it is a good conductor of heat and electricity.
Question 27.
Aluminium reacts with strong caustic alkalis forming aluminates.
Answer:
True

Question 28.
Conc. Nitric acid renders aluminium active due to the formation of nitride film on its surface.
Answer:
False.
Correct statement: Cone. Nitric acid renders aluminium passive due to the formation of oxide film on its surface.

Question 29.
Aluminium is a powerful reducing agent.
Answer:
True.

Question 30.
Duralumin alloy is light, having high tensile strength and corrosion-resistant.
Answer:
True.

Question 31.
Fe and Al₂O₃ are used in thermite welding.
Answer:
False.
Correct statement: Al powder and Fe₂O₃ is used in thermite welding.

Question 32.
The chief ore of copper is Ruby copper.
Answer:
False.
Correct statement: The chief ore of copper is copper pyrite.

Question 33.
Iron is a lustrous greyish white metal and can be magnetised.
Answer:
True.

Question 34.
The rust has the chemical formula as Fe₃O₄.
Answer:
False.
Correct statement: The rust has the chemical formula as Fe₂O₃. xH₂O.

V. Assertion and Reason.

Question 1.
Assertion (A): Nobel gas is unreactive.
Reason (R): They have unstable electronic configuration in their valence shells.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not 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 2.
Assertion (A): The nature of bond in NaI molecule is covalent.
Reason (R): The electronegativity difference between Na and I is 1.5
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(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 3.
Assertion (A): Haematite ore was purified by Hydraulic method.
Reason (R): Haematite is oxide ore.
(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): Corundum is a chief ore of aluminium.
Reason (R): Molecular formula of Corundum is Al₂O₃
(a) (A) and (R) are correct, (R) explains the (A)
(b) (A) is correct, (R) is wrong
(c) (A) is wrong, (R) is correct
(d) (A) and (R) are correct, (R) doesn’t explain (A).
Answer:
(c) (A) is wrong, (R) is correct

Question 5.
Assertion (A): The chemical properties of the elements in the same period are not similar.
Reason (R): As the electronic configuration changes across the period, the chemical properties of the elements are not similar.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(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): Copper, Silver and Gold are used in making coins and jewellery. So they are called coinage metals.
Reason (R): These metals release an enormous amount of nuclear energy.
(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): Metals like Titanium, Chromium, Manganese and Zirconium are called strategic metals!
Reason (R): They find their applications in the manufacturing of defence equipment.
(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): Gold, Silver and Platinum are the metals that are found in a free state.
Reason (R): Those metals have low chemical reactivity and are found in a free state or in the native state.
(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 9.
Assertion (A): Aluminium occurs in the combined state.
Reason (R): It is a reactive metal and so it occurs in combined state.
(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): In the aluminothermic process, Iron oxide is reduced to iron by igniting with Aluminium powder.
Reason (R): Aluminium is a powerful reducing agent.
(a) Both (A) and (R) are wrong
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are correct.
Answer:
(d) Both (A) and (R) are correct

Question 11.
Assertion (A): When iron is dipped in conc.HNO₃ it becomes chemically inert (or) passive.
Reason (R): Iron becomes passive when treated with nitric acid is due to the formation of a layer of iron oxide Fe₃O₄ on its surface.
(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 12.
Assertion (A): Duralumin is used in making aircraft, tools and pressure cookers.
Reason (R): Duralumin is an alloy that is light, strong, resistant to corrosion.
(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 13.
Assertion (A): Nickel steel is used in making cables, aircraft parts and propeller.
Reason (R): Nickel steel alloy is hard, brittle and polishable.
(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 14.
Assertion (A): Magnesium is used in a sacrificial protection method to prevent corrosion.
Reason (R): Magnesium is more reactive than iron. When it is coated on the articles made of steel, it sacrifices itself to protect steel.
(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 15.
Assertion (A): Electroplating method not only protects but also enhances the metallic appearance.
Reason (R): Electroplating is a method of coating one metal with another by passing current.
(a) (A) is right, (R) is wrong
(b) (A) is right, (R) is not relevant
(c) (A) is right, (R) is relevant
(d) Both (A) and (R) are wrong.
Answer:
(c) (A) is right, (R) is relevant

VI. Short Answer Questions.

Question 1.
State modern periodic law.
Answer:
Modem periodic law states that the physical and chemical properties of elements are the periodic function of their atomic numbers.

Question 2.
Write the flow chart of the long form of the periodic table.
Answer:

Question 3.
The distance between the adjacent copper atoms in solid copper is 2.56 Å. Calculate what is the metallic radius of Cu?
Answer:
Metallic radius of copper = \(\frac{2.56}{2}\) = 1.28 Å

Question 4.
Briefly write any four characteristics of a group in the periodic table.
Answer:

• The elements present in a group have the same valency.
• The elements present in a group have identical chemical properties.
• The physical properties of the elements in the group very gradually.
• The atomic radii of die elements present in a group increase downwards.

Question 5.
What is the principle behind Froth floatation?
Answer:
This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method. Eg: Zinc blende (ZnS).

Question 6.
What are minerals?
Answer:
A mineral may be a single compound or a complex mixture of various compounds of metals found in the earth, e.g. Clay Al₂O₃. 2SiO₂. 2H₂O is the mineral of Aluminium.

Question 7.
Write the increasing order of radii of the following species.
(a) Na, Na⁺, Cl, Cl^–
(b) Li, Na, K, Rb
Answer:
(a) Na⁺, Cl, Na, Cl^–
(b) Li, Na, K, Rb

Question 8.
Differentiate ore and mineral.
Answer:

Question 9.
Define metallurgy.
Answer:
The various steps involved in the extraction of metals from their ores as well as refining of crude metals are collectively known as metallurgy.

Question 10.
Elements a, b, c and d have the following electronic configurations
(a) ls² 2s² 2p⁶
(b) Is², 2s², 2p⁶, 3s², 3p¹
(c) Is², 2s², 2p⁶, 3s², 3p⁶
(d) Is², 2s², Ip¹
Answer:
(a) and (c), (b) and (d). Because the number of valence electrons are the same.

Question 11.
What is slag? Give an example.
Answer:
Slag is a fusible product formed when flux reacts with gangue during the extraction of metals.
Flux + gangue → slag
CaO + SiO₂ → CaSiO₃

Question 12.
Why the electron affinities of noble gases are zero?
Answer:
Noble gases show no tendency to accept electrons because the outer 5 and p orbitals of noble gases are completely filled. No more electrons can be added to them and hence their electron affinities are zero.

Question 13.
Why does gold, silver and platinum occur in free state?
Answer:
Gold, silver and platinum have low chemical reactivity and so they are found in the free state or in a native state.

Question 14.
How does Aluminium reacts with air?
Answer:
Reaction with air: It is not affected by dry air. On heating at 800°C, aluminium bums very brightly forming it’s oxide and nitride.
4Al + 3O₂ → Al₂O₃(Aluminium oxide)
2Al + N₂ → 2 AlN (Aluminium nitride)

Question 15.
How does Aluminium react with caustic soda? Give an equation.
Answer:
Aluminium reacts with caustic soda to give sodium meta aluminate with the liberation of H₂ gas.

Question 16.
Prove that aluminium is a powerful reducing agent.
Answer:
Aluminium is a powerful reducing agent When a mixture of aluminium powder and iron oxide is ignited, iron oxide is reduced to iron. This process is known as the aluminothermic process.

Question 17.
Write a note on Aluminothermic process.
Answer:
As reducing agent: Aluminium is a powerful reducing agent. When a mixture of aluminium powder and iron oxide is ignited, the latter is reduced to metal. This process is known as aluminothermic process.
Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat

Question 18.
What is the action of heat on copper?
Answer:
On heating at different temperatures in the presence of oxygen, copper forms two types of oxides CuO, Cu₂O.

Question 19.
Explain the action of dilute nitric acid with copper.
Answer:
Copper reacts with dil.HNO₃ with the liberation of Nitric oxide gas.

Question 20.
What happens when copper is treated with conc.HNO₃ and with conc.H₂SO₄?
Answer:

Question 21.
What do you mean by Ferrous and Non-Ferrous alloys? Given an example for each.
Answer:
Ferrous alloys : Contain iron as a major component.
Eg: Stainless steel, Nickel steel

Non-Ferrous alloys : These alloys do not contain iron as a major part.
Eg: Aluminium alloy, Copper alloy.

Question 22.
Explain the action of air with iron.
Answer:
3Fe + 2O₂ → Fe₃O₄ [Magnetic oxide (Black)].

Question 23.
What are amalgams? How are they prepared?
Answer:
An amalgam is an alloy of mercury with another metal.
Amalgams are formed by the metallic bonding with the electrostatic force of attraction between the electrons and the positively charged metal ions.

Question 24.
Explain the action of steam with iron.
Answer:

Question 25.
Mention the types of iron on the basis of carbon content.
Answer:

Question 26.
What is dry corrosion or chemical corrosion?
Answer:
The corrosive action in the absence of moisture is called dry corrosion. It is the process of a chemical attack on a metal by corrosive liquids or gases such as O₂, N₂, SO₂, H₂S etc… It occurs at high temperature, of all the gases mentioned above O₂ is the most reactive gas to impart the chemical attack.

Question 27.
What is an amalgam? Give one example with its use.
Answer:

• An amalgam is an alloy of mercury with metals such as sodium, gold and silver.
• Dental amalgam is an alloy of mercury with silver and tin and it is used in the dental filling.

Question 28.
Explain sacrificial protection.
Answer:
Magnesium is more reactive than iron. When it is coated on the articles made of steel, it sacrifices itself to protect steel.

Question 29.
Define corrosion.
Answer:
Corrosion is defined as the slow and steady destruction of a metal by the environment. It results in the deterioration of the metal to form metal compounds by means of chemical reactions with the environment.

Question 30.
What is the action of Air and moisture on copper?
Answer:
Copper gets covered with a green layer of basic carbonate in the presence of CO₂ and moisture.
2 Cu + O₂ + CO₂ + H₂O → CUCO₃.CU(OH)₂

Question 31.
Give any two uses of aluminium.
Answer:

• Aluminium metal is a corrosion – resistant and a good conductor of heat. So it is used in making utensils.
• Aluminium is used in welding as thermite and a very good reducing agent.

Question 32.
What is the modern periodic table?
Answer:
The modem periodic table is a tabular arrangement of elements in rows and columns, highlighting the regular repetition of properties of the elements.

Question 33.
Explain the smelting process of iron in the Blast furnace.
Answer:
The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top.

Question 34.
What are the periodic properties?
Answer:
Properties such as atomic radius, ionic radius, ionisation energy, electronegativity, electron affinity, show a regular periodicity and hence they are called periodic properties.

Question 35.
Define Atomic radius.
Answer:
The atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

Question 36.
Write the action of dil.HCl and dil.H₂SO₄ with iron.
Answer:
(i) Fe + 2HCl → FeCl₂ + H₂↑
(ii) Fe + H₂SO₄ → FeSO₄ + H₂↑

Question 37.
What is a covalent radius?
Answer:
It is defined as half the distance between the nuclei of two covalently bonded atoms of the same element in a molecule.

Question 38.
Define Ionisation energy.
Answer:
Ionisation energy is the minimum energy required to remove an electron from a gaseous atom in its ground state to form a cation.

Question 39.
What is an alloy?
Answer:
An alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements. The properties of alloys are often different from the component metals.

Question 40.
What is Electronegativity?
Answer:
Electronegativity of an element is the measure of the tendency of its atom to attract the shared pair of electrons towards itself in a covalent bond.

Question 41.
Write the steps involved in metallurgical process.
Answer:
A metallurgical process involves three main steps as follows:

• Concentration or Separation of the ore: It is the process of removal of impurities from the ore.
• Production of the metal: It is the conversion of the ore into metal.
• Refining of the metal: It is the process of purification of the metal.

Question 42.
Give a single term for each of the following:
(i) The process of extracting the ores from the Earth’s crust is called:
Answer:
Mining

(ii) The rocky impurities associated with an ore is called:
Answer:
Gangue or matrix

(iii) The substance added to the ore to reduce fusion temperature:
Answer:
Flux

(iv) Noble metals occur in this state:
Answer:
Native

Question 43.
How will you convert copper into copper carbonate?
Answer:
Copper reacts with oxygen in the presence of CO₂ and moisture to give copper carbonate.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂ (Copper Carbonate).

Question 44.
Mention the uses of iron.
Answer:
Uses of iron:

1. Pig iron (Iron with 2 – 4.5 % of carbon): It is used in making pipes, stoves, radiators, railings, manhole covers and drain pipes.
2. Steel (Iron with < 0.25 % of carbon): It is used in the construction of buildings, machinery, transmission cables and T.V towers and in making alloys.
3. Wrought iron (Iron with 0.25 – 2 % of wrought carbon): It is used in making springs, anchors and electromagnets.

Question 45.
What are the chemical properties of metals in terms of

1. valence electrons
2. Atomicity.

Answer:

1. Valence electrons: Metals usually have 1, 2 or 3 electrons in their outermost shell.
2. Atomicity: Metals are usually monoatomic in their vapour state.

Question 46.
Why alloys are said to solid solutions?
Answer:
Alloys can be considered solid solutions in which the metal with high concentration is solvent and other metals are solute.
Example: brass is a solid solution of zinc (solute) in copper (solvent).

Question 47.
Write a note on Dry corrosion.
Answer:

• The corrosive action in the absence of moisture is called dry corrosion.
• It is the process of a chemical attack on a metal by corrosive liquids or gases such as O₂, N₂, SO₂, H₂S etc. in which O₂ is more reactive.
• It occurs at high temperature

Question 48.
Explain Wet Corrosion.
Answer:

• The corrosive action in the presence of moisture is called wet corrosion.
• It occurs as a result of the electrochemical reaction of metal with water or an aqueous solution of salt or acids or bases.

Question 49.
What is electroplating?
Answer:
Electroplating is a method of coating one metal over another metal by passing an electric current.

VII. Long Answer Questions.

Question 1.
Explain the variation of ionisation energy along the group and period.
Answer:

• As the atomic size decreases from left to right in a period, more energy is required to remove the electrons. So, the ionisation energy increases throughout the period.
• Down the group, the atomic size increases and hence the valence electrons are loosely bound. They require relatively less energy for the removal. Thus, ionisation energy decreases down the group in the periodic table.

Question 2.
Explain the electrolytic refining of copper.
Answer:
(i) Blister copper contains 98% of pure Cu and 2 % of impurities and is purified by electrolytic refining.
For electrolytic refining of Cu we use:
Cathode: A thin plate of pure Cu.
Anode: A block of impure Cu
Electrolyte: CuSO₄ + dil H₂SO₄

(ii) When electric current is passed through the electrolytic solution, Pure Cu gets deposited at the cathode and the impurities settle at the bottom of the anode in the form of anode mud.

Question 3.
Explain Gravity separation method.
Answer:
Gravity Separation (or) Hydraulic method:
1. Principle: The difference in the densities or specific gravities of the ore and the gangue is the main principle behind this method. Oxide ores are purified by this method,
e.g., Haematite Fe₂O₃ the ore of iron.

2. Method: The ore is poured over a sloping, vibrating corrugated table with grooves and a jet of water is allowed to flow over it. The denser ore particles settle down in the grooves and lighter gangue particles are washed down by the water.

Question 4.
Discuss the magnetic separation methods.
Answer:
Magnetic separation method:
Principle: The magnetic properties of the ores from the basis of separation. When either the ore or the gangue is magnetic, this method is employed, e.g., Tinstone SnO₂, the ore of tin.

Method: The crushed ore is placed over a conveyer belt which rotates around two metal wheels, one of which is magnetic. The magnetic particles are attracted to the magnetic wheel and fall separately apart from the nonmagnetic particles.

Question 5.
Explain the froth floatation process.
Answer:
Froth floatation Process:
Principle: This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method, e.g., Zinc blende (ZnS).

Method: The crushed ore is taken in a large tank containing oil and water and agitated with a current of compressed air. The ore is wetted by the oil and gets separated from the gangue in the form of froth. Since the ore is lighter, it comes on the surface with the froth and the impurities are left behind, e.g., Zinc blende (ZnS).

Question 6.
How will you extract aluminium from its ore?
Answer:
The extraction of aluminium from bauxite involves two steps:
(i) Conversion of bauxite into alumina – Baeyer’s Process
The conversion of Bauxite into Alumina involves the following steps:
Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150° C to obtain sodium metal aluminate.
On diluting sodium meta aluminate with water, a precipitate of aluminium hydroxide is formed.
The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.
\(2 \mathrm{Al}(\mathrm{OH})_{3} \stackrel{1000^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}\)

(ii) Electrolytic reduction of alumina – Hall’s Process
Aluminium is produced by the electrolytic reduction of fused alumina (Al₂O₃) in the electrolytic cell.

Cathode: Iron tank linked with graphite
Anode: A bunch of graphite rods suspended in a molten electrolyte.
Electrolyte: Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)
Temperature: 900 – 950°C
The voltage used: 5 – 6 V
Aluminium is deposited at the cathode and oxygen gas is liberated at the anode. Oxygen combines with graphite to form CO₂.

Question 7.
Explain the extraction of copper from copper pyrites.
Answer:
Extraction of copper from copper pyrites involves the following steps:
(i) The concentration of ore: The ore is crushed and then concentrated by froth floatation process.

(ii) Roasting: The concentrated ore is roasted in excess of air. During the process of roasting, the moisture and volatile impurities are removed. Sulphur, phosphorus, arsenic and antimony are removed as oxides. Copper pyrite is partly converted into sulphides of copper and iron.
\(2 \mathrm{CuFeS}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{S}+2 \mathrm{FeS}+\mathrm{SO}_{2} \uparrow\)

(iii) Smelting: The roasted ore is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.

(iv) Bessemerisation: The molten matte is transferred to the Bessemer converter in order to obtain blister copper. Ferrous sulphide from matte is oxidized to ferrous oxide, which is removed as slag using silica.

(v) Refining: Blister copper contains 98% of pure copper and 2% of impurities and is purified A by electrolytic refining. This method is used to get metal of a high degree of purity. For electrolytic refining of copper, we use:
Cathode: A thin plate of pure copper metal.
Anode: A block of impure copper metal.
Electrolyte: Copper sulphate solution acidified with sulphuric acid.
When an electric current is passed through the electrolytic solution, pure copper gets deposited at the cathode and the impurities settle at the bottom of the anode in the form of sludge called anode mud.

Question 8.
Explain the metallurgy of iron.
Answer:
Iron is chiefly extracted from haematite ore (Fe₂O₃):
(i) Concentration by Gravity Separation: The powdered ore is washed with steam of water. As a result, the lighter sand particles and other impurities are washed away and the heavier ore particles settle down.

(ii) Roasting and Calcination: The concentrated ore is strongly heated in a limited supply of air in a reverberatory furnace. As a result, moisture is driven out and sulphur, arsenic and phosphorus impurities are oxidized off.

(iii) Smelting (in a Blast Furnace): The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top. There are three important regions in the furnace.

a. The Lower Region (Combustion Zone):
The temperature is at 1500°C. In this region, coke bums with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

b. The Middle Region (Fusion Zone):
The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

c. The Upper Region (Reduction Zone): The temperature prevails at 400°C. In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.
\(\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Fe}+3 \mathrm{CO}_{2}\)
The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

Question 9.
Explain the types of alloys.
Answer:
Based on the presence or absence of Iron, alloys can be classified into:
1. Ferrous alloys: Contain Iron as a major component.
A few examples of ferrous alloys are Stainless Steel, Nickel Steel etc.

2. Non – ferrous alloys: These alloys do not contain Iron as a major component.
For example, Aluminium alloy, Copper alloy etc.
Copper Alloys (Non – ferrous):

Aluminium Alloys (Non – ferrous):

Iron Alloys(Ferrous):

VIII. HOT Questions.

Question 1.
What would be the atomic number of the next

1. alkali metal
2. alkaline earth metal
3. Halogens
4. inert gas, if discovered in future.

Answer:

1. Alkali metal: 118 + 1 = 119
2. Alkaline earth metal: 120
3. Halogens: 117
4. Inert gas: 118

Question 2.
Explain the mechanism of rusting?
Answer:
Rust is chemically known as hydrated ferric oxide (it is formulated as Fe₂O₃.xH₂O).
Rusting results in the formation of scaling reddish – brown hydrated ferric oxide on the surface of iron and iron-containing materials.

Question 3.
All ores are minerals, but ait minerals are not ores. Why?
Answer:

• The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals whereas ore is a mineral from which the metal is profitably extracted.
• For example, aluminium exists in the two mineral forms, that is clay and bauxite. But aluminium is mainly extracted from bauxite which contains 70 % aluminium oxide. So, bauxite is an ore of aluminium whereas clay is not ore.
• So, all ores are minerals but all minerals need not be ores.

Question 4.
Why the graphite rods acting as anode in the Hall’s process should be replaced periodically?
Answer:
During Hall’s Process, the carbon electrodes get consumed, so they have to be replaced periodically.

Question 5.
Anionic radius Is higher than the corresponding neutral atom. Give reason.
Answer:
When an atom gains one or more electrons it forms an anion. During the formation of anion, the number of orbital electrons becomes greater than the nuclear charge. Hence, the electrons are not strongly attracted by the lesser number of nuclear charges. Hence anionic radius is higher than the corresponding neutral atom.

Question 6.
A reddish – brown metal A, when exposed to moist air, forms a green layer B. When A is heated at different temperatures in the presence of O₂, it forms two types of oxides – C (black) and D (red). Identify A, B, C, D and write the balanced equation.
Answer:
(i) A reddish – brown metal A is a copper (Cu).

(ii) When copper (A) is exposed to moist air it forms a green layer (B) is copper carbonate.

(iii) When copper is heated at different temperature in the presence of oxygen, it forms two types of oxides CuO and Cu₂O. (C and D)

Question 7.
A silvery – white metal on treatment with NaOH and HCl liberates H₂ gas to form B and C respectively. The metal A will not react with acid D due to the formation of a passive film on the surface. Hence it is used for transporting acid D. Identify A, B, C, D and support your answer with balanced equations.
Answer:
(i) A silvery – white metal (A) is Aluminium (Al).
(ii) Aluminium reacts with NaOH to form B which is known as sodium meta aluminate with the liberation of H₂ gas.

(iii) Aluminium reacts with HCl to form Aluminium chloride which is known as C with the liberation of H₂ gas.

(iv) Aluminium does not react with conc. nitric acid (HNO₃) which is known as D, due to the formation of a passive film on the surface.

Question 8.
Metal A belongs to period 4 and group 8. A in red hot condition reacts with steam to form B. A reacts with dilute HNO₃ to give C. A again reacts with conc. H₂SO₄ to give D. Find A, B, C and D with suitable reaction.
Answer:
(i) Metal (A) belongs to period 4 and group 8 is iron (Fe).

(ii) Iron (A) reacts with steam to form magnetic oxide (B)

(iii) Iron (A) reacts with dilute HNO₃ in cold condition to give ferrous nitrate (C).

(iv) Iron (A) reacts with conc.H₂SO₄ to form Ferric Sulphate (D).

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.2

10th Maths Exercise 2.2 Samacheer Kalvi Question 1.
For what values of natural number n, 4n can end with the digit 6?
Solution:
4ⁿ = (2 × 2)ⁿ = 2ⁿ × 2ⁿ
2 is a factor of 4ⁿ.
So, 4ⁿ is always even and end with 4 and 6.
When n is an even number say 2, 4, 6, 8 then 4ⁿ can end with the digit 6.
Example:
4² = 16
4³ = 64
4⁴ = 256
4⁵ = 1,024
4⁶ = 4,096
4⁷ = 16,384
4⁸ = 65, 536
4⁹ = 262,144

Ex 2.2 Class 10 Samacheer Question 2.
If m, n are natural numbers, for what values of m, does 2ⁿ × 5^m ends in 5?
Answer:
2ⁿ is always even for any values of n.
[Example. 2² = 4, 2³ = 8, 2⁴ = 16 etc]

5^m is always odd and it ends with 5.
[Example. 5² = 25, 5³ = 125, 5⁴ = 625 etc]
But 2ⁿ × 5^m is always even and end in 0.
[Example. 2³ × 5³ = 8 × 125 = 1000
2² × 5² = 4 × 25 = 100]
∴ 2ⁿ × 5^m cannot end with the digit 5 for any values of m.

Exercise 2.2 Class 10 Maths Samacheer Question 3.
Find the H.C.F. of 252525 and 363636.
Solution:
To find the H.C.F. of 252525 and 363636
Using Euclid’s Division algorithm
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0.
∴ Again by division algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0.
∴ Again by division algorithm.
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0.
∴ Again by division algorithm
30303 = 20202 × 1 + 10101
The remainder 10101 ≠ 0.
∴ Again using division algorithm
20202 = 10101 × 2 + 0
The remainder is 0.
∴ 10101 is the H.C.F. of 363636 and 252525.

10th Maths Exercise 2.2 Question 4.
If 13824 = 2^a × 3^b then find a and b.
Solution:
If 13824 = 2^a × 3^b
Using the prime factorisation tree

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2⁹ × 3³ = 2^a × 3^b
∴ a = 9, b = 3.

10th Maths Exercise 2.2 In Tamil Question 5.
If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400 where p₁, p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄ are integers, find the value of P₁, P₂, P₃, P₄ and x₁, x₂, x₃, x₄.
Solution:
If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400
p₁, p₂, p₃, P₄ are primes in ascending order, x₁, x₂, x₃, x₄ are integers.
using Prime factorisation tree.

113400 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7
= 23 × 34 × 52 × 7
= p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄
∴ p₁= 2, p₂ = 3, p₃ = 5, p₄ = 7, x₁ = 3, x₂ = 4, x₃ = 2, x₄ = 1.

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.
Solution:
408 and 170.

408 = 2³ × 3¹ × 17¹
170 = 2¹ × 5¹ × 17¹

∴ H.C.F. = 2¹ × 17¹ = 34.
To find L.C.M, we list all prime factors of 408 and 170, and their greatest exponents as follows.

∴ L.C.M. = 2³ × 3¹ × 5¹ × 17¹
= 2040.

10th Maths 2.2 Exercise Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?
Solution:
To find L.C.M of 24, 15, 36

24 = 2³ × 3
15 = 3 × 5
36 = 2² × 3²

∴ L.C.M = 2³ × 3² × 5¹
= 8 × 9 × 5
= 360
If a number has to be exactly divisible by 24, 15, and 36, then it has to be divisible by 360. Greatest 6 digit number is 999999.
Common multiplies of 24, 15, 36 with 6 digits are 103680, 116640, 115520, …933120, 999720 with six digits.
∴ The greatest number consisting 6 digits which is exactly divisible by 24, 15, 36 is 999720.

10th Maths Ex 2.2 Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Answer:
Find the L.C.M of 35, 56, and 91
35 – 5 × 7 56
56 = 2 × 2 × 2 × 7
91 = 7 × 13
L.C.M = 23 × 5 × 7 × 13
= 3640
Since it leaves remainder 7
The required number = 3640 + 7
= 3647
The smallest number is = 3647

10th Maths 2.2 Question 9.
Find the least number that is divisible by the first ten natural numbers.
Solution:
The least number that is divisible by the first ten natural numbers is 2520.
Hint:
1,2, 3,4, 5, 6, 7, 8,9,10
The least multiple of 2 & 4 is 8
The least multiple of 3 is 9
The least multiple of 7 is 7
The least multiple of 5 is 5
∴ 5 × 7 × 9 × 8 = 2520.
L.C.M is 8 × 9 × 7 × 5
= 40 × 63
= 2520

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14

Ex 3.14 Class 10 Samacheer Question 1.
Write each of the following expression in terms of α + β and αβ.

Solution:

10th Maths Exercise 3.14 Samacheer Kalvi Question 2.
The roots of the equation 2x² – 7x + 5 = 0 are α and β. Without solving the root find

Solution:
2x² – 7x + 5 = x² – \(\frac{7}{2} x+\frac{5}{2}\) = 0
α + β = \(\frac{7}{2}\)
αβ = \(\frac{5}{2}\)

Exercise 3.14 Class 10 Samacheer Question 3.
The roots of the equation x² + 6x – 4 = 0 are α, β. Find the quadratic equation whose roots are
(i) α² and β²
(ii) \(\frac{2}{\alpha} \text { and } \frac{2}{\beta}\)
(iii) α²β and β²α
Solution:
If the roots are given, the quadratic equation is x² – (sum of the roots) x + product the roots = 0.
For the given equation.
x² + 6x – 4 = 0
α + β = -6
αβ = -4
(i) α² + β² = (α + β)² – 2αβ
= (-6)² – 2(-4) = 36 + 8 = 44
α²β² = (αβ)² = (-4)² = 16
∴ The required equation is x² – 44x – 16 = 0.

(iii) α²β + β²α = αβ(α + β)
= -4(-6) = 24
α²β × β²α = α³β³ = (αβ)³ = (-4)³ = -64
∴ The required equation = x² – 24x – 64 – 0.

10th Maths Exercise 3.14 Question 4.
If α, β are the roots of 7x² + ax + 2 = 0 and if β – α = \(\frac{-13}{7}\) Find the values of a.
Solution:

10th Maths Exercise 3.14 Solutions Question 5.
If one root of the equation 2y² – ay + 64 = 0 is twice the other then find the values of a.
Solution:
Let one of the root α = 2β
α + β = 2β + β = 3β
Given

a² = 576
a = 24, -24

10th Maths Exercise 3.14 Solution Question 6.
If one root of the equation 3x² + kx + 81 = 0 (having real roots) is the square of the other then find k.
Solution:
3x² + kx + 81 = 0
Let the roots be α and α²

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

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Samacheer Kalvi 11th Chemistry Hydrogen Textual Evaluation Solved

I. Choose The Correct Answer:
11th Chemistry Chapter 4 Book Back Answers Question 1.
Which of the following statements about hydrogen is incorrect ? (NEET – 2016)
(a) Hydrogen ion, H₃O⁺ exists freely in solution.
(b) Dihydrogen acts a,s a reducing agent.
(c) Hydrogen has three isotopes of which tritium is the most common.
(d) Hydrogen never acts as cation in ionic salts.
Answer:
(c) Hydrogen has three isotopes of which tritium is the most common.
Hint:
Correct statement:
Hydrogen has three isotopes of which protium is the most common.

11th Chemistry Lesson 4 Book Back Answers Question 2.
Water gas is ………..
(a) H₂ O_(g)
(b) CO + H₂O
(C) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

11th Chemistry 4th Lesson Book Back Answers Question 3.
Which one of the following statements is incorrect with regard to ortho and para dihydrogen ?
(a) They are nuclear spin isomers
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
(c) The para isomer is favoured at low temperatures
(d) The thermal conductivity of the para isomer is 50% greater than that of the ortho isomer.
Answer:
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
Hints:
Correct statement:
Ortho isomer – one nuclear spin Para isomer – zero nuclear spin

11th Chemistry Unit 4 Book Back Answers Question 4.
Ionic hydrides are formed by …………….
(a) halogens
(b) chalogens
(c) inert gases
(d) group one elements
Answer:
(d) group one elements
e.g., Sodium hydride (Na⁺ H^– )

Chemistry Class 11 Samacheer Kalvi Question 5.
Tritium nucleus contains ……………..
(a) 1p + 0n
(b) 2p + 1n
(c) 1p + 2n
(d) none of these
Answer:
(c) lp + 2n
₁T³ (le^–, lp, 2n)

Class 11 Chemistry Solutions Samacheer Kalvi Question 6.
Non-stoichiometric hydrides are formed by……………..
(a) palladium, vanadium
(b) carbon, nickel
(c) manganese, lithium
(d) nitrogen, chlorine
Answer:
(a) palladium, vanadium

11th Chemistry Hydrogen Lesson Question 7.
Assertion : Permanent hardness of water is removed by treatment with washing soda.
Reason : Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates
(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:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
Ca²⁺ + Na₂CO₃ → CaCO₃↓ + 2Na⁺

11th Chemistry 4th Chapter Question 8.
If a body of a fish contains 1.2 g hydrogen in its total body mass, if all the hydrogen is replaced with deuterium then the increase in body weight of the fish will be ……………
(a) 1.2 g
(b) 2.4 g
(c) 3.6 g
(d) \(\sqrt{4.8}\) g
Answer:
(a) 1.2 g
Hints:
Mass of deuterium = 2 × mass of protium
If all the 1.2 g hydrogen is replaced with deuterium, the weight will become 2.4g. Hence the increase in body weight is (2.4 – 1.2 = 1.2 g).

Samacheer Kalvi 11th Chemistry Book Solutions Question 9.
The hardness of water can be determined by volumetrically using the reagent …………..
(a) sodium thio sulphate
(b) potassium permanganate
(c) hydrogen peroxide
(d) EDTA
Answer:
(d) EDTA

Hydrogen 11th Chemistry Question 10.
The cause of permanent hardness of water is due to ………….
(a) Ca(HCO₃)₂
(b) Mg(HCO_3k)₃
(c) CaCl₂
(d) MgCO₃
Answer:
(c) CaCl₂
Hints:
Permanent hardness if water is due to the presence of the chlorides, nitrates and sulphates of Ca²⁺ and Mg²⁺ ions.

Hydrogen Class 11 Questions And Answers Question 11.
Zeolite used to soften hardness of water is, hydrated ………….
(a) Sodium aluminium silicate
(b) Calcium aluminium silicate
(c) Zinc aluminium borate
(d) Lithium aluminium hydride
Answer:
(a) Sodium aluminium silicate
Zeolite is sodium aluminium silicate.
(NaAlSi₂O₆ .H₂O)

11th Chemistry Deleted Book Back Questions Question 12.
A commercial sample of hydrogen peroxide marked as 100 volume H₂O₂, it means that ……………
(a) 1 ml of H₂O₂ will give 100 ml O₂ at STP
(b) 1 L of H₂O₂ will give 100 ml O₂ at STP
(c) 1 L of H₂O₂ will give 22.4 L O₂
(d) 1 ml of H₂O₂ will give 1 mole of O₂ at STP
Answer:
(a) 1 ml of H₂O₂ will give 100 ml O₂ at STP

Samacheer Kalvi Guru 11th Chemistry Question 13.
When hydrogen peroxide is shaken with an acidified solution of potassium dichromate in presence of ether, the ethereal layer turns blue due to the formation of
(a) Cr₂ O₃
(b) CrO₄^2-
(c) CrO(O₂)₂
(d) none of these
Answer:
(c) CrO(O₂)₂ CrO(O₂)₂
Hints:
Cr₂O₇^2- + 2H⁺ + 4H₂O₂ → 2CrO(O₂)₂ + 5H₂O

Samacheer Kalvi 11th Chemistry Question 14.
For de colorization of 1 mole of acidified KMnO₄, the moles of H₂O₂ required is …………….
(a) \(\frac {1}{2}\)
(b) \(\frac {3}{2}\)
(c) \(\frac {5}{2}\)
(d) \(\frac {7}{2}\)
Answer:
(c) \(\frac {5}{2}\)
Hints:
2MnO₄^– + 5H₂O_2(aq) + 6H⁺ → 2Mn²⁺+ 5O₂ + 8H₂O

11th Chemistry Samacheer Kalvi Question 15.
Volume strength of 1.5 N H₂O₂ is ……………..
(a) 1.5
(b) 4.5
(c) 16.8
(d) 8.4
Answer:
(d) 8.4
Hints:
Volume strength of hydrogen peroxide = Normality of hydrogen peroxide × 5.6 = 1.5 x 5.6 = 8.4

Volume strength of hydrogen peroxide
= Normality × \(\frac {17 × 22.4}{68}\)
Volume strength of hydrogen peroxide = Normality x 5.6

Samacheer Kalvi Chemistry 11th Question 16.
The hybridization of oxygen atom is H₂O and H₂O₂ are respectively
(a) sp and sp³
(b) sp and sp
(c) sp and sp²
(d) sp3 and sp³
Answer:
(d) sp³ and sp³

Samacheer Kalvi 11th Chemistry Solutions Question 17.
The reaction H₃PO₂ + D₂O → H₂DPO₂ + HDO indicates that hypo-phosphorus acid is ……………
(a) tri basic acid
(b) di basic acid
(c) mono basic acid
(d) none of these

Answer:
(c) mono basic acid
Hints:
Hypophosphorus acid on reaction with D₂O, only one hydrogen is replaced P by deuterium and hence it is mono basic.

Samacheer Kalvi Class 11 Chemistry Solutions Question 18.
In solid ice, oxygen atom is surrounded
(a) tetrahedrally by 4 hydrogen atoms
(b) octahedrally by 2 oxygen and 4 hydrogen atoms
(c) tetrahedrally by 2 hydrogen and 2 oxygen atoms
(d) octahedrally by 6 hydrogen atoms
Answer:
(a) tetrahedrally by 4 hydrogen atoms

Samacheer Kalvi 11th Chemistry Guide Pdf Question 19.
The type of H-bonding present in ortho nitro phenol and p-nitro phenol are respectively ……………
(a) inter molecular H-bonding and intra molecular H-bonding
(b) intra molecular H-bonding and inter molecular H-bonding
(c) intra molecular H – bonding and no H – bonding
(d) intra molecular H – bonding and intra molecular H – bonding

Answer:
(A) intra molecular H-bonding and inter molecular H-bonding

11 Chemistry Samacheer Kalvi Question 20.
Heavy water is used as ……………
(a) modulator in nuclear reactions
(b) coolant in nuclear reactions
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (A)
Hints:
Heavy water is used as moderator as well as coolant in nuclear reactions.

Samacheer Kalvi.Guru 11th Chemistry Question 21.
Water is a ……………
(a) basic oxide
(b) acidic oxide
(c) amphoteric oxide
(d) none of these
Answer:
(c) amphoteric oxide

II. Write brief answer to the following questions

Question 22.
Explain why hydrogen is not placed with the halogen in the periodic table.
Answer:

• Hydrogen resembles alkali metals as well as halogens.
• Hydrogen resembles more alkali metals than halogens.
• Electron affinity of hydrogen is much less than that of halogen atom. Hence the tendency to form hydride ion is low compared to that of halogens.
• In most of its compounds hydrogen exists in +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Question 23.
the cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why ?
Answer:

• In an ice cube, each atom is surrounded tetrahedrally by four water molecules through hydrogen bond and its density is low.
• Liquid water at 0°C has the density as 999.82 kg/cm³. Maximum density is attained by water only at 4°C as 1000 kg/cm³.
• When the temperature changed from 4°C to 0°C, the density of water decreases rather than increases. This is called anomalous expansion of water.
• The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in water.
• At 0°C, ice cube sinks in liquid water at 0°C because of the lesser density and greater volume of water.

Question 24.
Discuss the three types of Covalent hydrides.
Answer:

1. They are the compounds in which hydrogen is attached to another element by sharing of electrons.
2. The most common examples of covalent hydrides are methane, ammonia, water and hydrogen chloride.
3. Molecular hydrides of hydrogen are further classified into three categories as,
   • Electron precise (CH₄, C₂ H₆ , SiH₄ , GeH₄ )
   • Electron-deficient (B₂ H₆ ) and
   • Electron-rich hydrides (NH₃ , H₂O)
4. Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

Question 25.
Predict which of the following hydrides is a gas on a solid (a) HCl (b) NaH. Give your reason.
Answer:

• At room temperature, HCl is a colourless gas and the solution of HCl in water is called hydrochloric acid and it is in liquid state.
• Sodium hydride NaH is an ionic compound and it is made of sodium cations (Na⁺) and hydride (H^–) anions. It has the octahedral crystal structure. It is an alkali metal hydride.

Question 26.
Write the expected formulas for the hydrides of 4th period elements. What is the trend in the formulas? In what way the first two numbers of the series different from the others ?
Answer:
The expected formulas for the hydrides of 4th period elements MH₄ (electron precise). M₂H₆ (electron deficient) and MH₃ (electron rich).
The trend in formula is –

• Electron precise hydrides – CH₄ C₂H₆, SiH₄, GeH₄
• Electron deficient hydrides – B₂H₆
• Electron rich hydrides – NH₃, H₂O

The first two members of the series KH, CaH₂ are ionic hydrides whereas the other members of the series CH₄, C₂H₆, SiH₄, B₂H₆, NH₃ are covalent hydrides.

Question 27.
Write chemical equation for the following reactions.

1. reaction of hydrogen with tungsten (VI) oxide NO₃ on heating.
2. hydrogen gas and chlorine gas.

Answer:

1. 3H₂ + WO₂ → W + 3H₂O
   Hydrogen reduces tungsten (VI) oxide. WO₃ to tungsten at high temperature.
2. H₂ + Cl₂ → 2HCl (Hydrogen Chloride)
   Hydrogen reacts with chlorine at room temperature under light to give hydrogen chloride.

Question 28.
Complete the following chemical reactions and classify them in to (a) hydrolysis (b) redox (c) hydration reactions.

1. KMnO₄ + H₂O₂ →
2. CrCl3+ H₄O →
3. CaO + H₂O →

Answer:

1. 2KMnO₄ + 3H₂O₂ → 2MnO₂ + 2KOH + 3H₂O + 3O_2(g)
   This reaction is a redox reaction.
2. CrCl₃ + 6H₂O₂ → [Cr(H₂O)₆)] Cl₃
   This reaction is a hydration reaction.
3. CaO + H₂O → Ca(OH)₂
   This reaction is a hydrolysis reaction.

Question 29.
Hydrogen peroxide can function as an oxidizing agent as well as reducing agent. Substantiate this statement with suitable examples.
Answer:
Hydrogen peroxide can function as an oxidizing agent as well as reducing agent.

• H₂O₂ act as oxidizing agent in acidic medium. For example,
  
• H₂O₂ act as reducing agent in basic medium. For example,
  >

Question 30.
Do you think that heavy water can be used for drinking purposes ?
Answer:

• Heavy water (D₂O) contains a proton and a neutron. This makes deuterium about twice as heavy as protium, but it is not radioactive. So heavy water is not radioactive.
• If you drink heavy water, you don’t need to worry about radiation poisoning. But it is not completely safe to drink, because the biochemical reaction in our cells are affected by the difference in the mass of hydrogen atoms.
• If you drink an appreciable volume of heavy water, you might feel dizzy because of the density difference. It would change the density of fluid in your inner ear. So it is unlikely to drink heavy water.

Question 31.
What is water-gas shift reaction?
Answer:
The carbon monoxide of water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalysts. This reaction is called water-gas shift reaction.
CO + H₂O → CO₂ + H₂↑

Question 32.
Justify the position of hydrogen in the periodic table?
Answer:
Hydrogen resembles alkali metals in the following aspects.

1. Electronic configuration Is¹ as alkali metals have ns¹.
2. Hydrogen forms unipositive H+ ion like alkali metals Na⁺, K⁺.
3. Hydrogen form halides (HX), oxides (H₂O) peroxide (H₂O₂) like alkali metals (NaX. Na₂O, Na₂O₂).
4. Hydrogen also acts as reducing agent like alkali metals. Hydrogen resembles halogens in the following aspects.
5. Hydrogen has a tendency to gain one electron to form hydride ion (H^–) as halogens to form halide ion. (X^–).
6. Comparing the properties of hydrogen with alkali metals and with halogens, we can conclude that hydrogen resembles more alkali metals. In most of the compounds hydrogen exist in +1 oxidation state.
7. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Question 33.
What are isotopes? Write the names of isotopes of hydrogen.
Answer:

1. Isotopes are atoms of the same element that have the same atomic number but having different mass numbers (or) Isotopes are atoms with the same number of protons and electrons but differ in number of neutrons.
2. Hydrogen has three naturally occuring isotopes namely Protium (₁H¹), Deuterium (₁H²) and Tritium (₁H³).
   

Question 34.
Give the uses of heavy water.
Answer:

1. Heavy water is used as moderator in nuclear reactors as it can lower the energies of fast moving neutrons.
2. D₂O is commonly used as an tracer to study organic reaction mechanisms and mechanism of metabolic reactions.
3. It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 35.
Explain the exchange reactions of deuterium.
Answer:
Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium.

Question 36.
How do you convert para hydrogen into ortho hydrogen ?
Answer:
Para hydrogen can be converted into ortho hydrogen by the following ways:

• By treating with catalysts platinum or iron.
• By passing an electric discharge
• By heating > 800°C.
• By mixing with paramagnetic molecules such as O₂, NO, NO₂.
• By treating with nascent/atomic hydrogen.

Question 37.
Mention the uses of deuterium.
Answer:

• Deuterium is used as a tracer element.
• Deuterium is used to study the movement of ground water by isotopic effect.

Question 38.
Explain preparation of hydrogen using electrolysis.
Answer:
High purity of hydrogen (>99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. This process is not economical for large scale production.
At anode : 2OH^– → H₂O + ½ O₂ + 2e^–
At cathode : 2H₂O + 2e^– → 2OH^– + H₂
Overall reaction : H₂O → H₂ + 14 O₂

Question 39.
A groups metal (A) which is present in common salt reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state. (B) on reaction with a gas (C) to give universal solvent (D). The compound (D) on reacts with (A) to give (B), a strong base. Identify A, B, C, D and E. Explain the reactions.
Answer:
1.Group (1) metal (A) is present in common salt NaCl. So, (A) is sodium – Na.
2. Sodium reacts with hydrogen (B) to give sodium hydride – NaH (C) in which hydrogen is in -1 oxidation state.

3. Hydrogen on reaction with oxygen (O₂) gas which is (C) to give a universal solvent water (D).

4. Water (D) reacts with sodium metal (A) to give a strong base sodium hydroxide NaOH which is (E).

Question 40.
An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a modulator in nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C₃H₆ to give (D). Identify A, B, C and D.
Answer:
1. An isotope of hydrogen Deuterium (A) reacts with diatomic molecule of element belongs to group number 16 and period number 2 oxygen O₂ to give a compound (B) which is heavy water D₂O. D₂O is used as a moderator in nuclear reaction.

2. Deuterium reacts with C₃H₆ propane (C) to give Deutero propane C₂D₆ (D).

Question 41.
NH₃ has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15- Explain.
Answer:

1. NH₃ has exceptionally high melting point and boiling point due to hydrogen bonding between NH₃ molecules.
2. Each molecule can form a maximum of 4 hydrogen bonds but on average 1 hydrogen bond per molecule as there is only one lone pair on NH₃ available for hydrogen bonding.
3. Hydrogen bonding is strong intermolecular attraction as H on NH₃ acts like a proton due to partial positive on it whole N has the partial negative charge. Thus when the very polarized H comes close to a N atom in another NH₃ molecule, a very strong hydrogen bond is formed.
4. Due to much strong intermolecular interactions compared to weaker permanent dipole-dipole interactions between other XH₃ molecules in group 15, large amount of energy are required to overcome the forces, giving it the highest boiling point and highest melting point.

Question 42.
Why interstitial hydrides have a lower density than the parent metal.
Answer:

• d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
• Hydrogen atom being small in size occupy some in the metallic lattice producing distortion without any change in its type.
• The densities of these hydrides are lower than those of metals from which they are formed since the crystal lattice expands due to the inclusion of dihydrogen.

Question 43.
How do you expect the metallic hydrides to be useful for hydrogen storage?
Answer:
In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

Question 44.
Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.
Answer:

• Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is
  HF>H₂O>NH₃
• The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.
• Among N, F nd O the increasing order of their electronegativities are
  N<O H₂O>NH₃.

Question 45.
Compare the structures of H₂O and H₂O₂.
Answer:
In water, O is sp³ hybridized. Due to stronger lone pair-lone pair repulsions than bond pair-bond pair repulsions, the HOH bond angle decreases from 109.5° to 104.5°. Thus water molecule has a bent structure.

H₂O₂ has a non-planar structure. The 0 – H bonds are in different planes. Thus, the structure of H₂O₂ is like an open book.

Samacheer Kalvi 11th Chemistry Hydrogen Additional Questions Solved

I. Choose the correct answer:

Question 1.
Which one of the following element mostly present in the sun and the stars?
(a) Hydrogen
(b) Lithium
(c) Helium
(d) Beryllium
Answer:
(a) Hydrogen

Question 2.
is the most abundant 90% of all atoms…………
(a) Lithium
(b) Hydrogen
(c) Oxygen
(d) Silicon
Answer:
(A) Hydrogen

Question 3.
At room temperature normal hydrogen consists of …………….
(a) 25% ortho form + 75% para form
(b) 50% ortho form + 50% para form
(c) 75% ortho form + 25% para form
(d) 60% ortho form + 40% para form
Answer:
(c) 75% ortho form + 25% para form

Question 4.
Which one of the metal is used to convert para hydrogen into ortho hydrogen?
(a) Copper
(b) Aluminium
(c) Sodium
(d) Platinum
Answer:
(d) Platinum

Question 5.
Consider the following statements…………….
(i) In ortho form of hydrogen molecule, the nuclear spins are opposed to each other
(ii) The magnetic moment of para hydrogen is twice that of ortho hydrogen
(iii) By passing an electric discharge, para hydrogen can be converted into ortho hydrogen. Which of the above statement is/are not correct?
(a) (i) only
(b) (iii) only
(c) (i) and (ii)
(d) (ii) and (iii)
Answer:
(c) (i) and (ii)

Question 6.
Which of the following is not used in the conversion of para hydrogen into ortho hydrogen?
(a) by heating more than 800°C
(b) by passing an electric discharge
(c) by mixing with atomic hydrogen
(d) by mixing with diamagnetic molecules
Answer:
(d) by mixing with diamagnetic molecules

Question 7.
The magnetic moment of para hydrogen is ……………
(a) one
(b) zero
(c) twice
(d) maximum
Answer:
(b) zero

Question 8.
Which one of the following does not contain neutron?
(a) ordinary hydrogen
(b) Heavy hydrogen
(c) Radioactive hydrogen
(d) Deuterium
Answer:
(a) ordinary hydrogen

Question 9.
The half life period of tritium is ……………..
(a) 123.3 year
(b) 12.33 years
(c) 1 year
(d) 1600 years
Answer:
(a) 12.33 years

Question 10.
Which of the following is used in illumination of wrist watches?
(a) Phosphorous
(b) Radon
(c) Tritium
(d) Deuterium
Answer:
(c) Tritium

Question 11.
Which one of the following is used to study the movements of ground water?
(a) Deuterium
(b) Protium
(c) Tritium
(d) HD
Answer:
(a) Deuterium

Question 12.
By which rays nuclear reactions are induced in upper atmosphere to produce tritium?
(a) α-rays
(b) β-rays
(c) γ-rays
(d) cosmic rays
Answer:
(d) cosmic rays

Question 13.
Which of the following is produced by bombardment of neutrons with lithium?
(a) Deuterium
(b) Protium
(c) Tritium
(d) Beryllium
Answer:
(c) Tritium

Question 14.
Consider the following statements ……………..
(i) Tritium is a beta emitting radioactive isotope of hydrogen.
(ii) Deuterium is known as heavy hydrogen.
(iii) Deuterium is used in emergency exit signs.
Which of the following statement is/are not correct?
(a) (i) only
(b) (iii) only
(c) (i) and (ii)
(d) (i) (ii) and (iii)
Answer:
(b) (iii) only

Question 15.
Which of the following mixture of gases is called water gas?
(a) CO_2(g)  + H_2(g)
(b) CO_2(g) + N_2(g)
(c) CO_(g) + H_2(g)
(d) N_2(g) + H_2(g)
Answer:
(c) CO_(g) + H_2(g)

Question 16.
Consider the following statements.
(i)Hydrogen is a colourless, odourless, tasteless heavy and highly inflammable gas.
(ii) Hydrogen is a good reducing agent.
(ii) Hydrogen can be liquefied under low pressure and high temperature.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only (c)
(c) (i) and (iii)
(d) (ii) and (Hi)
Answer:
(c) (i) and (iii)

Question 17.
The products formed during the cracking long chain hydrocarbon C₆H₁₂ are
(a) CO₂ + H₂O
(b) CO + H₂
(c) C₆H₁₀ + H₂
(d) C₆H₆ + 3H₂
Answer:
(d) C₆H₆ + 3H₂

Question 18.
Which one of the following is manufactured in Haber’s process?
(a) SO₃
(b) NH₃
(C) N₂
(d) H₂
Answer:
(b) NH₃

Question 19.
Hydrogen combines with carbon monoxide in the presence of copper catalyst will synthesise
(a) Ethanol
(b) Methane
(c) Methanol
(d) Methanal
Answer:
(c) Methanol

Question 20.
Match the List-I and List-II using the correct code given below the list.
List-I
A. Hydrogenation of unsaturated vegetable oils
B. Calcium hydride
C. Liquid hydrogen
D. Atomic hydrogen

List-II
1. Rocket fuel
2. Welding of metals
3. Desiccant
4. Margarine

Answer:

Question 21.
Statement-I: Hydrogen is placed at the top of the group which is in-line with the latest periodic table.
Statement-II: Hydrogen has a tendency to lose its electron to form H⁺, thus showing electropositive character like alkali metals. On the other hand, hydrogen has a tendency to gain an electron to yield H , thus showing electronegative character like halogens.
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.

Question 22.
Statement-I: The magnetic moment of para-hydrogen is zero.
Statement-II: The spins of two hydrogen atoms in para H₂ molecule neutralise each other.
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.

Question 23.
Which of the following process is important in food industry?
(a) Dehydration
(b) Dehalogenation
(c) Hydrogenation
(d) Carboxylation
Answer:
(c) Hydrogenation

Question 24.
Which of the following is used as desiccants to remove moisture from organic solvents?
(a) Calcium hydride
(b) LiAlH₄
(c) Sodium boro hydride
(d) Sodium hydride
Answer:
(a) Calcium hydride

Question 25.
Which of the following is used for cutting and welding?
(a) Atomic hydrogen and oxy hydrogen torches
(b) Liquid hydrogen
(c) Calcium hydride
(d) Sodium boro hydride
Answer:
(a) Atomic hydrogen and oxy hydrogen torches

Question 26.
Liquid hydrogen is used as
(a) a rocket fuel as well as in space research
(b) fuel cells for generating electrical energy
(c) cutting and welding torch
(d) desiccant to remove moisture from organic solvent
Answer:
(a) a rocket fuel as well as in space research

Question 27.
Which one of the following is a universal solvent?
(a) Alcohol
(b) Ether
(c) CCl₄
(d) Water
Answer:
(d) Water

Question 28.
At the temperature conditions of the earth (300K) the OPR of H₂O is ……………
(a) 2.5
(b) 3
(c) 1
(d) zero
Answer:
(b) 3

Question 29.
Water does not react with
(a) Sodium
(b) Magnesium
(c) Beryllium
(d) Calcium
Answer:
(c) Beryllium

Question 30.
Which of the following does not have any effect with water?
(a) Sodium
(b) Iron
(c) Lead
(d) Mercury
Answer:
(d) Mercury

Question 31.
Which set of the metals do not have any effect on water?
(a) Ag, Au, Pt
(b) Na, Mg, Al
(c) Fe, Ca, Zn
(d) Fe, Pb, Na
Answer:
(a) Ag, Au, Pt

Question 32.
Which of the following non-metal reacts with ordinary water?
(a) Carbon
(b) Sulphur
(c) Chlorine
(d) Phosphorous
Answer:
(c) Chlorine

Question 33.
Consider the following statements.
(i) Silver, Gold, Mercury and Platinum do not have any effect on water.
(ii) Carbon, Sulphur and Phosphorous do not react with water.
(iii) Beryllium reacts with water less violently.
Which of the following statements is/are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(c) (iii) only

Question 34.
Which one of the following is used as a bleach?
(a) Cl₂ water
(b) Br, water
(c) Water gas
(d) Liquid hydrogen
Answer:
(a) Cl₂ water

Question 35.
Water is an/a oxide.
(a) neutral
(b) acidic
(c) basic
(d) amphoteric
Answer:
(d) amphoteric

Question 36.
Permanent hardness of water is removed by
(a) boiling
(b) lime
(c) washing soda
(d) chlorine
Answer:
(c) washing soda

Question 37.
Which one of the following is used as water softener?
(a) Zeolites
(b) lime
(c) washing soda
(d) bleaching powder
Answer:
(a) Zeolites

Question 38.
In chelating method of softening of hard water is used.
(a) magnesia
(b) lime
(c) EDTA
(d) washing soda
Answer:
(c) EDTA

Question 39.
Which of the following is used to remove toxic and heavy metals from water?
(a) zeolites
(b) magnesia
(c) Bleaching agent
(d) lime
Answer:
(a) zeolites

Question 40.
Match the List-I and List-Il using the code given below the list.
List-I
A. Heavy water
B. Hydrogen peroxide
C. Heavy hydrogen
D. Lithium Aluminium hydride

List-Il
1. Antiseptic
2. Moderator
3. Reducing agent
4. Tracer

Answer:

Question 41.
Consider the following statements.
(i) H₂O₂ is a powerful oxidising agent.
(ii) H₂O₂ is stored in dark coloured bottles
(iii) H₂O₂ is used as moderator in nuclear reactors.
Which of the above statements is/are not correct.
(a) (i) only
(b) (i) and (ii)
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(c) (iii) only

Question 42.
Statement-I: Heavy water has been widely used as moderator in nuclear reactors.
Statement-Il: Heavy water can lower the energies of fast moving neutrons.
(a) Statements-I and II are correct and Statement-Il is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-IT is not the correct explanation of statement-I;
(c) Statement-I is correct but Statement-Il is wrong.
(d) Statement-I is wrong but Statement-Il is correct.
Answer:
(a) Statements-I and II are correct and Statement-Ills the correct explanation of statement-I.

Question 43.
Which one of the following is prepared in industry by the auto oxidation of 2-alkylanthraquional?
(a) Heavy water
(b) Deuterium
(c) Hydrogen peroxide
(d) Tritium
Answer:
(c) Hydrogen peroxide

Question 44.
Which one of the following is an ionic or saline hydride?
(a) SiH₄
(b) GeH₄
(c) B₂H₆
(cl) LiH
Answer:
(d) LiH

Question 45.
Which one of the following is an electron deficient hydride?
(a) C₂H₆
(b) B₂H₆
(c) GeH₄
(d) CH₄
Answer:
(b) B₂H₆

Question 46.
Which of the following pair is an electron rich hydride?
(a) NH₃, H₂O
(b) CH₄, C₂H₆
(c) B₂H₆, GeH₄
(d) CH₄, SiH₄
Answer:
(a) NH₃, H₂O

Question 47.
Which one of the following is not a covalent hydride?
(a) Ammonia
(b) Methane
(c) Lithium hydride
(d) water
Answer:
(e) Lithium hydride

Question 48.
Metallic hydrides are otherwise called …………….
(a) Salt hydrides
(b) Saline hydrides
(c) molecular hydrides
(d) Interstitial hydrides
Answer:
(d) Interstitial hydrides

Question 49.
Which one of the following is used for hydrogen storage applications?
(a) Saline hydrides
(b) Interstitial hydrides
(c) covalent hydrides
(d) molecular hydrides
Answer:
(b) Interstitial hydrides

Question 50.
Which one of the following is known as Hydrogen sponge?
(a) Lithium hydride
(b) Diborane
(c) Palladium hydride
(d) Ammonia
Answer:
(c) Palladium hydride

Question 51.
Which of the following is the correct order of stability of bonds?
(a) Hydrogen bond < CovaLent bond < Vanderwaals bond
(b) Vanderwaals bond < Hydrogen bond <Covalent, bond
(c) Vanderwaals bond > Hydrogen bond > Covalent bond
(d) Covalent bond < Hydrogen bond <Vanderwaals bond
Answer:
(b) Vanderwaals bond < Hydrogen bond <Covalent, bond

Question 52.
Which one of the following does not have intramolecular hydrogen bonding?
(a) water
(b) o-nitrophenol
(c) Salicylaldehyde
(d) Salicylic acid
Answer:
(a) water

Question 53.
Which of the following contains intramolecular hydrogen bonding?
(a) Acetic acid
(b) o-nitrophenol
(c) Hydrogen fluoride
(d) water
Answer:
(b) o-nitrophenol

Question 54.
Consider the following statements.
(i) In ice, each oxygen atom is surrounded by hydrogen atoms tetrahedrally to four water molecules.
(ii) Acetic acid exists as dimer due to intra molecular hydrogen bonding.
(iii) Strong hydrogen bonds lead to an increase in the melting and boiling points.
Which of the above statements is/are not correct? ,
(a) (ii) only
(b) (i) and (iii)
(c) (i) (ii) and (iii)
(d) (i) only
Answer:
(a) (ii) only

Question 55.
Which one of the following is an example for Clatharate hydrate?
(a) CuSO₄.5H₂O
(b) Na₂CO₃. 10H₂O
(c) CH₄. 20 H₂O
(d) FeSO₃.7H₂O
Answer:
(c) CH₄. 20 H₂O

Question 56.
Which one of the following is not a crystalline hydrate?
(a) CH₄. 20H₂O
(b) Na₂,CO₃. 10H₂O
(c) CuSO₄.5H₂O
(d) FeSO₄.7H₂0
Answer:
(a) CH₄. 20H₂O

Question 57.
Statement-I: Hydrogen can be used as a clean burning fuel.
Statement-II: Hydrogen on combustion give only water as end product and it is free from pollutants.
(a) Statements-I and li are correct and Statement-lI is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-Il is not the correct explanation of statement-I.
(e) Statement-I is correct but Statement-Il is wrong.
(d) Statement-I is wrong but Statement-lI is correct.
Answer:
(a) Statements-I and II are correct and Statement-Il is the correct explanation of statement-I.

Question 58.
Which isotope of hydrogen is radioactive?
(a) Protium
(b) Deuterium
(c) Tritium
(d) H
Answer:
(c) Tritium

Question 59.
Which type of elements form interstitial hydrides ………..
(a) s-block and p-block
(b) p-block only
(c) d-block and f-block
(d) s-block only
Answer:
(c) d-block and f-block

Question 60.
Which of the following is named as perhydrol and used as an antiseptic?
(a) D₂O
(b) H₂O₂
(c) NaH
(d) B₂H₆
Answer:
(b) H₂O₂

Question 61.
Which of the following causes temporary hardness of water?
(a) MgCl₂
(b) Na₂SO₄
(c) Mg(HCO₃)₂
(d) NaCl
Answer:
(c) Mg(HCO₃)₂

Question 62.
Which of the following can oxidise Hydrogen peroxide?
(a) acidified KMnO₄
(b) Cu
(c) dil. HNO₄
(d) CrO₂Cl₂
Answer:
(a) acidified KMnO₄

Question 63.
Which type of hydrides are generally non-stoichiometric in nature?
(a) Metallic hydride
(b) Covalent hydrides
(c) Ionic hydride
(d) Saline hydride
Answer:
(a) Metallic hydride

Question 64.
Hydrogen gas is generally prepared by the ………….
(a) reaction of granulated zinc with dilute H₂SO₄
(b) reaction of zinc with cone, H₂SO₄
(c) reaction of pure zinc with dil. H₂SO₄
(d) action of stream on red hot coke
Answer:
(a) reaction of granulated zinc with dilute H₂SO₄

Question 65.
The higher density of water than that of ice is due to ……………
(a) dipole-dipole interaction
(b) dipole-induced dipole interaction
(c) hydrogen bonding
(d) all of these
Answer:
(c) hydrogen bonding

Question 66.
Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect?
(a) Its tendency to lose an electron to form a cation.
(b) Its tendency to gain an electron to attain stable electronic configuration.
(c) Its low negative electron gain enthalpy value.
(d) Its small size.
Answer:
(b) Its tendency to gain an electron to attain stable electronic configuration.

Question 67.
Metal hydrides are ionic, covalent or molecular in nature. Among L1H, NaH, KH, RbH, CsH, the correct order of increasing ionic character is ………….
(a) LiH>NaH>CsH>KH>RbH
(b) LiH<NaH<KH<RbHCsH>NaH>Kil>LIH
(d) NaH>CsH>RbH>LiH>KH
Answer:
(b) LiH<NaH<KH<RbH<CsH

Question 68.
Statement-I: Permanent hardness of water is removed by treatment with washing soda ………..
Statement-II: Washing soda reacts with soluble magnesium and calcium sulphate to form insoluble carbonate.
(a) Statements-I and II are correct and Statement-Il is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-Il is not the correct explanation of statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statements-I and II are correct and Statement-Ills the correct explanation of statement-I.

Question 69.
Which of the following is a saline hydride?
(a) HCl
(b) NH₃
(c) NaH
(d) PbH
Answer:
(e) NaH

Question 70.
Which metal does not liberate H₂ gas from dilute aqueous hydrochloric acid at 298 K?
(a) Mg
(b) Zn
(c) Al
(d) Cu
Answer:
(d) Cu

Question 71.
When zeolite is treated with hard water, the sodium ions are exchanged with …………
(a) H⁺ ions and Cl^– ions
(b) Ca²⁺ ions
(c) Cl^– ions
(d) Ca²⁺ ions and Mg²⁺ ions
Answer:
(d) Ca²⁺ ions and Mg²⁺ ions

Question 72.
The most abundant element in the universe is …………..
(a) Carbon
(b) Nitrogen
(c) Silicon
(d) Hydrogen
Answer:
(d) Hydrogen

Question 73.
Which of the following can effectively remove all types of hardness of water?
(a) Soap
(b) Slaked lime
(c) Washing soda
(d) Zeolite
Answer:
(a) Soap

Question 74.
A commercial sample of H₂O₂ is labelled as 100 volume. Its percentage strength is nearly
(a) 10%
(b) 30%
(c) 100%
(d) 90%
Answer:
(b) 30%

Question 75.
Which of the following will not produce di hydrogen gas?
(a) Cu + dil (HCl)
(b) CH_2(g) + H₂O_(g)
(c) Zn + dil. HCl
(cl) C_(s) + H₂O_(g)
Answer:
(a) Cu + dil (HCl)

Samacheer Kalvi 11th Chemistry Hydrogen 2-Mark Questions
Question 1.
Draw and define ortho and para hydrogen molecule.
Answer:
Molecular hydrogen have oriho and para form in which the nuclear spins are aligned or opposed, respectively.

Question 2.
What is the nuclear reaction that take place in the sun and other stars?
Answer:
The sun and other stars are composed mainly of 85 – 95% hydrogen which generates their energy by nuclear fusion of hydrogen nuclei into helium.

Question 3.
Mention the uses of tritium.
Answer:

• Tritium has replaced radium in application such as emergency exit sign.
• Tritium is used in illumination of wrist watches.

Question 4.
Draw the structures of three isotopes of hydrogen, Hydrogen Deuterium
Answer:

Question 5.
What is the half life period of tritium? How is it undergoes radioactive disintegration?
Answer:

• The half life of tritium = 12.33 years.
• Tritium is a beta-emitting radioactive isotope of hydrogen.
  

Question 6.
Why hydrogen gas is used as fuel?
Answer:
Hydrogen burns in air, virtually free from pollution and produces significant amount of energy. This reaction is used in fuel cells to generate electricity.
2H_2(g) + O_2(g) → 2H₂O_(l) + energy

Question 7.
How ammonia is manufactured from Hydrogen? Give the uses of ammonia.
Answer:

• Ammonia is manufactured by Habe?s prôcess in which the largest consumer of hydrogen is used.
  N_2(g) + 3H_2(g) ⇌ 2 NH_3(g)
• Ammonia is employed for the manufacture of nitric acid, fertilizers and explosives.

Question 8.
how ¡s methanol synthesized from hydrogen? Give the uses of methanol.
Answer:

• Huge quantities of hydrogen are used for the synthesis of methanol from carbon monoxide in presence of copper catalyst.
  CO_(g) + 2H_2(g) → CH₃OH_(l)
•  Methanol is an industrial solvent and a starting material for the manufacture of formaldehyde used in makihg plastics.

Question 9.
What is hydrogenation? Give one example.
Answer:
Hydrogenation is a reaction in which addition of hydrogen to an alkene /alkyne. Compounds containing multiple bonds.

Question 10.
How is hydrogen used in metallurgy? Prove it with an example.
Answer:
In metallurgy, for the extraction of pure metals from their mineral sources, hydrogen is used to reduce many metal oxides to metals at high temperatures.
CuO_(s) +H_2(g) → Cu_(s) +H₍₂₎O_(g)

Question 11.
How alkali metals react with water? Give an equation?
Answer:
The most reactive alkali metals decompose water in the cold with the evolution of hydrogen and leaving an alkali solution.
2Na_(s) +2H₂O_(l) → 2NaOH_(aq) +H_2(g)

Question 12.
What happens when steam is passed over red hot iron?
Answer:
When steam is passed over red hot iron, iron oxide will be formed with the release of hydrogen.
3Fe_(s) + 4H₂O_(l) → Fe₃O + 4H_4(g)

Question 13.
Explain the action of chlorine with water.
Answer:
Chlorine reacts with the water to form hydrochloric acid and hypochiorous acid.
Cl_2(g) + H₂O_(l) → HCl_(aq) + HOCl_(aq)

Question 14.
What is temporary hardness of water? How is it removed?
Answer:
Temporary hardness of water is due to the presence of soluble bicarbonates of magnesium and calcium. By heating / boiling, these salts decomposed into insoluble carbonate and hydroxide, respectively. The resulting precipitates can be removed by filtration.
Ca (HCO₃)_2(aq) → CaCO_3(s) + H₂O_(l) + CO_2(l)
Mg(HCO₃)_2(aq) → Mg(OH)_2(s) + CO_2(g)

Question 15.
What ¡s permanent hardness of water? How ¡t will be removed?
Answer:
Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and suiphates in water. It can be removed by adding washing soda which reacts with these metal chlorides and suiphates in hard water to form insoluble carbonates.
MCl_2(aq) + Na₂CO_3(aq) → MCO_3(s) + 2NaCl_(aq)
MSO_4(aq) + Ni₂ CO_3(aq) → MCO_3(s) + Na₂SO_4(s)
Where M = Ca or Mg.

Question 16.
How would you prepare Hydrogen peroxide?
Answer:
Hydrogen peroxide can be prepared by adding a metal peroxide to dilute acid.
BaO_2(s)+ H₂SO_4(aq) → BaSO_4(s) + H₃O_2(aq)

Question 17.
H₂O₂ is always stored in plastic bottles. Why?
Answer:
The aqueous solution of hydrogen peroxide is spontaneously disproportionate to give oxygen. The reaction is slow but it is explosive when it is catalyzed by metal or alkali dissolved from glass. For this reason, its solution are stored in plastic bottles.
H₂O_2(aq) → H₂O_(l) + ½ O_2(g)

Question 18.
Why H₂O₂ ¡s used as mild antiseptic?
Answer:
The oxidising property of hydrogen peroxide and harmless nature of its products such as water and oxygen, leads to oxidation of pollutants in water and act as a mild antiseptic.

Question 19.
Why the bond angle in solid phase of H₂O₂ is reduced when compared to gas phase of H₂O₂?
Answer:
Both the gas phase and solid phase, H₂O₂ adopt a skew configuration due to the repulsive interaction of the OH bonds with lone pairs of electrons on each oxygen atoms. Indeed, it is the smallest molecule known to show hindered rotation about a single bond. In the solid phase, the dihedral angle is sensitive and hydrogen bonding decreases from 111.50 in the gas phase to 90.2 in the solid phase.

Question 20.
What is meant by 100 – volume hydrogen peroxide?
Answer:
A 30% solution is marketed as 100 – volume hydrogen peroxide indicating that at STP, 100 volumes of oxygen are liberated per millimeter of the solution.

Question 21.
Prove that Hydrogen peroxide is a vigorous oxidizing agent and the solution of H₂O₂ is slightly acidic.
Answer:
H₂ O_2(aq) + 2 H₂O_(l) ⇌ H₃ O⁺_(aq) + HO⁺_2(aq)

H₃O⁺ Hydronium ion formation proves that solution of H₂O₂ is acidic. Because it donates H⁺ to H₂O to form H₃O⁺ ion.
H₂O₂ oxidizes Ferrous sulphate to Ferric sulphate in acidic medium.
2FeSO_4(aq) + H₂SO_4(aq) + H₄O_2(aq) ⇌ Fe₂(SO₄)_3(aq)( + 2H₂O_(l)

Question 22.
What is meant by binary hydride? Give example.
Answer:
A binary hydride is a compound formed by hydrogen with other electropositive elements including metals and non-metals, e.g., LiH or MgH₂.

Question 23.
What are ternary hydrides? Give example?
Answer:
Ternary hydrides are compounds in which the molecule is constituted by hydrogen and two types of elements, e.g., LiBH₄ or LiAlH₄.

Question 24.
What arc the different types of hydrides?
Answer:
The hydrides are classified as Ionic, Covalent and Metallic Hydrides.
Ionic hydride – LiH
Covalent hydride – CH₄
Metallic hydride -TiH

Question 25.
Why metallic hydrides are called interstitial hydrides? Give one example.
Answer:
Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides. e.g., PdH.

Question 26.
What is hydrogen bonding?
Answer:
When a hydrogen atom (H) is covalently bonded to a highly electronegative atom such as fluorine (F) or oxygen (O) or nitrogen (N), the bond is polarized in such a way that the hydrogen atom is able to form a weak bond between the hydrogen atom of a molecule and the electronegative atom of a second molecule. The bond thus formed is a hydrogen bond and it is denoted by dotted lines (……)

Question 27.
What are the types of hydrogen bonding? Give example.
Answer:
There are two types of hydrogen bonding.

• Intramolecular hydrogen bonding: It can occur with a molecule. e.g., o – nitrophenol.
• Intermolecular hydrogen bonding: It is formed between two molécules of same type or different type. e.g., H₂O.

Question 28.
Explain about the type of bonding present in hydrogen fluoride?
Answer:
In hydrogen fluoride (HF), for example, one molecule is strongly attracted to the fluorine on its neighboring hydrogen. In both liquid and solid, hydrogen fluoride forms long hydrogen bonded zig-zag chains as a consequence of the orientation of the lone pairs on the fluorine atoms.

Question 29.
Ice is less dense than water at 0°C. Justify this statement.
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three – dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 30.
Draw the structure of –

1. Acetic acid
2. Water.

Answer:
1. Acetic acid:

2. Water:

Question 31.
Write a note about gas hydrates.
Answer:
Gas hydrates are a kind of inclusion compounds, where gas molecules arc trapped in the crystal lattice having voids of right size, without being chemically bonded. An interesting hydrate is that of the hydronium ion (H₂O) in the gas-phase, similar to methane hydrate. Each water molecule is bonded to three others in the dodecahedron.

Question 32.
Give the advantages of future fuel – Hydrogen.
Answer:
Hydrogen is considered as a potential candidate for this purpose as it is a clean burning fuel. Hence, hydrogen can directly be used as a fuel and can replace existing gasoline (petrol) diesellkerosene powered engines, and indirectly be used with oxygen in fuel cells to generate electricity. One major advantage of using hydrogen is that the combustion product is essentially free from pollutants; the end product formed in both cases is water.

Question 33.
What do you understand by the term non-stoichiometric hydrides’? Do you expect this type of hydrides to be formed by alkali metals? Justify your answer.
Answer:
Those hydrides which do not have fixed composition are called non-stoichiometric hydrides, and the composition varies with temperature and pressure. This type of hydrides are formed by d and f-block elements. They cannot be formed by alkali metals because alkali thetal hydrides form ionic hydrides.

Question 34.
How does the atomic hydrogen or oxy – hydrogen torch function for cutting and welding purposes? Explain.
Answer:
When hydrogen is burnt in oxygen the reaction is highly exothermic, it produces very high temperature nearly 4000°C which is used for cutting and welding purposes.

Question 35.
How does H₂O₂ behave as bleaching agent?
Answer:
Bleaching action of H₂O₂ is due to the oxidation of colouring matter by nascent oxygen.
H₂ O₂ → H₂O_(l) + O_(g)

Question 36.
Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?
Answer:
Cone. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂
Zn + 2H₂SO₄ (Conc.) → ZnSO₄ + 2H₂O + SO₂
Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

Question 37.
Write the chemical reactions to show the amphoteric nature of water.
Answer:
Water is amphotenc in nature and it behaves both as an acid as well as base. With acids stronger than itself (e.g., H₂S) it behaves as a base and with bases stronger than itself (e.g., NH₃) it acts as an acid.

1. As a base : H₂O_(l) + H₂S_(aq) → H₃ O_(aq) + HS^–_(aq)
2. As an acid : H₂O_(l) + NH_3(aq) → OH^–_(aq) + NH⁺_4(aq)

Question 38.
Why is hydrogen peroxide stored ¡n wax-lined plastic coloured bottles?
Answer:
The decomposition of H₂O₂ occurs readily in the presence of rough surface (acting as catalyst). It is also decomposed by exposure of light. Therefore, wax-lined smooth surface and coloured bottles retard the decomposition of H₂O₂.

Samacheer Kalvi 11th Chemistry Hydrogen 3 – Mark Questions

Question 1.
Compare the properties of ortho and para hydrogen.
Answer:

Question 2.
Compare the properties of isotopes of hydrogen.
Answer:

Question 3.
Draw the structure of the isotopes of hydrogen and distinguish them.
Answer:

Question 4.
Explain the different methods of preparation of Tritium with equation.
Answer:
It occurs naturally as a result of nuclear reactions induced by cosmic rays in the upper atmosphere.

Question 5.
How would you prepare hydrogen in the laboratory?
Answer:
Small amounts of hydrogen are conveniently prepared in laboratory by the reaction of metals, such as zinc, iron and tin, with dilute acid.
Zn_(s)  + 2HCl_(aq) → ZnCl_2(s) + H_2(g)
In principle, any metal with a negative standard reduction potential will react with an acid to generate hydrogen.

Question 6.
What happens when hydrogen reacts with –

1. O₂
2. Cl₂
3. Na ?

Answer:

1. 2 H_2(g) + O_2(g) → 2 H₂O_(l) – Water
2. H_2(g) + Cl_2(g) → 2 HCl_(g) – Hydrogen Chloride
3. 2 Na_(s) + H_2(g) → 2 NaH_(s) – Sodium hydride

Question 7.
Write a note about ortho water and para water.
Answer:

1. Water exists in space in the interstellar clouds, ¡n proto-planetary disks, in the comets and icy satellites of the solar system, and on the Earth.
2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho-H₂O and para-H₂O, in which the directions are antiparallel.
   
3. At the temperature conditions of the Earth (300 K), the OPR of H₂O is 3.
4. At low temperatures below (< 50 K) the amount of para-H₂O increases. It is known that the OPR of water in interstellar clouds and comets has more para-H₂O (OPR = 2.5) than on Earth.

Question 8.
Water ¡s an amphoteric oxide. Justify this statement.
Answer:

1. Water is an amphoteric oxide. It has the ability to act an acid as well as a base. That is, water shows this behavior when it reacts with hydrogen chloride and ammonia.
2. When water reacts with ammonia, it behaves as an acid.
   
3. When water reacts with hydrogen chloride, behaves as a base.
   
   So water is an amphoteric oxide.

Question 9.
Distinguish between Hard water and Soft water.
Answer:
Hard water:

• Presence of magnesium and calcium in the form of bicarbonate, chloride and sulphate in water makes hard water.
• Cleaning capacity of soap is rcduced when used in hard water.
• When hard water is boiled deposits of insoluble carbonates of magnesium and calcium are obtained.

Soft water:

• Presence of soluble salts of calcium and magnesium in water makes it soft water.
• Cleaning capacity of soap is more when used in soft water.
• When soft water is boiled, there is no deposition of salts.

Question 10.
Explain the action of soap with hard water.
Answer:

• The cleaning capacity of soap is reduced when used in hard water.
• Soaps are sodium or potassium salts of long chain fatty acids.
• When soap is added to hard water, the divalent magnesium and calcium cations present in hard water react with soap.
• The sodium salts present in soaps are converted to their corresponding magnesium and calcium salts which are precipitated as scum or precipitate.
  M²⁺ + 2RCOONa (RCOO)₂M_(s) + 2Na⁺_(aq);
  Where, M Ca or Mg;
  R = C₁₇H₃₅.

Question 11.
Describe about ion exchange method of softening water (or) Explain Zeolite (or) Permutit process.
Answer:

1. Hardness can be removed by passing through an ion-exchange bed like zeohtes or resin containing column. Thus, the zeolites work as water softener.
2. Zeolites are hydrated sodium alumini-silicates with a general formula, NaO.Al₂O₃.xSiO₂.yH₂O (x = 2 – 10, y = 2 – 6). They have high ion exchange capacity.
3. The complex structure can represented as Na₂ – Z with sodium as exchangeable cations. This method is called zeolite or permutit process.
4. Zeolites have porous structure in which the monovalent sodium ions are loosely held and can be exchanged with hardness producing divalent metal ions (Ca or Mg) in water.
   Na₂ – Z_(s) + M²⁺_(aq) → M-Z_(s) + 2Na⁺_(aq)
5. When exhausted, the materials can be regenerated by treating with aqueous sodium chloride. Hard minerals caught in the zeolite are released and they get replenished with sodium ions.
   M-Z_(s) + 2NaCl_(aq) → Na_(s)-Z_(s) + MCl_2(aq)

Question 12.
Explain about the exchange reactions of deuterium oxide.
Answer:
When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for Deuterium.

Question 13.
Complete the following reactions.
A1₄C₃ + D₂O → ?
CaC₂ + D₂O →?
Mg₃N₂ + D₂O →?
Ca₃P₂ + D₂O →?
Answer:

Question 14.
What are the uses of hydrogen peroxide?
Answer:

• H₂O₂ is used in water treatment to oxidize pollutants.
• H₂O₂ is used as a mild antiseptic.
• H₂O₂ is used as a bleach in textile, paper and hair-care industry.

Question 15.
Prove that H₂O₂ act as reducing agent ¡n alkaline medium.
Answer:
In alkaline conditions, H₂O₂ act as a reducing agent.
2KMnO_4(aq)(Potassium permanganate) + 3 H₂SO_4(aq) + 5H₂O_2(aq) → K₂ SO₄ + 2MnSO₄ + 8H₂ O_(l) + 5O_2(g)

Question 16.
Write a note about saline (or) ionic hydride.
Answer:
Ionic hydrides composed of an electropositive metal, generally, an alkali or alkaline-earth metal (except beryllium and magnesium) formed by transferring of electrons from metal to hydrogen atoms. They can be prepared by the reaction of elements at about 400°C. These are salt-like, high-melting, white, crystalline solids having hydride ions (H^–) and metal cations (Mⁿ⁺).
2Li_(s) + H_2(g) → 2LiH_(s) (Lithiuinhydride)
2 Ca_(s) + H_2(g) → 2 CaH_2(s) (Calcium hydride)

Question 17.
What are metallic hydrides? Explain about it.
Answer:

• Metallic hydrides are obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides.
• The hydrides show properties similar to parent metals and hence they are also known as metallic hydrides.
• They arc mostly non-stoichiometric with variable composition (TiH_1.5-1.818 and PdH_0.6-0.8).
• Some are relatively light, inexpensive and thermally unstable which makes them useful for hydrogen storage applications. Example, TiH₂, ZrH₂, ZnH₂.

Question 18.
What are intra molecular hydrogen bonding? Explain with an example.
Answer:

1. Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs between two functional groups within a molecule.
   
2. An intramolecular hydrogen bond (dashed lines) joins the OH group to the doubly bonded oxygen atom of the carboxyl group on the same molecule. e.g., Salicylic acid.
   
3. Salicylic acid act as an analgesic and antipyretic.

Question 19.
What are intermolecular hydrogen bonds? Explain with example.
Answer:

1. Intermolecular hydrogen bonds occur between two separate molecules. They can occur between any numbers of like or unlike molecules as long as hydrogen donors and acceptors are present in positions in which they can interact.
2. For e.g., Intermolecular hydrogen bonds can occur between ammonia molecules alone, between water molecules alone or between ammonia and water.
   
   

Question 20.
Explain about the importance of hydrogen bonding ¡n proteins.
Answer:

• Hydrogen bonds occur in complex biomolecules such as proteins and in biological systems.
• For example, hydrogen bonds play an important role in the structure of deoxyribonucleic acid (DNA), since it holds together the two helical nucleic acid chains.
• In these systems, hydrogen bonds are formed between specific pairs, for example. with a thymine unit in one chain bonding to an adenine unit in another; similarly, a cytosine unit in one chain bonds to a guanine unit in another.
• Intramolecular hydrogen bonding also plays an important role in the structure of polymers, both synthetic and natural.

Question 21.
What are Clatharate hydrate? Explain it with suitable example.
Answer:

• Gas hydrates ¡n which the guest molecules are not bonded chemically but retained by the structure of host is called Clatharates.
• Water forms clatharate hydrates, e.g., methane hydrate (CH₄ 20H₂O) which arc a type of ice that will bum when a lit match is held to it.
• The structure of methane hydrate is made of linked polyhedra that contains 20 hydrogen bonded water molecules forming a cage in which methane molecule is trapped.
• Deposits of methane clatharates occur naturally in deep sea bed.
• Hydrates are commonly obtained when water is frozen in presence of a gas such as argon, methane, etc.
• Most gases form hydrates under high pressure.

Question 22.
What are crystalline hydrates? Explain it with example.
Answer:

1. In these, hydrogen bonding is very important. Often the water molecules serve to fill in the interstices and bind together structure.
2. A specific example is CuSO₄  5H₂O.
3. Although there are five water molecules for every divalent copper ion, only four are coordinated to the cation, it’s six-coordination being completed from sulphate anions. The fifth water molecule is held in place of hydrogen bonds, O – H – O, between it and two coordinated water molecules and then coordinate sulphate anion.
4. Water forms hydrated salts during crystallization. Examples, Na₂ CO₃ . 10H₂O, FeSO₄ .7H₂O.
5. The water present in the hydrates is called as water of hydration.

Question 23.
What do you understand by –

1. Electron-deficient
2. Electron-precise
3. Electron-rich compounds of hydrogen? Provide justification with suitable examples.

Answer:

1. Electron deficient hydrides:
   Compounds in which central atom has incomplete octet, are called electron deficient hydrides. For example, BeH₂ , BH₃  are electron deficient hydrides
2. Electron precise hydrides:
   Those compounds in which exact number of electrons are present in central atom or the central atom contains complete octet are called precise hydrides e.g., CH₄ , SiH₄, GeH₄ etc. are precise hydrides.
3. Electron rich hydrides:
   Those compounds in which central atom has one or more lone pair of excess electrons are called electron rich hydrides. e.g., NH₃, H₂O.

Question 24.
What causes the temporary and permanent hardness of water?
Answer:
Temporary hardness of water is due to the presence of bicarbonates of calcium and magnesium in water i.e., Ca(HCO₃)₂ and Mg(HCO₃) in water. Permanent hardness of water is due to the presence of soluble chlorides and suiphates of calcium and magnesium i.e., CaCl₂, CaSO₄, MgCl₂ and MgSO₄.

Question 25.
Write chemical reaction to show the amphoteric nature of water.
Answer:
Water is amphoteric in nature because it acts as an acid.

Question 26.
What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer:
Hydrolysis is a chemical reaction in which a substance reacts with water under neutral, acidic or alkaline conditions.

Hydration on the other hand is the property of a chemical compound to take up molecules of water of crystallization and get hydrated.

Samacheer Kalvi 11th Chemistry Hydrogen 5-Mark Questions

Question 1.
Explain about the different industrial preparation of hydrogen.
Answer:
In the large scale, hydrogen is produced currently by steam reforming of hydrocarbons. Steam and methane reacts with each other in the presence of nickel catalyst at 35 atm and at a temperature of 800°C gives hydrogen.
CH_4(g) + H₂O_(g) → CO_(g) +3H_2(g)
Steam is passed over a red hot coke to produce carbon monoxide and hydrogen.

Water is reduced to hydrogen with carbon monoxide by passing over iron oxide catalyst at 400°C.
CO_(g) + H₂O_(g) → CO_2(g) + H_2(g)
Hydrogen is produced as a by-product in oil refining industry during the cracking of long chain hydrocarbons.
C₆H_12(g) → CH_66(g) + 3H_2(g)
Hydrogen is also obtained in the manufacture of chlonne and sodium hydroxide via electrolysis of a concentrated solution of sodium chloride.

Question 2.
Explain about the uses of hydrogen compounds.
Answer:

• The hydrogen compounds such as sodium borohydride (NaBH₄) and lithium aluminium hydride (LiAlH₄) are commonly used as reducing agents in organic chemistry.
• Hydrides such as sodium hydride (NaH) and potassium hydride (KH) are used as strong bases in organic synthesis.
• Calcium hydride is used as desiccant to remove moisture from organic solvents.
• Hydride complexes are catalysts.
• Atomic hydrogen and oxy-hydrogen torches for cutting and welding.
• Hydrogen is used in fuel cells for generating electrical energy.
• Liquid hydrogen is used as a rocket fuel as well as in space research.
• Metallic hydrides are used in battery applications.

Question 3.
Describe the process of water softening and purification.
Answer:

1. An idealised image of water softening process involves replacement of cations such as Mg, Ca and Fe in water with sodium ions.by a cation exchange zeolite. The ion exchange zeolites or resins are used to replace the Mg and Ca ions found in hard water with sodium ions.
2. They can be recharged by washing it with a solution containing a high concentration of sodium ions.
3. The calcium and magnesium ions migrate from the zeolite or resin being replaced by sodium ions from the solution until a new equilibrium is reached. That is, the salt is used to recharge an ion exchange medium, which itself is used to soften the water.
4. A couple of other methods, namely chelating method and reverse osmosis are also used to soften hard water. Chelating method employs a polydentate ligand such as EDTA, while reserve osmosis uses high pressure to force the water through a semi-permeable membrane.
5. In the case of water purification application, ion exchange zeolites or resins are used to remove toxic (eg., copper) and heavy metal (e.g., cadmium or lead) ions from solution, replacing them with harmless sodium or potassium ions.

Question 4.
(a) How is H₂O₂ prepared?
(b) Explain about the structure of H₂O₂.
Answer:
(a) Hydrogen peroxide can be made by adding a metal peroxide to dilute acid.
BaO_2(s) + H_(s)SO₄ → BaSO_4(s) + H₂O_2(aq)
(b) Structure of H₂O₂

1. H₂O₂ has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in the figure.

2. Both in gas phase and solid phase, the H₂O₂ molecule adopt a skew configuration due to repulsive interaction of the – OH bonds with lone pairs of electrons on each oxygen atom.

3. Indeed, it is the smallest molecule known to show hindrance rotation about a single bond. in solid phase, the dihedral angle is sensitive and hydrogen bonding decreasing from 111.50 in the gas phase to 90.2°, in the solid phase.

4. Structurally, H₂O₂ is represented by the dihydroxyl formula in which the two O-H groups do not lie in the same plane. In the solid phase of molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°.

5. One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spin.

Question 5.
Explain about Hydrogen sponge.
Answer:
1. Hydrogen sponge (or) Metal hydride e.g., palladium-hydrogen system is a binary hydride (PdH).

2. Upon heating, H atoms diffuse through the metal to the surface and recombine to form molecular hydrogen. Since no other gas behaves this way with palladium, this process has been used to separate hydrogen gas from other gases:
2Pd_(s) + H_2(g) ⇌ 2PdH_(s)

3. The hydrogen molecule readily adsorb on the palladium surface, where it dissociates into atomic hydrogen. The dissociated atoms dissolve into the interstices or voids (octahedral or tetrahedral) of the crystal lattice.

4. Technically, the formation of metal hydride is by chemical reaction but it behaves like a physical storage method, i.e., it is absorbed and released like a water sponge. Such a reversible uptake of hydrogen in metals and alloys is also attractive for hydrogen Storage and for rechargeable metal hydride battery applications.

Question 6.
How are reducing agents in synthetic organic chemistry prepared?
Answer:
Hydrogen has a tendency to react with reactive metals like lithium, sodium to give corresponding hydrides.
2Li+H₂ → 2LiH
2Na+H₂ → 2NaH
These hydrides are used as reducing agents in synthetic organic chemistry. It is also used to prepare important hydrides such as lithium aluminium hydride and sodium boro hydride (organic reducing agents).
4 LiH + AlCl₃ → Li[AIH₄] + 3 LiCl
4 NaH + B(OCH₃)₃ → Na[BH₄] + 3 CH₃ONa

Question 7.
How does water react with –
1. SiCl₄
2. P₄O₁₀
Answer:
1. Water reacts with SiCl₄ to give silica.
SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl

2. Water reacts with P₄O₁₀ to give ortho phosphoric acid.

Question 8.
Explain about the structure of CuSO₄.5H₂O
Answer:
Copper sulphate pentahydrate CuSO₄.5H₂O. In this compound, 4 water molecules form coordinate bonds while the fifth water molecule present outside the coordination can form intermolecular hydrogen bond with another molecule as [Cu(H₂O)₂] SO₂.H₂O.

Question 9.
How is hydrogen peroxide prepared on industrial scale?
Answer:
On an industrial scale, hydrogen peroxide is prepared by auto oxidation of of 2-alkyl antliraquinol.

Question 10.
How is hydrogen peroxide is used to restore the white colour of old paintings.
Answer:
Hydrogen peroxide is used to restore the white colour which was lost due to the reaction of hydrogen suiphide in air with the white pigment Pb₃(OH)₂(CO₃)₂ to form black colored lead suiphide (PbS) Hydrogen peroxide oxidises black coloured lead suiphide to white coloured lead sulphate, there by restoring the colour.
PbS + 4H₂O₂ → PbSO₂ + 4H₂O

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.7

10th Maths Exercise 2.7 Samacheer Kalvi Question 1.
Which of the following sequences are in G.P?
(i) 3, 9, 27, 81, ……..
(ii) 4, 44, 444, 4444, ………
(iii) 0.5, 0.05, 0.005, ……..
(iv) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 6 } \),\(\frac { 1 }{ 12 } \)………
(v) 1, -5, 25, -125, …….
(vi) 120, 60, 30, 18, …….
(vii) 16, 4, 1, \(\frac { 1 }{ 4 } \), ……
Solution:
(i) 3, 9, 27, 81
r = Common ratio




10th Maths Exercise 2.7 Question 2.
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
(ii) a = \(\sqrt { 2 }\) , r = \(\sqrt { 2 }\)
(iii) a = 1000, r = \(\frac { 2 }{ 5 } \)
Solution:
(i) a = 6, r = 3
t_n = ar^n-1
t₁ = ar^1-1 = ar⁰ = a = 6
t₂ = ar^2-1 = ar¹ = 6 × 3 = 18
t₃ = ar^3-1 = ar² = 6 × 3² = 54
∴ The 3 terms are 6, 18, 54, ….

The 3 terms are 1000, 400, 160, ……………

Ex 2.7 Class 10 Samacheer Question 3.
In a G.P. 729, 243, 81,… find t₇.
Solution:
G.P = 729, 243, 81 ……
t₇ = ?

Ex 2.7 Class 10 Question 4.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.
Answer:
\(\frac{t_{2}}{t_{1}}\) = \(\frac { x+12 }{ x+6 } \),\(\frac{t_{3}}{t_{2}}\) = \(\frac { x+15 }{ x+12 } \)
Since it is a G.P.
\(\frac { x+12 }{ x+6 } \) = \(\frac { x+15 }{ x+12 } \)
(x + 12)² = (x + 6) (x + 15)
x² + 24x + 144 = x² + 21x + 90
3x = -54 ⇒ x = \(\frac { -54 }{ 3 } \) = -18

Exercise 2.7 Class 10 Maths Question 5.
Find the number of terms in the following G.P.
(i),4, 8, 16, …, 8192
(ii) \(\frac { 1 }{ 3 } \),\(\frac { 1 }{ 9 } \),\(\frac { 1 }{ 27 } \),……\(\frac { 1 }{ 2187 } \)
Solution:
(i) 4, 8, 16, …… 8192

10th Maths 2.7 Exercise Question 6.
In a G.P. the 9^th term is 32805 and 6^th term is 1215. Find the 12th term.
Solution:
In a G.P
t_n = ar^n-1
t₉ = 32805
t₆ = 1215
t₁₂ = ?
Let
t₉ = ar⁸ = 32805 ………(1)
t₆ = ar⁵ = 1215 ………. (2)

10th Math Exercise 2.7 Solution Question 7.
Find the 10th term of a G.P. whose 8^th term is 768 and the common ratio is 2.
Answer:
Here r = 2, t₈ = 768
t₈ = 768 (t_n = ar^n-1)
a. r^8-1 = 768
ar⁷ = 768 …..(1)
10^th term of a G.P. = a.r 10^-1
= ar⁹
= (ar⁷) × (r²)
= 768 × 2² (from 1)
= 768 × 4 = 3072
∴ 10^th term of a G.P. = 3072

Class 10 Maths Exercise 2.7 Solutions Question 8.
If a, b, c are in A.P. then show that 3^a,3^b,3^c are in G.P.
Solution:
If a, b, c are in A.P
t₂ – t₁ = t₃ – t₂
b – a = c – b
2b = c + a
To prove that 3^a, 3^b, 3^c are in G.P
⇒ 3^2b = 3^c+a + a [Raising the power both sides]
⇒ 3^b.3^b = 3^c.3^a
\(\Rightarrow \frac{3^{b}}{3^{a}}=\frac{3^{c}}{3^{b}}\)
\(\Rightarrow \frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{1}}\)
⇒ Common ratio is same for 3^a,3^b,3^c
⇒ 3^a, 3^b, 3^c forms a G.P
∴ Hence it is proved .

Exercise 2.7 Class 10 Maths Solution Question 9.
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is \(\frac { 57 }{ 2 } \). Find the three terms.
Solution:
Let the three consecutive terms in a G.P are \(\frac { a }{ r } \), a, ar.
Their Product = \(\frac { a }{ r } \) × a × ar = 27
a³ = 27 = 3³
a = 3
Sum of the product of terms taken two at a time is \(\frac { 57 }{ 2 } \)

2.7 Exercise Class 10 Question 10.
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Solution:
Starting salary = ₹ 60,000
Increase per year = 5%
∴ At the end of 1 year the increase
= 60,0,00 × \(\frac { 5 }{ 100 } \)
₹ 3000
∴ At the end of first year his salary
= ₹ 60,000 + 3000
I year salary = ₹ 63,000
II Year increase = 63000 × \(\frac { 5 }{ 100 } \)
At the end of II year, salary
= 63000 + 3150
= ₹ 66150
III Year increase = 66150 × \(\frac { 5 }{ 100 } \)
= 3307.50
At the end of III year, salary = 66150 + 3307.50
= ₹ 69457.50
IV year increase = 69457.50 × \(\frac { 5 }{ 100 } \)
= ₹ 3472.87

Exercise 2.7 Class 10 Question 11.
Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹ 20,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
Offer B: ₹ 22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4th year with respect to the offers A and B?
Solution:
Offer A
Starting salary ₹ 20,000
Annual increase 6%

At the end of
III year ,salary = 22472 + 1348 = 23820
∴ IV year salary = ₹ 23820
Offer B
Starting salary = ₹ 22,000

Salary as per Option A = ₹ 23820
Salary as per Option B = ₹ 24040
∴ Option B is better.

10 Class Math Exercise 2.7 Solution Pdf Question 12.
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that x^b-c × y^c-a × z^a-b = 1.
Solution:
a, b, c are three consecutive terms of an AP.
∴ Let a, b, c be a, a + d, a + 2d respectively ………… (1)
x, y, z are three consecutive terms of a GP.
∴ Assume x, y, z as x, x.r, x.r² respectively ……… (2)
PT : x^b-c , y^c-a , z^a-b = 1
Substituting (1) and (2) in LHS, we get
LHS = x^a+d-a-2d × (xr)^a+2d-a × (xr²)^a-a-d
= (x)^-d . (xr)^2d (xr²)^-d
= \(\frac{1}{x^{d}}\) × x^2d . r^2d × \(\frac{1}{x^{d} r^{2 d}}\) = 1 = RHS

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.2

9th Maths Exercise 1.2 Samacheer Kalvi Question 1.
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {v : v = \(\frac { 4 }{ 3n }\) ,n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integers, x ∈ Z and -5 ≤ x < 5}
(v) S = The set of all leap years between 1882 and 1906.
Solution:
(i) n(M) = 6
(ii) W = {0, 1, 2, 3, ……. }
if n = 0, x = 3(0) + 2 = 2
if n = 1, x = 3(1) + 2 = 5
if n = 2, x = 3(2) + 2 = 8
if n = 3, x = 3(3)+ 2 =11
if n = 4, x = 3(4) + 2=14
∴ P= {2, 5, 8, 11, 14}
n(P) = 5

(iii) N = {1,2, 3, 4, …..}
n ∈ {3, 4, 5}

n(Q) = 3

(iv) x ∈ z
R = {-5, – 4, -3, -2, -1, 0, 1, 2, 3, 4}
n(R)= 10.

(v) S = {1884, 1888, 1892, 1896, 1904}
n (S) = 5.

9th Maths Set Language Exercise 1.2 Question 2.
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x² – 5x + 6 = 0, x ∈ N}
Solution:
(i) Finite set
(ii) Infinite set
(iii) A = { ……. , -2, -1, 0, 1, 2, 3, 4}
∴ Infinite set

(iv) x² – 5x + 6 = 0
(x – 3) (x – 2) = 0
B = {3, 2}
∴ Finite set.

9th Maths Exercise 1.2 Question 3.
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = A = { x : x is a letter in the word “LIFE”}
Y = {F, I, L, E}
(iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution:
(i) A = {a, e, i, o, u}
B = {V, O,W, E, L}
The sets A and B contain the same number of elements.
∴ Equivalent sets

(ii) C ={2, 3, 4, 5}
D = {2, 3, 4}
∴ Unequal sets

(iii) X = {L, I, F, E}
Y = {F, I, L, E}
The sets X and Y contain the exactly the same elements.
∴ Equal sets.

(iv) G = {5, 7, 11, 13, 17, 19}
H = {1, 2, 3, 6, 9, 18}
∴ Equivalent sets.

9th Maths Set Language Exercise 1.2 Solutions Question 4.
Identify the following sets as null set or singleton set.
(i) A = (x : x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2.
(iii) C = {0}
(iv) D = The set of all triangles having four sides.
Solution:
(i) A = { } ∵ There is no element in between 1 and 2 in Natural numbers.
∴ Null set

(ii) B = { } ∵ All even natural numbers are divisible by 2.
∴ B is Null set

(iii) C = {0}
∴ Singleton set

(iv) D = { }
∵ No triangle has four sides.
∴ D is a Null set.

9th Maths Exercise 1.2 In Tamil Question 5.
State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s)
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
(iii) E = {x: x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
Solution:
(i) A = {f, i, a, s}
B = {a, n, f, h, s}
A ∩ B ={f, i, a, s} ∩ {a, n,f h, s} = {f, a, s}
Since A ∩ B ≠ ϕ , A and B are overlapping sets.

(ii) C = {3, 5, 7, 11, ……}
D = {2}
C ∩ D = {3, 5, 7, 11, …… } ∩ {2} = { }
Since C ∩ D = Ø, C and D are disjoint sets.

(iii) E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27}
E ∩ F = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {3, 6, 9, 12, 15, 18, 21, 24, 27}
= {3, 6, 12, 24}
Since E ∩ F ≠ ϕ, E and F are overlapping sets.

9th Standard Maths Exercise 1.2 In Tamil Question 6.
If S = {square,rectangle,circle,rhombus,triangle}, list the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°
(iv) The set of shapes which have 5 sides.
Solution:
(i) {Square, Rhombus}
(ii) {Circle}
(iii) {Triangle}
(iv) Null set.

9th Standard Maths Exercise 1.2 Question 7.
If A = {a, {a, b}}, write all the subsets of A.
Solution:
A= {a, {a, b}} subsets of A are { } {a}, {a, b}, {a, {a, b}}.

9th Std Maths Exercise 1.2 Question 8.
Write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = Ø
Solution:
(i) The subsets of A are Ø, {a}, {b}, {a, b}
The power set of A
P(A ) = {Ø, {a}, {b}, {a,b}}

(ii) The subsets of B are ϕ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
The power set of B
P(B) = {Ø, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

(iii) The subset of D are Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s},{p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}}
The power set of D
P(D) = {Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}

(iv) The power set of E
P(E) = { }.

9th Maths 1.2 In Tamil Question 9.
Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red,blue, yellow}
(ii) X = { x² : x ∈ N, x² ≤ 100}.
Solution:
(i) Given W = {red, blue, yellow}
Then n(W) = 3
The number of subsets = n[P(W)] = 2³ = 8
The number of proper subsets = n[P(W)] – 1 = 2³ – 1 = 8 – 1 = 7

(ii) Given X ={1,2,3, }
X² = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(X) = 10
The Number of subsets = n[P(X)] = 2¹⁰ = 1024
The Number of proper subsets = n[P(X)] – 1 = 2¹⁰ – 1 = 1024 – 1 = 1023.

9th Maths 1.2 Exercise Question 10.
(i) If n(A) = 4, find n[P(A)].
(ii) If n(A) = 0, find n[P(A)].
(iii) If n[P(A)] = 256, find n(A).
Solution:
(i) n( A) = 4
n[ P(A)] = 2n = 2⁴ = 16
(ii) n(A) = 0
n[P(A)] = 2⁰ = 1
(iii) n[P(A)] = 256

n[P(A)] = 28
∴ n(A) = 8.

Enhance your subject knowledge with Tamilnadu State Board for Chapter 6 Gaseous State and learn all the underlying concepts easily. Make sure to Download Samacheer Kalvi 11th Chemistry Book Solutions, Notes Pdf Chapter 6 Gaseous State Questions and Answers PDF on a day to day basis and score well in your exams. Are given after enormous research by people having high subject knowledge. You can rely on them and prepare any topic of Chemistry as per your convenience easily.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Students looking for Chapter 6 Gaseous State Concepts can find them all in one place from our Tamilnadu State Board Solutions. Simply click on the links available to prepare the corresponding topics of Chemistry easily. Samacheer Kalvi 11th Chemistry Chapter wise Questions and Answers are given to you after sample research and as per the latest edition textbooks. Clarify all your queries and solve different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your speed and accuracy in the final exam.

Samacheer Kalvi 11th Chemistry Gaseous State Textual Evaluation Solved

I. Choose the correct answer from the following:

11th Chemistry Lesson 6 Book Back Answers Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement (s) is correct for non – ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the inter molecular interactions become significant
Answer:
(d) at high pressure the inter molecular interactions become significant

11th Chemistry Chapter 6 Book Back Answers Question 2.
Rate of diffusion of a gas is …………
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Gaseous State 11th Chemistry Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?

Answer:

11th Chemistry Unit 6 Book Back Answers Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules ………….
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) colide without loss of energy
Answer:
(b) exert no attractive forces on each other

11th Chemistry Gaseous State Question 5.
Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen ………..
(a) 1/3
(b) 1/2
(c) 2/3
(d) 1/3 × 273 × 298
Answer:
(a) 1/3
Hint:
mass of methane = mass of oxygen = a
number of moles of methane = \(\frac {a}{16}\)
number of moles of Oxygen = \(\frac {a}{32}\)
mole fraction of Oxygen =
Partial pressure of oxygen = mole fraction x Total Pressure = \(\frac {1}{3}\)P

Gaseous State Class 11 Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called …………
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature
Hint:
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called Boyle temperature

Gaseous State Class 11 Notes Pdf Question 7.
In a closed room of 1000 m³ a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Samacheer Kalvi Guru 11th Chemistry Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be ………….
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle
Hint:
Rate of diffusion α 1/√m
m_NH3  = 17
m_HCl = 36.5
γ_NH3  > γ_HCl
Hence white fumes first formed near hydrogen chloride.

Samacheer Kalvi Class 11 Chemistry Solutions Question 9.
The value of universal gas constant depends upon ………..
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Samacheer Kalvi.Guru 11th Chemistry Question 10.
The value of the gas constant R is …………
(a) 0.082 dm³ atm.
(b) 0.987 cal mol^-1 K^-1
(c) 8.3 J mol^-1K^-1
(d) 8 erg mol^-1K^-1
Answer:
(c) 8.3 J mol^-1K^-1

Samacheerkalvi.Guru 11th Chemistry Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Class 11 Chemistry Solutions Samacheer Kalvi Question 12.
The table indicates the value of van der Waals constant ‘a’ in (dm³)² atm. mol^-2

The gas which can be most easily liquefied is ………….
(a) O₂
(b) N₂
(c) NH₃
(d) CH₄
Answer:
(c) NH₃
Hint:
Higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquefaction. Option (c) is correct

Gaseous State Questions And Answers Pdf Question 13.
Consider the following statements.
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Class 11 Gaseous State Question 14.
Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. The molar volume of CO₂ under these conditions is ………..
(a) 22.04 dm³
(b) 2.24 dm³
(c) 0.41 dm³
(d) 19.5 dm³
Answer:
(c) 0.41 dm³
Compressibility factor (z) = \(\frac {Pv}{nRT}\)
V = \(\frac {z x nRT }{p}\)

V = 0.41 dm³

Samacheer Kalvi 11th Chemistry Solution Question 15.
If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes ………….
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(c) P
Hint:

P₂ = P₂ Option (c)

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3\(\sqrt{3}\) times that of a hydrocarbon having molecular formula What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(b) 4.
Hint:

Squaring on both sides and rearranging
27 x 2 = m_CnH_2n-2
54 = n(12) + (2n-2)(l)
54 = 12n+2n – 2
54 = 14n – 2
n = (54 + 2)/14 = 56/14 = 4

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase 1)
(a) 3/8
(b) 1/2
(c) 1/8
(d) 1/4
Answer:
(c) 1/8
Hint:

The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is 1/8

Question 18
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\)For an ideal gas α is equal to ………..
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(b) 1/T
Hint:

Question 19.
Four gases P, Q, R and S have almost same values of ‘b’ but their ‘a’ values (a, b are Van der Waals Constants) are in the order Q < R < S < P. At a particular temperature, among the four gases the most easily liquefiable one is ………….
(a) P
(b) Q
(c) R
(d) S
Answer:
(a) P
Hint:
Greater the ‘a’ value, casier the liquefaction

Question 20.
Maximum deviation from ideal gas is expected from (NEET)
(a) CH_4(g)
(b) NH_3(g)
(c) H_2(g)
(d) N_2(g)
Answer:
(b) NH_3(g)

Question 21.
The units of Van der Waals constants ‘b’ and ‘a’ respectively
(a) mol L^-1 and L atm² mol^-1
(b) mol L and L atm mol²
(c) mol^-1 L and L² atm mol^-1
(d) none of these
Answer:
(c) mol^-1 L and L² atm mol^-1
Hint:
an²/V² atm
a = atm L²/mol² = L² mol^-2 atm
nb = L
b = L /mol = L mol^-1

Question 22.
Assertion : Critical temperature of CO₂ is 304 K, it can be liquefied above 304 K.
Reason : For a given mass of gas, volume is to directly proportional to pressure at constant temperature.
(a) both assertion and reason arc 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:
(d) both assertion and reason are false
Hint:
Correct Statement: Critical temperature of CO₂ is 304 K. It means that CO₂ cannot be liquefied above 304 K, whatever the pressure may applied. Pressure is inversely proportional to volume.

Question 23.
What is the density of N, gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^-1 mol^-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L
Hint:
Density = \(\frac {Mass}{Volume}\)

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas ? (T is measured in K)

Answer:

For a fixed mass of an ideal gas V α T
P α 1/V
and PV = Constant

Question 25.
25 g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI
Hint:
At a given temperature and pressure
Volume α number of moles
Volume α Mass / Molar mass
Volume α 28 / Molar mass
i.e. if molar mass is more , volume is less. Hence Hl has the least volume.

II. Answer these questions briefly.

Question 26.
State Boyle’s law.
Answer:
Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperalure can be used as a model for Charles’ law.
Answer:
Charles’ law:

• V T at constant P and n (or) \(\frac {V}{T}\) = Constant
• A balloon filled with air at room temperature and cooled to a much lower temperature. the size of the balloon is reduced. Because if the temperature of the gas decreases, the volume also decreases in a direct proportion.
• When temperature is reduced, the gas molecules inside in over slower due to decreased temperature and hence the volume decreases.

Question 28.
Name two items that can serve as a model for Gay Lussac’s law and explain.
Answer:
Gay Lussac’s law:
1. P α T at constant volume (or) = \(\frac {V}{T}\)
2. Example – 1:
You fill the car type completely full of air on the hottest day of summer. The type cannot change it shape and volume. But when winter comes, the pressure inside the lyre is reduced and the shape is also reduced. This confirms that pressure and temperature are direct related to each other.
3. Example – 2:
The egg in the bottle experiment.

A glass bottle is taken, inside the bottle put some pieces of cotton with fire. Then place a boiled egg (shell removed) at the top of the bottle. The temperature inside the bottle increases from the fire, rising (he pressure. By scaling the bottle with egg, the fire goes on, dropping the temperature and pressure. This causes the egg to be sucked into the bottle.
P α T is proved (or) = \(\frac{P_{1}}{V_{1}}=\frac{P_{2}}{V_{2}}\)

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means.
Answer:

1. The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
2. \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
   Where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is dirctly proportional to the number of the moles of the gas.

Question 30.
What are ideal gases? In what way real gases differ from ideal gases.
Answer:

1. Ideal gases are the gases that obey gas laws or gas equation PV = nRT.
2. Real gases do not obey gas equation. PV = nRT.
3. The deviation of real gases from ideal behaviour is measure in terms of a ratio of PV to nRT. This is termed as compression factor (Z). Z = \(\frac {PV}{nRT}\)
4. For ideal gases Z = 1.
5. For real gases Z > 1 or Z < 1. For example, at high pressure real gases have Z >1 and at intermediate pressure Z < 1.
6.  Above the Boyle point Z> 1 for real gases and below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.
7. So, it is clear that at low pressure and high temperature, the real gases behave as ideal gases.

Question 31.
Can a Van der Waals gas with a = 0 be liquefied? Explain.
Answer:

• a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
• If a = 0, there is a very less interaction between the molecules of gas.
• ‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
• If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Question 32.
Suppose there ¡s a tiny sticky area on the wan of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:

• Molecules hitting the tiny sticky area on the wall of the container of gas moves faster as they get closer to adhesive surface, but this effect is not permanent.
• The pressure on the sticky wall is greater than on the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The type of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude
Answer:
(a) In aerated water bottles, CO₂ gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more of gas will be present above the liquid surface in the glass bottle.

In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept under water. As a result, the temperature is likely to decrease and the solubility of CO₂ is likely to increase in aqueous solution resulting in decreased pressure.

(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes accident.

However, if the bottle is cooled under tap water for sometime, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will corne out of the bottle at a slower rate, reduces the chances of accident.

(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.

(d) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Question 34.
Give suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container
(b) Gases diffuse through all the space available to them and
(c) Explain with an increase in temperature
Answer:
(a) Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

(b) When a sample of a gas introduced to one part of a closed container, its molecules very quickly disperse throughout the container, this process by which molecules disperse in space in response to differences in concentration is called diffusion. For e.g., you can smell perfume in a room, because it difluses into the air totally inside the room.

(c) Diffusion is faster at higher temperature because the gas molecules have greater kinetic energy. Since heat increase the motion, then diffusion happens faster.

Question 35.
Suggest why there ¡s no hydrogen (H₂) in our atmosphere. Why does the moon have no atmosphere?
Answer:
1. Hydrogen is the lightest element thus when produced in free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There it literally leaks from the atmosphere to the empty space. Hydrogen easily gains velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O₂ , in its way to produce H₂O. So majority portion of H₂ reacts and very less amount of it present in the upper level of atmosphere and gains velocity to escape the atmosphere.

2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon have thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere in the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that moon has no atmosphere.

Question 36.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if –
(a) it is compressed to a smaller volume at constant temperature
(b) the temperature is raised while keeping the volume constant
(c) more gas is introduced into the same volume and at the same temperature
Answer:
(a) it a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) if more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. if the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 37.
Which of the following gases would you expect to deviate from ¡deal behaviour under conditions of low temperature F Cl₂, or Br₂? Explain.
Answer:
1. Bromine deviates (Br₂) from the ideal gas maximum than Cl₂ and F₂. Because Br₂ has biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.

2. Br₂ deviates from ideal behaviour because it has largest atomic radii compared to Cl₂ and F₂. So it contains more electrons than other two, and the Vander Waals forces are stronger in Br₂ than in Cl₂ and F₂. So Br₂ deviates from ideal behaviour.

Question 38.
Distinguish between diffusion and effusion.
Answer:
Diffusion:

• Diffusion is the spreading of molecules of a substance throughout a space or a second substance.
• Diffusion refers to the ability of the gases to mix with each other.
• E.g.. Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

• Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
• Effusion is a ability of a gas to travel through a small pin-hole.
• E.g., pouring out something like the soap studs bubbling out from a bucket of water.

Question 39.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions. So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire or leave it in the direct sunlight. even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

• The gas pressure increases.
• More of the liquefied propellant turns into a gas.

Question 40.
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
1. When the driver of an automobile applies brake, the passengers are pushed toward the front of the car due to inertia of the body, but a helium balloon pushed toward back of the car. Helium balloon responds to the air around it. Helium molecules are lighter than air of our atmosphere, and so they move toward back by gravity as a result of the accelerating frame.

2. Upon forward acceleration, the passengers arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. Helium balloon is going to move opposite to this pseudo gravitational force.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn’t enough atmospheric pressure to push the water up in the straw all the way to the mouth.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.
Answer:
Van der Waals equation of state for real gases is –
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT

Correction term for pressure:
\(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\) is the pressure correction. It represents the intermolecular interaction that causes the non ideal behaviour.

Correction term for Volume:
V – nb is the volume correction. it is the effective volume occupied by real gas.

Question 43.
Derive the values of critical constants from the Van der Waals constants.
Answer:
Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT for 1 mole
From this equation, the values of critical constant P_CV_C and T_C arc derived in terms of a and b the Van der Waals constants.
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V – b)\) = RT ………..(1)
On expanding the equestion (1)
P V + \(\frac {a}{V}\) – pb – \(\frac{\mathrm{ab}}{\mathrm{V}^{2}}\) – RT = 0 ………(2)
Multiplying eqestion (2) by \(\frac{V^{2}}{P}\),

equation (3) is rearranged in the powers of V
V³ – \(\left[\frac{\mathrm{RT}}{\mathrm{P}}+\mathrm{b}\right]\) V² + \(\frac {aV}{P}\) –
= 0 ………..(4)
The above equation (4) is an cubic equation of V, which can have three roots. At the critical point. all the three values of V are equal to the critical volume V_C.
i.e. V = V_C.
V – V_C = O ……….(5)
(V – V_C)³ = O ………(6)
(V³ – 3V_CV² + 3V_C³V – V_C³ = 0 ………(7)
As the equation (4) is identical with equation (7), comparing the ‘V’ ternis in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

substituting the values of Vc and Pc in equation (9)

Critical constant a and b can be calculated using Van der Waals Constant as follows:

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
In space, there is no pressure, if we do wear a pressurised suit, our body will die. In space, we have to wear a pressurised suit, otherwise our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurised suit (protective suits).

Question 45.
When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. Why do more funies appear near HCl?
Answer:

1. When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. The reaction takes place in neutralization between a weak base and a strong acid.
2. The property of the gas is diffusion.
3. Diffusion of gases Ammonia and hydrogen chloride. Concentrated ammonia solution is placed on a pad in one end of a tube and concentrated HCl on the pad at the other. After about a minute, the gases diffuses far enough to meet and a ring of solid ammonium chloride is formed near the HCl end.

Question 46.
A sample of gas at 15°C at 1 atm has a volume of 2.58 dm³. Vhen the temperature is raised to 38°C at I atm does the volume of the gas increase? if so, calculate the final volume.
Answer:

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.

Question 47.
A sample of gas has a volume of 8.5 dm³ at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 6.37 dm³. What ¡s its initial temperature?
Answer:

T₁ = 364.28 K

Question 48.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume of 37.6 dm³ at 298K, and the sample B is in a vessel of volume 16.5 dm³ at 298 K. Calculate the number of moles in sample B.
Answer:
n_A = 1.5 mol n_B = ?
V_A = 37.6 dm³ V_B = 16.5 dm³
(T = 298 K constant)

Question 49.
Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 dm³ at 69.5°C, assuming ¡deal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P = ?
PV = nRT
P = \(\frac {nRT}{V}\)

P = 94.25 atm.

Question 50.
Argon is an inert gas used in light bulbs to retard the vapourlzation of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P₁ = 1.2 atm
T₁ = 18°C + 273 = 291 K
T₂ = 85°C + 273 = 358 K
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 51.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is 1.5 mL.
Answer:
T₁ = 6°C + 273 = 279 K
P₁ = 4 atm V₁ = 1.5m
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm V ₁ = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V₂ = 6.41 mol.

Question 52.
Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 x 10^-3 dm³ of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 x 10^-3dm³
P = 742 mm of Hg
T = 298 K m = ?
n = \(\frac{PV}{RT}\) =
= 0.006 mol
n = \(\frac{PV}{RT}\)
n = \(\frac{Mass}{Molar Mass}\)
Mass = n x Molar mass
= 0.006 x 2.016
= 0.0121 g = 12.1 mg.

Question 53.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?
Answer:

Question 54.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure in the tank ¡s 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
m_O2 = 52.5 g
P_O2 = ?
m_CO2 = 65.1 g
P_CO2 = ?
T = 300 K P = 9.21 atm
P_O2 = X_O2 x total pressure

P_O2 = X_O2 x Total pressure
= 0.53 x 9.21 atm = 4.88 atm
P_CO2 = X_CO2 x Total pressure
= 0.47 x 9.21 atm = 4.33 atm

Question 55.
A combustible gas Is stored in a metal tank at a pressure of 2.98 atm at 25 °C. The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = 1100 K).
Answer:
T₁ = 298 K;
P₁ = 2.98 atm;
T₂ = 1100K;
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
P₂ = \(\frac{P_{1}}{T_{1}} \times T_{2}\)
= \(\frac{2.98 \mathrm{atm}}{298 \mathrm{K}} \times 1100 \mathrm{K}\) = 11 atm
At 1100 K, the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

In Text Questions – Evaluate Yourself

Question 1.
Freon-I 2, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now It has been replaced by eco-friendly compounds. Consider 1.5 dm³ sample of gaseous freon at a pressure of 0.3 atm. If the pressure is changed to 1.2 atm. at a constant temperature, what will be the volume of the gas increased or decreased?
Answer:
Volume of freon (V₁) = 1.5 dm³
Pressure (P₁) = 0.3 atm
‘T’ is constant
P₂ = 1.2 atm
V₂ = ?
P₁V₁ = P₂V₂
V₂ = \(\frac{P_{1} V_{1}}{P_{2}}\)

= 0.375 dm³
Volume decreased from 1.5 dm³ to 0.375 dm³

Question 2.
Inside a certain automobile engine, the volume of air in a cylinder is 0.375 dm³, when the pressure is 1.05 atm. When the gas is compressed to a volume of 0.125 dm³ at the same temperature, what is the pressure of the compressed air?
Answer:
V₁ = 0.375dm³
V₂ = 0.125 dm³
P₁ = 1.05 atm
P₂ = ?
T – Constant
P₁V₁ = P₂V₂
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10.5 \times 0.375}{0.125}\)
= 3.15 atm.

Question 3.
A sample of gas has a volume of 3.8 dm³ at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 2.27 dm³. What is its initial temperature?
Answer:
V₁ = 3.8 dm³
T₂ = 0°C = 273K
T₁ = ? V₂ = 2.27dm³
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

T₁ = 457 K

Question 4.
An athlete in a kinesiology research study has his lung volume of 7.05 dm³ during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his Jung decreases to 2.35 dm³ How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)
Answer:
V₁ = 7.05 dm³
V₂ = 2.35 dm³
n₁ = 0.312 mol
n₁ =?
‘P’ and ‘T’ are constant
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\)

n₂ = 0.104 mol
Number of moles exhaled = 0.312 – 0.104 = 0.208 moles

Question 5.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 8° C and 6.4 atm. to the water surface, where the temperature ¡s 25°C and pressure is 1 atm. Calculate the final volume in (ml) of the bubble, if its initial volume is 2.1 ml.
Answer:
T₁ = 8°C = 8 + 273 = 281K
P₁ = 6.4atm V₁ = 2.1 mol
T₂ = 25°C = 25 + 273 = 298 K
P₂ = 1 atm
V₂ = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V₂ = 14.25 ml

Question 6.
(a) A mixture of He and O₂ were used ¡n the ‘air’ tanks of underwater divers for deep dives. For a particular dive 12 dm³ of O₂ at 298 K, I atm. and 46 dm3 of He, at 298 K, 1 aim. were both pumped into a 5 dm³ tank. Calculate the partial pressure of each gas and the total pressure In the tank at 298 K
Answer:

P = 1 atm
V_total = 5 dm³
P_O2 = X_O2 x P_total

P_He = 3.54 atm

Question 6.
(b) A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction 2KClO₃ → 2KCl_(s) + 3O₂ The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mn of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:
2KCl_3(s) → 2KCl_(s) 3O_3(g)
P_total = 772 mm Hg
P_H2O = 26.7 mm Hg
P_total = P_O2 + P_H2O
P_O2 = P_total – P_H2O
P₁ = 26.7 mm Hg
T₁ = 300 K
T₂ = 295 K
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P₂ = 26.26 mm Hg
∴ P_O2 = 772 – 26.26
= 745.74 mm Hg

Question 7.
A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes 4.73 min to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be, (Given that molar mass of bromine = 159.8 g/ mole)
Answer:
t₁ = 1.5 minutes (gas)_hydrocarbon
t₂ = 4.73 minutes (gas)_Bromi

n (12) + (2n + 2) 1 = 16 (general formula for hydrocarbon C_nH_2n+2)
12n + 2n + 2 = 16
14n = 16 – 2
14n = 14
n = 1
The hydrocarbon is C₁H_{2(1) + 2} = CH₄

Question 8.
Critical temperature of H₂O, NH₃ and CO₂ are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?
Answer:
Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
∴ When cooling starts from 700 K, H₂O vill liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.

In-Text Example Problems

Question 9.
In the below figure, let us find the missing parameters [volume in (b) and pressure in (c)]
P₁ = 1 atm P₂ = 2 atm P₃ = ? atm
V₁ I dm3  V₂ =? dm3 V₃ = 0.25 dm³
T₁ = 298 K T₂ = 298 K T₃ = 298 K
Solution:
According to Boyle’s law, at constant temperature for a given mass of gas at constant temperature,
P₁V₁ = P₂V₂ = P₃V₃
I atm x 1 dm³ = 2 atm x V₂ = P₃x 0.25 dm³
∴ 2 atm x V₂ = 1 atm x 1 dm³

V₂ = 0.5 dm³
and P₃ x 0.25 dm³ = 1 atm x 1 dm³

P₃ = 4atm

Question 10.
In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]
P₁ = 1 atm P₂ = 1 atm P₃ = 1 atm
V₁ = 0.3dm³ V₂ = ?dm³ V₃ = 0.15dm³
T₁ = 200K T₂ = 300 K T₃ = ? K
Answer:

According to Charles law,


T₃ = 100k

Question 11.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm³ at 70°C assuming It is an ideal gas.
Answer:
We will use the ideal gas equation for this calculation as below:
P = \(\frac {nRT}{V}\)

= 9.39 atm.

Question 12.
A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm. at a fixed temperature. Solve this problem using Dalton’s law.
Answer:
P_Ne = X_Ne P_Total

P_Ne = X_Ne P_Total = 0.595 x 2 = 1.19 atm.
P_Ne = X_Ne P_Total = 0.093 x 2 = 0. 186 atm.
P_Ne = X_Ne P_Total = 0.312 x 2 = 0.624 atm.

Question 13.
An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.
Answer:

Question 14.
If a scuba diver takes a breath at the surface filling his lungs with 5.82 dm3 of air what volume will the air in his lungs occupy when he drives to a depth, where the pressure ¡s 1.92 atm. (assume temperature is constant and the pressure at the surface is exactly 1 atm.)
Solution :
Temperature = Constant
Pressure at the surface = 1 atm – P₁
Pressure at the depth = 1.92 atm – P₂
Vdlume of air breathing at the surface of the air = 5.82 dm³ – V₁
Volume of air breathing at the depth = V₂ = ?

V₁ = 3.03 dm³
The volume of air scuba diver’s lung occupy = 3.03 dm³

Question 15.
Inside a certain automobile engine, the volume of air in a cylinder is 0.475 dm³, when the pressure is 1.05 atm. When the gas is compressed, the pressure increased to 5.65 atm. at the same temperature. What is the volume of compressed air?
Solution:
Volume of air in the cylinder 0.475 dm³ – V₁
Pressure of air P₁ = 1.05 atm
Increased pressure P₂ = 5.65 atm
Volume of air compressed V₂ = ?
P₁V₁ = P₂V₂

V₂ = 0.08827 dm³
Compressed volume of air = 0.08 827 dm³

Samacheer Kalvi 11th Chemistry Gaseous State Additional Questions Solved

I. Choose the correct answer.

Question 1.
For one mole of a gas, the ideal gas equation is ………..
(a) PV = \(\frac {1}{2}\) RT
(b) PV = RT
(c) PV = \(\frac {3}{2}\)RT
(d) PV = \(\frac {5}{2}\)RT
Answer:
(b) PV= RT

Question 2.
The average kinetic energy of the gas molecule is …………
(a) inversely proportional to its absolute temperature
(b) directly proportional to its absolute temperature
(c) equal to the square of its absolute temperature
(d) All of the above
Answer:
(b) directly proportional to Its absolute temperature

Question 3.
Which of the following is the correct mathematical relation for Charles’ law at constant pressure?
(a) V ∝ T
(b) V ∝ t
(c) V ∝ – \(\frac {1}{T}\)
(d) all of above
Answer:
(a) V ∝ T

Question 4.
At constant temperature, the pressure of the gas is reduced to one-third, the volume
(a) reduce to one-third
(b) increases by three times
(c) remaining the same
(d) cannot be predicted
Answer:
(b) increases by three times

Question 5.
With rise in temperature, the surface tension of a liquid …………
(a) decreases
(b) increases
(c) remaining the same
(d) none of the above
Answer:
(a) decreases

Question 6.
Viscosity of a liquid is a measure of ……………
(a) repulsive forces between the liquid molecules
(b) frictional resistance
(c) intermolecular forces between the molecules
(d) none of the above
Answer:
(b) frictional resistance

Question 7.
The cleansing action of soaps and detergents is due to …………..
(a) internal friction
(b) high hydrogen bonding
(c) viscosity
(d) surface tensions
Answer:
(d) surface tensions

Question 8.
In Vander Waals equation of state for a non-ideal gas the net force of attraction among the molecules is given by ………..
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(b) P + \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(c) P – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(d) – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
Answer:
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

Question 9.
The compressibility factor, z for an ideal gas is ………….
(a) zero
(b) less than one
(c) greater than one
(d) equal to one
Answer:
(d) equal to one

Question 10.
Which of the following gases will have the lowest rate of diffusion?
(a) H₂
(b) N₂
(c) F₂
(d) O₂
Answer:
(c) F₂

Question 11.
Which of the following is a mono atomic gas in nature?
(a) Oxygen
(b) Hydrogen
(c) Helium
(d) Ozone
Answer:
(c) Helium

Question 12.
Which of the following is a diatomic gas in nature?
(a) Oxygen
(b) Ozone
(c) Helium
(d) Radon
Answer:
(a) Oxygen

Question 13.
Which one of the following is not a monoatomic gas?
(a) Neon
(b) Xenon
(c) Argon
(d) Oxygen
Answer:
(d) Oxygen

Question 14.
Among the following groups which contains monoatomic gases?
(a) Group 17
(b) Group 18
(c) Group 1
(d) Group 15
Answer:
(b) Group 18

Question 15.
Which of the following is a tri atomic gas at room temperature?
(a) Oxygen
(b) Helium
(c) Ozone
(d) Nitrogen
Answer:
(c) Ozone

Question 16.
Which of the following gas is essential for our survival?
(a) N₂
(b) H₂
(c) O₂
(d) He
Answer:
(c) O₂

Question 17.
Among the following, which is deadly poison?
(a) CO₂
(b) HCN
(c) HCl
(d) NH₃
Answer:
(b) HCN

Question 18.
Which of the following is not chemically inert?
(a) Helium
(b) Oxygen
(c) Argon
(d) Krypton
Answer:
(b) Oxygen

Question 19.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 20.
Pressure of a gas is equal to ………..
(a) \(\frac {F}{a}\)
(b) F x a
(c) \(\frac {a}{F}\)
(d) F – a
Answer:
(a) \(\frac {F}{a}\)

Question 21.
The SI unit of pressure is ………..
(a) Nm^-2  Kg^-1
(b) Pascal
(c) bar
(d) atmosphere
Answer:
(b) Pascal

Question 22.
Statement-I: The pressure cooker takes more time for cooking at high altitude.
Statement-II: Air is subjected to Earth’s gravitational force. The pressure of air gradually decreases from the surface of the Earth to higher altitude.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(e) Statement-I is wrong but Statement-II is correct
(ð) Statement-I is correct but Statement-II is wrong
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 23.
The instrument used for measuring the atmospheric pressure is …………….
(a) lactometer
(b) barometer
(c) electrometer
(d) ammeter
Answer:
(b) barometer

Question 24.
The standard atmospheric pressure at sea level at 0°C is equal to ……………..
(a) 1 mm Hg
(b) 76 mm Hg
(c) 760 mm Hg
(d) 680 mm Hg
Answer:
(c) 760 mm Hg

Question 25.
Mathematical expression of Boyle’s law is ………….
(a) P₁V₁ = P₂V₂
(b) \(\frac {P}{V}\) = Constant
(c) \(\frac {V}{T}\) = Constant
(J) \(\frac {P}{T}\) = Constant
Answer:
(a) P₁V₁ = P₂V₂

Question 26.
Statement-I: If the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
Statement-II: If the volume is halved, the density of the gas is doubled.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(a) Statement-I and liare correct and Statement-II is the correct explanation of Statement-I

Question 27.
Which one of the following represents the Charles’ law?
(a) PV = Constant
(b) \(\frac {V}{T}\) = Constant
(c) VT Constant
(d) \(\frac {T}{V}\) = R
Answer:
(b) \(\frac {V}{T}\) = Constant

Question 28.
Which one of the following is absolute zero?
(a) 293 K
(b) 273 K
(c) – 273.15°C
(d) 0°C
Answer:
(c) – 273.15°C

Question 29.
\(\frac {P}{T}\) = Constant is known as …………..
(a) Boyle’s law
(b) Charles’ law
(c) Gay Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay Lussac’s law

Question 30.
The ideal gas equation is …………..
(a) PV = RT for 1 mole
(b) P₁ V₁ = P₂V₂
(c) \(\frac {P}{T}\) = R
(d) P = P₂ + P₂ + P₂
Answer:
(a) PV = RT for 1 mole

Question 31.
The value of Universal gas constant in a ideal gas equation is equal to ………….
(a) 8.3 14 KJ
(b) 0.082057 dm³ atm mol^-1 K^-1
(c) 1 Pascal
(d) 8.314 x 10^-2Pascal
Answer:
(b) 0.082057 dm³ atm mol^-1 K^-1

Question 32.
Mathematical expression of Graham’ s law is ……………….

Answer:

Question 33.
Which law is used in the isotopic separation of deuterium and protium?
(a) Boyle’s law
(b) Charles’ law
(c) Graham’ s law
(d) Gay Lussac’s law
Answer:
(c) Graham’ s law

Question 34.
The value of compression factor Z is equal to ………….
(a) \(\frac {nRT}{PV}\)
(b) \(\frac {PV}{RT}\)
(c) PV x nRT
(d) \(\frac {PV}{nRT}\)
Answer:
(d) \(\frac {PV}{nRT}\)

Question 35.
The value of critical volume is equal in terms of Vander Waals constant is ……….
(a) 3b
(b) \(\frac{8a}{27 Rb}\)
(c) \(\frac{a}{27 b^{2}}\)
(d) \(\frac{2a}{Rb}\)
Answer:
(a) 3b

Question 36.
The value of critical temperature of carbon dioxide is …………
(a) 273 K
(b) 303.98 K
(c) 373 K
(d) – 80°C
Answer:
(b) 303.98 K

Question 37.
Match the List-I and List-II using the correct code given below the list.


Answer:

Question 38.
The value of critical pressure of CO₂ is ……………….
(a) 173 atm
(b) 73 atm
(c) 1 atm
(d) 22.4 atm
Answer:
(b) 73 atm

Question 39.
The temperature below which a gas obey Joule Thomson effect is called …………..
(a) critical temperature
(b) standard temperature
(c) inversion temperature
(d) normal temperature
Answer:
(c) Inversion temperature

Question 40.
The substance used in adiabatic process of liquefaction is ……………
(a) liquid helium
(b) gadolinium sulphate
(c) iron sulphate
(d) liquid ammonia
Answer:
(b) Gadolinium sulphate

Question 41.
The temperature produced in adiabatic process of liquefaction is …………..
(a) zero kelvin
(b) -273 K
(c) 10^-4 K
(d) 10⁴ K
Answer:
(c) 10^-4 K

Question 42.
The molecules of a gas A travel four times faster than the molecules of gas B at same temperature. The ratio of molecular weight M_A/ M_B is ………….
(a) 1/16
(b) 4
(c) 1/4
(d) 16
Answer:
(a) 1/16

Question 43.
The compressibility factor for an ideal gas is …………….
(a) 1.5
(b) 2
(c) 1
(d) x
Answer:
(c) 1

Question 44.
Which of the following pair will diffuse at the same rate?
(a) CO₂ and N₂O
(b) CO₂ and NO
(c) CO₂ and CO
(d) N₂O and NO
Answer:
(a) CO₂ and N₂O

Question 45.
The value of Vander Waals constant “a” is maximum for ……………….
(a) helium
(b) nitrogen
(c) methane
(d) ammonia
Answer:
(d) Ammonia

Question 46.
A person living in Shimla observed that cooking food with using pressure cooker takes more time. The reason for this observation is that at high altitude …………..
(a) pressure increases
(b) temperature decreases
(c) pressure decreases
(a) temperature decreases
Answer:
(c) pressure decreases

Question 47.
Statement-I : At constant temperature PV vs V plot for real gases is not a straight line.
Statement-II : At high pressure, all gases have Z >1, but at intermediate pressure most gases have Z <1.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-Il is wrong
(d) Statement-I is wrong but Statement-Il is correct .
Answer:
(a) Statement-I and liare correct and Statement-II is the correct explanation of Statement-I

Question 48.
Statement-I: Gases do not liquefy above their critical temperature, even on applying high press ure.
Statement-II: Above critical temperature, the molecular speed is high and intermolecular
attractions cannot hold the molecules together because they escape because of high speed.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(c) Statement-I and II are correct and Statement-IIs the correct explanation of Statement-I

Question 49.
The rate of diffusion of a gas is ………….
(a) directly proportional to its density
(b) directly proportional to its molecular mass
(c) directly proportional to its square root of its molecular mass
(d) inversely proportional to its square root of its molecular mass
Answer:
(d) inversely proportional to its square root of its molecular mass

Question 50.
In a closed flask of 5 liters, 1.0 g of H₂ is heated from 300 to 600 K, which statement is not correct’?
(a) pressure of the gas increases
(b) the rate of the collusion increase
(c) the number of moles of gas increases
(d) the energy of gaseous molecules increases
Answer:
(c) the number of moles of gas increases

Question 51.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 52.
Consider the following statements.
(i) All the gases have higher densities than liquids and solids.
(ii) All gases occupy zero volume at absolute zero.
(iii) At very low pressure all gases exhibit ideal behaviour.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
Answer:
(a) (i) only

Question 53.
Which of the following gas is present maximum in atmospheric air?
(a) O₂
(b) N₂
(c) H₂
(d) radon
Answer:
(b) N₂

Question 54.
Which law is used in the process of enriching the isotope of U²³⁵ from other isotopes?
(a) Boyle’s law
(b) Dalton’s law of partial pressure
(c) Graham’s law of diffusion
(d) Charles’ law
Answer:
(c) Graham’s law of diffusion

Samacheer Kalvi 11th Chemistry Gaseous State 2 – Mark Questions

Question 1.
Identify the elements that are in gaseous state under normal atmospheric conditions.
Answer:

• Hydrogen, nitrogen, oxygen, fluorine and chlorine exist as gaseous diatomic molecules.
• Another form of oxygen namely ozone tr iatomic molecule exist as a gas at room temperature.
• Noble gases namely helium, neon, argon, krypton, xenon and radon are mono atomic gases.

Question 2.
Distinguish between a gas and a vapour.
Answer:

• Gas : A substance that is normally in a gaseous state at ordinary temperature and pressure. .,e.g., Hydrogen.
• Vapou r: The gaseous form of any substance that is a liquid or solid at normal temperature and pressure. e.g., At 298 K and 1 atm, water exist as water vapour.

Question 3.
Define pressure. Give its units.
Answer:

• Pressure is defined as the force exerted by a gas on unit area of the wall. Force F
• Pressure = \(\frac {Force}{Area}\) = \(\frac {F}{a}\)
• The SI unit of pressure is Pascal (Pa)

Question 4.
Define atmospheric pressure. What is its value?
Answer:

• The pressure exerted on a unit area of Earth by the colunm of air above it is called atmospheric pressure.
• The standard atmospheric pressure = 1 atm.
• 1 atm = 760 mm Hg.

Question 5.
Deep sea divers ascend slowly and breath continuously by time they reach the surface. Give reason.
Answer:

• For every 10 m of depth, a diver experiences an additional 1 atm of pressure due to the weight of water surrounding him.
• At 20 m, the diver experiences a total pressure of 3 atm. So the most important rule in diving is never hold breath.
• Divers must ascend slowly and breath continuously allowing the regulator to bring the air pressure in their lungs to 1 atm by the time they reach the surface.

Question 6.
Most aeroplanes cabins are artificially pressurized. Why?
Answer:
The pressure decreases with the increase in altitude because there are fewer molecules per unit volume of air. Above 9200 m (30,000 ft), for example, where most commercial aeroplanes fly, the pressure is so low that one could pass out for lack of oxygen. For this reason most aeroplanes cabins arc artificially pressurized.

Question 7.
What is the reason behind the cause of ear pain while climbing a mountain? How it can be rectified?
Answer:

• When one ascends a mountain in a plain, the external pressure drops while the pressure within the air cavities remains the same. This creates an imbalance.
• The greater internal pressure forces the eardrum to bulge outward causing pain.
• With time and with the help of a yawn or two, the excess air within your ear’s cavities escapes thereby equalizing the internal and external pressure and relieving the pain.

Question 8.
State Charles’ law.
Answer:
Charles’ law:
For a fixed mass of a gas constant pressure, the volume is directly proportional to temperature (K).
Mathematically V – T at constant P and n. (or) \(\frac {V}{T}\) = Constant (or) \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) = Constant

Question 9.
What are the applications of Charles’ law?
Answer:

• A hot air inside the balloon rises because of its decreased density and causes the balloon to float  inside the balloon rises because of its decreased density and causes the balloon to float.
• If you take a helium balloon outside on a chilly day, the balloon will crumble. Once you get back into warm area, the balloon will return to its original shape. This is because, in accordance with Charles’ law, a gas like helium takes up more space when it is warm.

Question 10.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. Mathematically V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant

Question 11.
Define Dalton’s law of partial pressure.
Answer:
Dalton’s law of partial pressure:
It states that the total pressure of a mixture of gases is the sum of partial pressures of the gases present.
P_total = P₁ + P₂ + P₃

Question 12.
What are the applications of Dalton’s law of partial pressure?
Answer:
1. Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab, the values are reported as partial pressures.
Gas – Normal range
P_CO2 35 – 45mm of Hg
P_O2  80 – 100 mm of Hg

2. When gas is collected by downward displacement of water, the pressure of dry vapour collected is computed using Dalton’s law

Question 13.
How can you identify a heavy smoker with the help of Dalton’s law?
Answer:
Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab. The values are reported as partial pressures.

A heavy smoker may be expected to have low O₂ and huge CO₂ partial pressures.

Question 14.
Define Graham’s law of diffusion.
Answer:
Graham’s law of diffusion:
The rate of diffusion or effusion is inversely proportional to the square root of molecular mass of a gas through an orifice.
\(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}}\)
r_Ar_B = rate of diffusion of gases A, B
M_A, M_B = Molecular mass of gases A, B