Category: Class 11

Students can Download Accountancy Chapter 1 Introduction to Accounting Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 1 Introduction to Accounting

Samacheer Kalvi 11th Accountancy Introduction to Accounting Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Samacheer Kalvi Guru 11th Accountancy Question 1.
The root of financial accounting system is ……………..
(a) Social accounting
(b) Stewardship accounting
(c) Management accounting
(d) Responsibility accounting
Answer:
(b) Stewardship accounting

11th Accountancy Chapter 1 Question 2.
Which one of the following is not a main objective of accounting?
(a) Systematic recording of transactions
(b) Ascertainment of the profitability of the business
(c) Ascertainment of the financial position of the business
(d) Solving tax disputes with tax authorities
Answer:
(d) Solving tax disputes with tax authorities

11th Accountancy 1st Chapter Question Answer Question 3.
Which one of the following is “hot a branch of accounting?
(a) Financial accounting
(b) Management accounting
(c) Human resources accounting
(d) None of the above
Answer:
(d) None of the above

Samacheer Kalvi 11th Accountancy Guide Question 4.
Financial position of a business is ascertained on the basis of ……………..
(a) Journal
(b) Trial balance
(c) Balance Sheet
(d) Ledger
Answer:
(c) Balance Sheet

Samacheer Kalvi Accountancy 11th Question 5.
Who is considered to be the internal user of the financial information?
(a) Creditor
(b) Employee
(c) Customer
(d) Government
Answer:
(b) Employee

II. Very Short Answer Questions

11th Accountancy Samacheer Kalvi Question 1.
Define accounting.
Answer:
American Accounting Association has defined accounting as “the process of identifying, measuring and communicating economic information to permit informed judgements and decisions by users of the information.”

Samacheer Kalvi 11th Accountancy Book Question 2.
List any two functions of accounting.
Answer:
The main functions of accounting:

1. Measurement
2. Forecasting

1. Measurement: The main function of accounting is to keep systematic record of transactions, post them in the ledger and ultimately prepare the final accounts.

2. Forecasting: With the help of the various tools of accounting, future performance and financial position of the business enterprises can be forecasted.

Accountancy Class 11 Pdf Samacheer Kalvi Question 3.
What are the steps involved in the process of accounting?
Answer:
Accounting is the systematic process of identifying, measuring, recording, classifying, summarising, interpreting and communicating financial information.

11th Accountancy 1st Chapter Question 4.
Who are the parties interested in accounting information?
Answer:

1. Internal users: Owners, Management and Employees.
2. External users: Creditors, Investors, Customers, Tax authorities, Government, Researchers and General Public.

Samacheer Kalvi 11th Accountancy Question 5.
Name any two basis of recording accounting information.
Answer:
There are three basis of accounting in common usage.

1. Cash basis
2. Accrual or mercantile basis

III. Short Answer Questions

Samacheer Kalvi 11th Accountancy Book Back Answers Question 1.
Explain the meaning of accounting.
Answer:
Accounting is the systematic process of identifying, measuring, recording, classifying, summarising, interpreting and communicating financial information. Accounting gives information on:

1. The resources available
2. How the available resources have been employed and
3. The results achieved by their use.

11th Accountancy Solutions Samacheer Kalvi Question 2.
Discuss briefly the branches of accounting.
Answer:
The main branches of accounting are:
1. Financial Accounting: It involves recording of financial transactions and events.

2. Cost Accounting: It involves the collection, recording, classification and appropriate allocation of expenditure for the determination of the costs of products or services and for the presentation of data for the purpose of cost control and managerial decision making.

3. Management Accounting: It is concerned with the presentation of accounting information in such a way as to assist management in decision making and in the day – to – day operations of an enterprise.

4. Social Responsibility Accounting: It is concerned with presentation of accounting information by business entities and other organisations from the view point of the society by showing the social costs incurred such as environmental pollution by the enterprise and social benefits such as infrastructure development and employment opportunities created by them.

5. It is concerned with identification, quantification and reporting of investments made in human resources of an enterprise.

11th Accountancy Book Samacheer Kalvi Question 3.
Discuss in detail the importance of accounting.
Answer:
The importance of accounting is:
1. Systematic records: All the transactions of an enterprise which are financial in nature are recorded in a systematic way in the books of accounts.

2. Preparation of financial statements: Results of business operations and the financial position of the concern can be ascertained from accounting periodically through the preparation of financial statements.

3. Assessment of progress: Analysis and interpretation of financial data can be done to assess the progress made in different areas and to identify the areas of weaknesses.

4. Aid to decision making: Management of a firm has to make routine and strategic decisions while discharging its functions.

5. Satisfies legal requirements: Various legal requirements like maintenance of provident fund (PF) for employees, Tax deducted at source (TOS), filing of tax returns and properly fulfilled with the help of accounting.

6. Information to interested groups: Accounting supplies appropriate information to different interested groups like owners, management, creditors, employees, financial institutions, tax authorities and the Government.

7. Legal evidence: Accounting records are generally accepted as evidence in courts of law and other legal authorities in the settlement of disputes.

8. Computation of tax: Accounting records are the basic source for computation and settlement of income tax and other taxes.

9. Settlement during mergers: When two or more business units decide to merge, accounting records provide information for deciding the terms of merger and any compensation payable as a consequence of merges.

Account 11th 1st Chapter Question 4.
Why are the following parties interested in accounting information?

1. Investors
2. Government

Answer:
1. Investors: Persons who are interested in investing their funds in an organisation should know about the financial condition of a business unit while making their investment decisions. They are more concerned about future earnings and risk bearing capacity of the organisation which will affect the return to the investors.

2. Government: The scarce resources of the country are used by business enterprises. Information about performance of business units in different industries helps the government in policy formation for development of trade and industry, allocation of scarce resources, grant of subsidy, etc.

11th Accounts Samacheer Kalvi Solutions Question 5.
Discuss the role of an accountant in the modem business world.
Answer:
The important role of an accountant is:
1. Record keeper: The accountant maintains a systematic record of financial transactions.

2. Provider of information to the management : The accountant assists the management by providing financial information required for decision making and for exercising controls.

3. Protector of business assets: The accountant maintains records of assets owned by the business which enables the management to protect and exercise control over these assets.

4. Financial advisor: The accountant analysis financial information and advises the business managers regarding investment opportunities, strategies for cost savings, capital budgeting, provision for future growth and development, expansions of enterprise, etc.

5. Tax managers: The accountant ensures that tax returns are prepared and filed correctly on time and payment of tax is made on time

6. Public relation officer: The accountant provides accounting informations to various interest users for analysis as per their requirements.

Textbook Case Study Solved

Samacheer Kalvi 11th Accountancy Solutions Question 1.
How do SHGs maintain their accounting?
Answer:
They are maintaining single entry systems, so the cash book and personal accounts are being maintained by them. The nominal accounts are not being maintained by them, so we can not find out the accurate profit or loss on the business. We can find out the estimated value only.

11th Accountancy Guide Samacheer Kalvi Question 2.
Do you think that financial accounting system is suitable for all businesses?
Answer:
This system is not suitable for all business because the actual profit or loss cannot be found out by this system. If it is a small business only then we can follow this system.

Samacheer Kalvi 11th Accountancy Introduction to Accounting Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

11th Accounts Chapter 1 Question 1.
In 1494, ……………. an Italian developed double – entry book – keeping system.
(a) Luca Pacioli
(b) Kautilya
(c) Wheeler
(d) R.N.Cart
Answer:
(a) Luca Pacioli

Accountancy Class 11 Samacheer Kalvi Question 2.
Who is considered to be the external user of the financial information?
(a) Employee
(b) Owners
(c) Management
(d) Creditor
Answer:
(d) Creditors

11th Accountancy Chapter 1 Book Back Answers Question 3.
Which one is not a role of an accountant?
(a) Record keeper
(b) Tax manager
(c) PRO
(d) Owner
Answer:
(d) Owner

11th Accounts Book Samacheer Kalvi Question 4.
The first step of accounting cycle.
(a) Transactions
(b) Journlising
(c) Profit & loss account
(d) Trading account
Answer:
(a) Transactions

Samacheer Kalvi 11th Accountancy Book Pdf Question 5.
Original entry is otherwise called ………………
(a) Journal
(b) Ledger
(c) Trial balance
(d) Balance sheet
Answer:
(a) Journal

Accountancy 1st Chapter Question 6.
……………… is the language of business.
(a) Accounting
(b) Book – keeping
(c) Trade
(d) Banking
Answer:
(a) Accounting

Accountancy Book Class 11 Samacheer Kalvi Question 7.
Financial information for managerial decision making caused emergence of ……………… accounting.
(a) Management
(b) Cost
(c) Financial
(d) Corporate
Answer:
(a) Management

Samacheer Kalvi 11th Accounts Question 8.
Transferring the entries from the journal to the ledger ……………….
(a) Posting
(b) Journal
(c) Ledger
(d) Transaction
Answer:
(a) Posting

Question 9.
The balance in the trading account is the gross profit or ……………….
(a) Net profit
(b) Net loss
(c) gross loss
(d) balance
Answer:
(c) gross loss

Question 10.
A statement showing the balances of assets and liabilities is called as ……………….
(a) Profit & loss A/c
(b) Trading A/c
(c) Balance sheet
(d) Final A/c
Answer:
(c) Balance sheet

Question 11.
Two or more business units forming a single entity is known as ……………….
(a) Joint
(b) Merger
(c) Link
(d) Compound
Answer:
(b) Merger

Question 12.
………………. is irrecoverable debt ……………….
(a) debtor
(b) creditor
(c) bad debt
(d) loan
Answer:
(c) bad debt

Question 13.
Unsold goods lying in a business on a particular date are known as ……………….
(a) Stock
(b) creditor
(c) debtor
(d) cash
Answer:
(a) Stock

Question 14.
………………. is the incapability of a person or an enterprises to pay the debts
(a) Asset
(b) Liability
(c) Insolvency
(d) Sales
Answer:
(c) Insolvency

Question 15.
………………. is the amount incurred in order to produce and sell the goods and services.
(a) Creditor
(b) Debtor
(c) Stock
(d) Expense
Answer:
(d) Expense

II. Very Short Answer Questions

Question 1.
Write any two objectives of Accounting.
Answer:

1. To keep a systematic record of financial transactions and events.
2. To ascertain the profit or loss of the business enterprise.

Question 2.
Who are researchers?
Answer:
Researchers who carry out their research can use accounting information and make use of the published financial statements for analysis and evaluation.

Question 3.
Who is Public Relation Officer? (PRO)
Answer:
The accountant provides accounting information to various interested users for analysis as per their requirements.

Question 4.
What do you mean by legal evidence?
Answer:
Accounting records are generally accepted as evidence in courts of law and other legal authorities in the settlement of disputes.

Question 5.
Who is a creditor of business?
Answer:
A person who gives a benefit without receiving money or money’s worth immediately but to claim in future.

Question 6.
What is a bad debt?
Answer:
It is a loss to the business arising out of failure of a debtor to pay the dues. It is irrecoverable debt.

Question 7.
What is capital?
Answer:
It is the amount invested by the owner or proprietor in an organisation.

Question 8.
What is drawings?
Answer:
It is the amount of cash or value of goods, assets, etc., withdrawn from the business by the owner for the personal use of the owner.

Question 9.
What is a voucher?
Answer:
Any written or printed document in support of a business transaction is called a voucher.

Question 10.
What is depreciation?
Answer:
It refers to the gradual reduction in the value of fixed assets due to usage and passage of time.

III. Short Answer Questions

Question 1.
Write a note on

1. Debtor
2. Creditor

Answer:

1. Debtor: A person who receives a benefit without giving money or money’s worth immediately, but liable to pay in future or in due course of time.
2. Creditor: A person who gives a benefit without receiving money or money’s worth immediately but to claim in future.

Question 2.
Write a note on

1. Purchases
2. Sales.

Answer:

1. Purchases: Buying of goods with the intention of resale is called purchase.
2. Sales: When goods meant for resale are sold, it is called sales.

Question 3.
What is the difference between cash transaction and credit transactions?
Answer:
Cash transaction:
It is a transaction which involves immediate cash receipt or immediate cash payment.

Credit transaction:
It is a transaction in which cash is not received or paid immediately, but will be received or paid later.

Question 4.
What is the difference between voucher and invoice?
Answer:
Voucher:
Any written or printed document in support of business transaction is called a voucher. Examples: cash receipts, bank pay – in – slip, etc.

Invoice:
It is a statement prepared by a seller of goods to be sent to the buyer. It shows details of quantity, price, value, etc., of the goods and any discount given, finally showing the net amount payable by the buyer.

Students who are in search of 11th Bio Zoology exam can download this Tamilnadu State Board Solutions for 11th Bio Zoology Chapter 1 The Living World from here for free of cost. These cover all Chapter 1 The Living World Questions and Answers, PDF, Notes, Summary. Download the Samacheer Kalvi 11th Biology Book Solutions Questions and Answers by accessing the links provided here and ace up your preparation.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 1 The Living World

Kickstart your preparation by using this Tamilnadu State Board Solutions for 11th Bio Zoology Chapter 1 The Living World Questions and Answers get the max score in the exams. You can cover all the topics of Chapter 1 The Living World Questions and Answers easily after studying the Tamilnadu State Board 11th Bio Zoology Textbook solutions pdf.

Samacheer Kalvi 11th Bio Zoology The Living World Text Book Back Questions and Answers

Multiple Choice Question and Answers
Choose the correct answer
Zoology 11th Class First Chapter Question 1.
A living organism is differentiated from non-living structure based on
(a) Reproduction
(b) Growth
(c) Metabolism
(d) AH the above
Answer:
(c) Metabolism

11th Zoology 1st Chapter Question 2.
A group of organisms having similar traits of a rank is
(a) Species
(b) Taxon
(c) Genus
(d) Family
Answer:
(a) Species

Samacheer Kalvi Guru 11th Zoology Question 3.
Every unit of classification regardless of its rank is
(a) Taxon
(b) Variety
(c) Species
(d) Strain
Answer:
(a) Taxon

Zoology Class 11 Chapter 1 Question 4.
Which of the following is not present in same rank?
(a) Primata
(b) Orthoptera
(c) Diptera
(d) Insecta
Answer:
(a) Primata

Zoology Chapter 1 Class 11 Question 5.
What taxonomie aid gives comprehensive information about a taxon?
(a) Taxonomic Key
(b) Herbarium
(c) Flora
(d) Monograph
Answer:
(a) Taxonomic Key

Samacheer Kalvi 11 Zoology Solutions Question 6.
Who coined the term biodiversity?
(a) Walter Rosen
(b) AG Tansley
(c) Aristotle
(d) AP de Candole
Answer:
(a) Walter Rosen

Samacheer Kalvi 11 Bio Zoology Solutions Question 7.
Cladogram considers the following characters –
(a) Physiological and Biochemical
(b) Evolutionary and Phylogenetic
(c) Taxonomie and systematic
(d) None of the above
Answer:
(b) Evolutionary and Phylogenetic

Samacheer Kalvi 11th Zoology Solutions Question 8.
Molecular taxonomic tool consists of –
(a) DNA and RNA
(b) Mitochondria and Endoplasmic reticulum
(c) Cell wall and Membrane proteins
(d) All the above
Answer:
(a) DNA and RNA

Samacheer Kalvi 11th Zoology Question 9.
Differentiate between probiotic and pathogenic bacteria.

Samacheer Kalvi Guru 11th Bio Zoology Question 10.
Why mule is sterile in nature?
Answer:
Mule gets one set of chromosomes (32) from male parent, horse and one set of chromosomes (31) from female parent, donkey. These two sets of chromosomes do not match with each other and cannot produce gametes by meiosis. Hence mule is sterile in nature.

Class 11 Zoology Chapter 1 Notes Question 11.
List any five salient features of the family Felidae.
Answer:

• The species in the Felidae family are carnivores or omnivores.
• Felids are generally solitary with a few exceptions.
• They have sharp vision, hearing and a strong sense of smell.
• They have short faces and their paws have toes in the 5 forefeet and 4 toes in the hind feet.
• Most Felids live in the wild e.g. cat, tiger, lion, cheetah.

The Living World Class 11 Pdf Question 12.
What is the role of Charles Darwin in relation to concept of species?
Answer:
Charles Darwin’s book on Origin of Species explains the evolutionary connections of species by the process of natural selection.

Class 11 Zoology Chapter 1 Question 13.
Why elephants and other wild animals are entering into human living area?
Answer:
Elephants and other wild animals enter into human living area because of the loss of their habitat, deforestation, mono-culture vegetation by destroying forests

11th Bio Zoology Samacheer Kalvi Question 14.
What is the difference between a Zoo and wild life sanctuary?
Answer:

Zoology Chapter 1 Question 15.
Can we use recent molecular tools to identify and classify organisms?
Answer:
The recent molecular taxonomical tools can be used to identify and classify the organism. The following molecular techniques and approaches are used in molecular tools.

(a) DNA bar coding – Short genetic marker in an organism’s DNA to identify whether it belongs to a particular species.

(b) DNA hybridization – Measures the degree of genetic similarity between pools of DNA sequences.

(c) DNA finger printing – to identify an individual from a sample of DNA by looking at unique patterns in their DNA.

(d) Restriction Fragment Length Polymorphism (RFLP) Analysis- difference in homologous DNA sequences can be detected by the presence of fragments of different lengths after digestion of DNA samples.

(e) Polymerase chain reaction (PCR) sequencing- to amplify a specific gene, or portion of gene.

Samacheer Kalvi 11th Bio Zoology Solutions Question 16.
Explain the role of Latin and Greek names in Biology.
Answer:
Aristotle (384 to 322 BC) was the first to classify all animals in his Historia Animalium in Latin. He classified the living organisms into plants and animals. Animals were classified as walking (terrestrial), flying (birds), and swimming (aquatic) based in their locomotion.

He classified the animals with red blood cells as Enaima and those without red blood cells as Anaima. Though his method of classification had limitation, his contribution to biology was remarkable. Theophrastus did his research on the classification of plants. He was known as the Father of Botany.

In-Text Questions Solved

11th Zoology Solutions Question 1.
If you find an animal with four legs, with two eyes, paired ear pinna, covered with fur, possessing mammary gland, which class will you position it? How will you give a binomial name, if you are the first person to discover and report that animal.
Answer:
Do it yourself.

Question 2.
What may be the reasons for the extinction of Dinosaurs? If you know the reasons for their extinction, why Sparrows are listed as endangered species?
Answer:
The extinction of the dinosaurs is an enigma that has captivated scientists for well over a century. We find the fossilized remains of giant reptiles all over the earth.

Yet we do not see any of the creatures alive today. If sparrows are not there the population of birds of prey may also be affected. Apart from this, every constitute of an ecosystem is important from an ant to an elephant. We are eliminating species by species which are important links which make the web of life. Today it’s these species which are getting extinct.

Textbook Activity Solved

The main objective of this activity is to check the students understanding about animals and its characteristics before learning the lesson. Observe the picture given below, identify the animals and classify them according to you own understanding write one character about each class of animals.

Take the students to the school ground and ask them to observe and identify few invertebrates (insects, earthworm, spiders etc). Ask the students to write few characteristics of each animal which they have observed.

Entrance Examination Questions Solved
Choose the correct answer:
Question 1.
The smallest taxon among the following is ………. (PMT-94)
(a) class
(b) order
(c) species
(d) genus
Answer:
(c) species

Question 2.
Taxonomically a species is ……….. (PMT-94)
(a) A group of evolutionary related population
(b) A fundamental unit in the phylogeny of organisms
(c) Classical evolutionary taxonomy
(d) A community taken into consideration. An evolutionary base.
Answer:
(b) A fundamental unit in the phylogeny of organisms

Question 3.
Species is
(a) not related to evolution
(b) specific class of evolution
(c) specific unit of evolution
(d) fertile specific unit in the evolutionary history of a race
Answer:
(d) fertile specific unit in the evolutionary history of a race

Question 4.
Two words comprising the binomial nomenclature are ………… (DPMT-96)
(a) Family & genus
(b) order & family
(c) genus & species
(d) species & variety
Answer:
(c) genus & species

Question 5.
A group of plants or animals with similar traits of any rank is kept under ………. (PMT-96)
(a) species
(b) genus
(c) order
(d) taxon
Answer:
(d) taxon

Question 6.
Which of the following is the correct sequence in the increasing order of complexity? (PMT-97)
(a) molecules, tissues, community, population
(b) cell, tissues, community, population
(c) tissues, organisms, population, community
(d) molecules, tissues, community, cells
Answer:
(c) tissues, organisms, population, community

Question 7.
New systematic and the concept of life was given by ……….. (BHU-98)
(a) Huxley
(b) Odom
(c) Elton
(d) Linnaeus
Answer:
(c) Elton

Question 8.
Two organisms of same class but different families will be kept under the same ……… (CET-98)
(a) genera
(b) species
(c) order
(d) family
Answer:
(c) order

Question 9.
Which of the following will form a new species? (PMT-98)
(a) inter breeding
(b) variations
(c) differential reproduction
(d) none of the above
Answer:
(b) variations

Question 10.
A community includes ……….. (CET-98)
(a) a group of same genera
(b) a group of same population
(c) a group of individuals from same species
(d) different populations interacting with each other
Answer:
(d) different populations interacting with each other

Question 11.
Binomial nomenclature was given by …………(BHU-97)
(a) Huxley
(b) Ray
(c) Darwin
(d) Linnaeus
Answer:
(d) Linnaeus

Question 12.
In classification the category below the level of family is ……….. (CET-98)
(a) class
(b) species
(c) phylum
(d) genus
Answer:
(d) genus

Question 13.
Taxon is …………. (CET-2000)
(a) species
(b) unit of classification
(c) highest rank in classification
(d) group of closely related
Answer:
(b) unit of classification

Question 14.
One of the following includes most closely linked organisms ………… (PMT-2001)
(a) species
(b) genus
(c) family
(d) class
Answer:
(a) species

Question 15.
Which of the following taxons cover a greater number of organisms?(PMT-2001)
(a) order
(b) family
(c) genus
(d) phylum
Answer:
(d) phylum

Question 16.
Inbreeding is possible between two members of ………… (AMU-2005)
(a) order
(b) family
(c) genus
(d) species
Answer:
(d) species

Question 17.
Which of these is correct order of hierarchy? (WARDHA-2002)
(a) kingdom, division, phylum, genus & species
(b) phylum, division, genus & class
(c) kingdom, genus, class, phylum & division
(d) phylum, kingdom, genus, species &class
Answer:
(a) kingdom, division, phylum, genus & species

Question 18.
Which is not a unit of taxonomic category? (BVP-2002)
(a) series
(b) glumaceae
(c) class
(d) phylum
Answer:
(b) glum aceae

Question 19.
Which is the first step of taxonomy? (MGIMS-2002)
(a) nomenclature
(b) classification
(c) identification
(d) hierarchical arrangement
Answer:
(c) identification

Question 20.
The five kingdom classification was given by ………… (BYP-2002)
(a) Whittaker
(b) Linnacus
(c) Copeland
(d) Haeckel
Answer:
(a) Whittaker

Question 21.
Taxon includes ………… (PMT-2002)
(a) Genus and species
(b) kingdom and division
(c) all ranks of hierarchy
(d) none of the above
Answer:
(c) all ranks of hierarchy

Question 22.
Binomial nomenclature refers to …………. (CET – 2000)
(a) Two names of a species
(b) one specific and one local name of a species
(c) two words for the name of a species
(d) two life cycles of a organism
Answer:
(c) two words for the name of a species

Question 23.
Carl Linnaeus is famous for ………… (GGSPU-2002)
(a) coining the term ‘systematics’
(b) introducing binomial nomenclature
(c) giving all natural system of classification
(d) all of these
Answer:
(b) introducing binomial nomenclature

Question 24.
True species are …………
(a) interbreeding
(b) sharing the same niche
(c) feeding on the same food
(d) reproductively isolated
Answer:
(d) reproductively isolated

Question 25.
The smallest unit of classification is ………… (GGSPU-2002)
(a) species
(b) sub – species
(c) class
(d) genus
Answer:
(a) species

Question 26.
Who coined the term ‘taxonomy’? (BVP-2003)
(a) Candolle
(b) Waksman
(c) Leuwenhoek
(d) Louis Pasteur
Answer:
(a) Candolle

Question 27.
Basic unit of classification of organisms is ………… (CET-2003)
(a) species
(b) population
(c) class
(d) family
Answer:
(a) species

Question 28.
The unit of classification containing concrete biological entities is ……….. (WARDHA-2003)
(a) taxon
(b) species
(c) category
(d) order
Answer:
(a) taxon

Question 29.
Species are considered as …………..
(a) real basic units of classification
(b) the lowest units of classification
(c) artificial concept of human mind which cannot be defined in absolute terms
(d) real units of classification devised by taxonomists
Answer:
(a) real basic units of classification

Question 30.
The living organisms can be unexceptionally distinguished from the non-living things on the basis of their ability for …………
(a) interaction with the environment and progressive evolution
(b) reproduction
(c) growth and movement
(d) responsiveness to touch
Answer:
(b) reproduction

Question 31.
Taxonomic category arrange in descending order ……….. (MH-01)
(a) key
(b) hierarchy
(c) taxon
(d) taxonomic category
Answer:
(d) taxonomic category

Question 32.
In which of the animal dimorphic nucleus is found? (PMT 2002).
(a) Amoebaproteus
(b) Trypanosoma gambiense
(c) Plasmodium vivax
(d) Paramecium caudatum
Answer:
(d) Paramecium caudatum

Question 33.
When a fresh-water protozoan possessing a contractile vacuole, is placed in a glass containing marine water, the vacuole will ………. (PMT 2004)
(a) increase in number
(b) disappear
(c) increase in size
(d) decrease in size
Answer:
(d) decrease in size

Question 34.
Which form of reproduction is correctly matched? (AIIMS 2007)
(a) Euglena transvers binary fission
(b) Paramecium longitudinal binary fission
(c) Amoeba multiple fission
(d) Plasmodium binary fission
Answer:
(c) Amoeba multiple fission

Question 35.
The presence of two types of nuclei, a macronucleus and a micronucleus, is characteristic of protozoans are grouped under the class ……….. (BHU 1994, 1999)
(a) sporozoa
(b) flagellate
(c) sarcodina
(d) ciliata
Answer:
(d) ciliata

Question 36.
Which class of protozoa is totally parasitic? (BHU 1994)
(a) Sporozoa
(b) Mastigophora
(c) Ciliate
(d) Sarcodina
Answer:
(a) Sporozoa

Question 37.
Reproduction in Paramecium is controlled by ………. (BHU 1999).
(a) flagella
(b) cell wall
(c) micronucleus
(d) macronucleus
Answer:
(c) micronucleus

Question 38.
In the life cycle of Plasmodium exflagellation occurs in ……….. (BHU 2007)
(a) sporozoties
(b) microgametes
(c) macrogametes
(d) signet ring
Answer:
(b) microgametes

Question 39.
Excretion in Amoeba occurs through ………… (DPMT 1997)
(a) lobopodia
(b) plasma membrane
(c) uroid portion
(d) contractile vacuole
Answer:
(d) contractile vacuole

Question 40.
Method of dispersal in Amoeba is ……….. (DPMT 1995)
(a) locomotion
(b) encystment
(c) sporulation
(d) binary fission
Answer:
(b) encystment

Question 41.
Mode of feeding in free living protozoans is …………. (DPMT 2007).
(a) holozoic
(b) saprozoic
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)

Question 42.
Infection of Entamoeba is caused ………… (UP- CPMT 1996, 1999).
(a) by kissing
(b) by wearing clothes of patient
(c) by contaminated food
(d) none of these
Answer:
(c) by contaminated food

Question 43.
Choose the correct statement.
(a) All reptiles have a three chambered heart.
(b) All Pisces have gills covered by a operculum
(c) All mammals are viviparous
(d) All cyclostomes do not posses jaws and paired fin
Answer:
(d) All cyclostomes do not posses jaws and paired fin

Question 44.
Which of the following characteristics is mainly responsible for diversification of insects on land?
(a) Segmentation
(b) Bilateral symmetry
(c) Exoskeleton
(d) Eyes
Answer:
(c) Exoskeleton

Question 45.
The primitive prokaryotes responsible for the production of biogas from the ruminant animals. Include the …………. (2016)
(a) Thermoacidophiles
(b) methanogens
(c) Eubacteria
(d) Halophiles.
Answer:
(b) methanogens

Question 46.
Methanogens belong to ………. (2016)
(a) Dinoflagellates
(b) Slime moulds
(c) Eubacteria
(d) Archaebacteria
Answer:
(d) Archaebacteria

Samacheer Kalvi 11th Bio Zoology The Living World Additional Questions Solved

Multiple Choice Questions
Choose the correct answer:
Question 1.
Which of the following is not correct with regard to ecosystem?
(a) It includes living and non-living things.
(b) It shows interrelationship among living things.
(c) It shows interrelationship among living and non-living things.
(d) It has a large number of species.
Answer:
(b) It shows interrelationship among living things.

Question 2.
Who was known as the father of Botany?
(a) Aristotle
(b) Carolus Linnaeus
(c) John Ray
(d) Theophrastus
Answer:
(d) Theophrastus

Question 3.
Who developed the scientific system of Taxonomy and Binomial nomenclature?
(a) R.H. Whittaker
(b) Carolus Linnaeus
(c) John Ray
(d) Charles Darwin
Answer:
(b) Carolus Linnaeus

Question 4.
The system of classification of organisms based on evolutionary and genetic relationship among organisms is called as –
(a) Natural system of classification
(b) Phylogenetic system of classification
(c) Binomial classification
(d) Trinomial classification
Answer:
(b) Phylogenetic system of classification

Question 5.
The bacteria which live in salty environment are called –
(a) Theromoacidophiles
(b) Methanogens
(c) Halophiles
(d) Pathogens
Answer:
(c) Halophiles

Question 6.
Which of the following is the characteristics of the domain Bacteria?
(a) They are eukaryotic organisms
(b) They have true nucleus and membrane bound organelles
(c) The ribosomes are of 80 s type in the cytoplasm
(d) They have membrane bound 70 s type ribosomes.
Answer:
(d) They have membrane bound 70 s type ribosomes.

Question 7.
Which of the following is not correct with regard to species?
(a) They have similar morphological features
(b) They are reproductively isolated
(c) They produce viable young ones
(d) They have similar anatomical features
Answer:
(b) They are reproductively isolated

Question 8.
The cross between male lion and female tiger results in the production of –
(a) Mule
(b) Tigon
(c) Liger
(d) Hinny
Answer:
(c) Liger

Question 9.
The related families constitute –
(a) Clans
(b) Phylum
(c) Genus
(d) Order
Answer:
(d) Order

Question 10.
Which of the following is not related to scientific names of organisms?
(a) There is only one name for a species.
(b) They are universally accepted.
(c) They are named based on the guidelines of the ICZN.
(d) They are not unique to an organism.
Answer:
(d) They are not unique to an organism.

Question 11.
Naja naja is the zoological name of –
(a) Indian house crow
(b) Indian cobra
(c) Emerald dove
(d) Indian pea fowl
Answer:
(b) Indian cobra

Question 12.
Origin of species is the book written by –
(a) John Ray
(b) Charles Darwin
(c) Linnaeus
(d) Aristotle
Answer:
(b) Charles Darwin

Question 13.
Taxonomical keys are useful to study –
(a) Similarities and dissimilarities of animals
(b) Preserved plants and animals
(c) Animals kept in protected environments
(d) Plant taxonomy
Answer:
(a) Similarities and dissimilarities of animals

Question 14.
DNA hybridization is helpful to –
(a) Identify an individual from a sample of DNA
(b) Detect difference in homologous DNA sequences
(c) Measure the degree of genetic similarity between pools of DNA sequences
(d) Amplify a specific gene
Answer:
(c) Measure the degree of genetic similarity between pools of DNA sequences

Question 15.
DAISY, the cyber tool is –
(a) Digital automated identification system
(b) Digital automated information system
(c) Direct automated identification system
(d) Digital automated identification science
Answer:
(a) Digital automated identification system

Question 16.
INOTAXA is an e-Taxonomic resource useful for –
(a) Studying electron microscopic images to study molecular structures of cell organelles.
(b) Digital images and description about the species.
(c) Studying behaviour of organisms
(d) Identifying digital resources of animals.
Answer:
(b) Digital images and description about the species.

II. Give Reasons

Question 1.
Classification of organisms is necessary.
Answer:
Classification of organisms is necessary to recognise, identify them and differentiate closely related species.

Question 2.
Extremophiles inhabiting volcanic vents prepare food.
Answer:
Extremophiles inhabiting volcanic vents prepare food without sunlight and oxygen by utilizing hydrogen peroxide and other chemicals through chemosynthesis.

Question 3.
The mating between difference species produces sterile off springs.
Answer:
The maternal and paternal chromosomes of the off springs produced by the mating between different species are not identical and hence gametes are not produced by meiotic division.

Question 4.
Nomenclature of organisms is necessary.
Answer:
The unique nomenclature for each organisms is necessary as it is uniform in all countries and in all languages. A scientific name refers to only one organism.

Question 5.
Why are molecular tools used now to study taxonomy?
Answer:
Molecular tools are accurate and authentic. Hence they are used to study taxonomy.

III. Match the following
Question 1.

Answer:

1. (d) A.G. Tansley
2. (c) AP de Candolle
3. (a) Walter Rosen
4. (e) John Ray
5. (b) Carolus Linnaeus

Question 2.

Answer:

1. (d)  Extreme conditions
2. (a)  Beneficial
3. (e) disease causing
4. (c)  membrane bound organelles
5. (b)  70 s type of ribosomes

Question 3.

Answer:

1. (d)  Jungle cat
2. (e) Emerald dove
3. (a) Pea fowl
4. (b) Red Panda
5. (c)  Tiger

Question 4.

Answer:

1. (d) Aristotle
2. (c) John Ray
3. (a)  Charles Darwin
4. (e) Carl Woese
5. (b) R.H. Whittaker

Question 5.

Answer:

1. (d) Preserved plants and animals
2. (b) Similarities and dissimilarities of organisms.
3. (e) Plant Taxonomy
4. (c) Difference in homologous DNA sequence
5.  (a) Analysis of DNA

Short Answer Questions

Question 1.
Define ecosystem.
Answer:
Ecosystem is defined as a community of living organisms (plants and animals), non-living things (minerals, climate, soil, sunlight and water) and their interrelationships, e.g. Forest and grassland.

Question 2.
What is biodiversity?
Answer:
The presence of a large number of species in a particular ecosystem is called biodiversity.

Question 3.
Distinguish between living and non-living things.
Answer:

Question 4.
What is the need for classification?
Answer:
The basic need for classification are:

• To identify and differentiate closely related species.
• To know the variation among the species.
• To understand the evolution of the species.
• To create a phylogenetic tree among the different groups.
• To conveniently study living organisms.

Question 5.
Define Taxonomy.
Answer:
It is the branch of science of identifying, describing, naming and classifying organisms.

Question 6.
What are the features of systematics?
Answer:

• Identifying, describing, naming, arranging, preserving and documenting the organisms.
• Investigating evolutionary history of the species, their adaptations to the environment and the interrelationship among species.

Question 7.
How did Aristotle classify organisms?
Answer:
Aristotle classified living organisms into plants and animals. Animals were classified into walking (terrestrial), flying (birds) and swimming (aquatic) based on their locomotion. Based on the presence or absence of red blood animals were classified into Enaima and Anaima.

Question 8.
What are the limitations of Aristotle’s classification?
Answer:
Many organisms were not fitting into his classification. Frogs have lungs and they are amphibians while their larva, tadpole is aquatic and respires through gills. It is difficult to classify frogs according to his method. All flying organisms such as birds, bats, flying insects were grouped together. Ostrich, emu and penguin are flightless birds and hence they cannot be classified by his method.

Question 9.
What is natural system of classification?
Answer:
Linnaeus considered a few characters of organisms for classifying them. Later many characters were considered for classifying organisms. Morphological, anatomical and embryological characters were considered.

Question 10.
What is numerical taxonomy?
Answer:
The evaluation of resemblances and differences of organisms through statistical methods followed by computer analysis to establish the numerical degree of relationship among them is known as numerical taxonomy.

Question 11.
What is phylogenetic classification or cladistics?
Answer:
The classification of organisms based on evolutionary and genetic relationship among them is known as phylogenetic classification.

Question 12.
What is cladogram?
Answer:
A tree diagram which represents the evolutionary relationship among organisms is known as cladogram.

Question 13.
What is cladistic classification?
Answer:
Cladistic classification is the method of classifying the organisms based on genetic differences among all species in a phylogenetic tree.

Question 14.
What is the significance of cladistic classification?
Answer:
Cladistic classification takes into account ancestral characters (traits commons for the entire group) and derived characters (traits whose structure and function differ from the ancestral characters). The accumulation of derived characters resulted in the formation of new subspecies.

Question 15.
Explain five kingdom classification of Whittaker.
Answer:
R.H. Whittaker (1969) proposed the Five kingdom classification. They included Monera, Protista, Fungi, Plantae and Animalia. He classified organisms based on cell structure, nutrition, reproduction and phylogenetic relationships.
Kinds of kingdom:

Question 16.
Describe three domain classification.
Answer:
Three domain classification was proposed by Carl Woese (1977) and his co-workers. They classified organisms based on the difference in 16s rRNA genes. This adds the taxon domain higher than the kingdom. In this system, prokaryotes are divided into two domains-bacteria and Arachaea.

All eukaryotes are placed under the domain Eukarya. Archae appears to have common features with Eukarya. Archaea differ from bacteria in cell wall composition and differ from bacteria and eukaryotes in membrane composition and RNA types.

Question 17.
Distinguish Archaea, Bacteria and Eukarya.
Answer:

Question 18.
Classify organisms on the basis of seven kingdom system.
Answer:

Question 19.
Define species.
Answer:
Species is a group of organisms that have similar morphological and physiological characters which can interbreed to produce fertile off springs.

Question 20.
Show the mating with closely related species.
Answer:

• Male Donkey – Female Horse – Mule
• Male Horse – Female donkey – Hinny
• Male lion – female tiger – Liger
• Male tiger – Female lion – Tiger

Question 21.
Distinguish between species and genus.
Answer:
Species:

• Species is a group of interbreeding population having similar characters.
• It is the basic unit of classification e.g. Felis domestica, genus species.

Genus:

• Genus is a group of related species.
• It is the second level in classification, e.g., Felis margarita genus species.

Question 22.
Distinguish between Family and Order.
Answer:
Family:

• Family is a group of related genera.
• It is the lower taxon e.g. felidae

Order:

• Order is a group of related families.
• It is the higher taxon e.g. Carnivora

Question 23.
Distinguish between Class and Phylum.
Answer:
Class:

1. The one or more related orders with some common characters is the class
2. It is the lower taxon e.g. Mammalia.

Phylum:

1. The related orders constitute the phylum.
2. It is the higher taxon, e.g. Chordata.

Question 24
What are the advantages of assigning nomenclature to organisms?
Answer:
Assigning nomenclature or scientific name to organisms have advantages.

• They are universally accepted.
• Each organism has unique nomenclature.
• It avoids confusion in naming the organisms.

Question 25.
What is binomial system of nomenclature?
Answer:
The system of naming the organism with two names, generic name and specific (species) name is known as binomial system of nomenclature, e.g. Pavo cristatus – Indian pea fowl.

Question 26.
What is trinomial system of nomenclature?
Answer:
The system of naming the organism with three names, generic name, specific name (species) and sub-species name is known as trinomial system of nomenclature, e.g. Corvus splendens -Indian house crow.

Question 27.
What are the rules to be followed in the nomenclature of organisms?
Answer:
The scientific name should be italicized in printed form and the generic name and specific name should be underlined separately if it is handwritten.

• The first alphabet of the generic name should be of uppercase.
• The specific name (species) should be in lower case letters.
• The name or abbreviated name of the scientist who first published the scientific name may be written after the specific (species) name along with the year of publication, e.g. Felis Leo Linn., 1958.
• If the specific (species) name is framed after any person’s name, the name of the species shall end with i, ii or ae. e.g. Ground – dwelling lizard Cyrtodactylus varadgirii

Question 28.
What are the taxonomical tools used for the study of plants and animals?
Answer:
Herbarium and Botanical garden may be used as tools for the plant taxonomy. Museum, Taxonomical keys and Zoological parks are classical tools for animal studies. Field visits, survey, identification, classification, preservation and documentation are the important components of taxonomical tools.

Question 29.
Explain the classical taxonomical tools.
Answer:
Taxonomical tools are the tools for the study of classification of organisms. They include- Taxonomical keys: Keys are based on comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories. Museum: Biological Museums have collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.

Zoological parks:
These are places where wild animals are kept in protected environments under human care. It enables us to study their food habits and behavior.

Marine parks:

• Marine organisms are maintained in protected environments.
• Printed taxonomical tools: It consist of identification cards, description, field guides and manuals.

Question 30.
Name some Automated species identification tools or cybertools.
Answer:
ALIS : Automated Leafhopper Identification System.
DAISY : Digital Automated Identification System.
ABIS : Automatic Bee Identification System.
SPIDA : Species Identified Automatically (spiders, wasp, bee wing characters).
Draw wing : Honey bee wing identification.

Question 31.
What are neo – taxonomical tools?
Answer:
The new taxonomical tools which are based on electron microscopic images to study the molecular structure of cell organelles are neo-taxonomical tools.

Question 32.
What is INOTAXA?
Answer:
INOTAXA is an electronic resource for digital images and description about species. It was developed by Natural History Museum, London. INOTAXA means Integrated Open Taxonomic Access.

Question 33.
Scientists and their contribution for taxonomy.
Answer:

1. Ecosystem – A.G. Tansley, 1935
2. Biodiversity – Walter Rosen, 1985
3. Taxonomy – AP de Candolle
4. Father of Taxonomy (classical) – Aristotle
5. Father of modem taxonomy, Founder of modem systematics – Carolus Linnaeus
6. Aristotle – Historia Animalium
7. Theophrastus – Father of Botany
8. Species – John Ray (1627 – 1708)
9. Five kingdom classification – R.H. Whittaker (1969)
10. Three domain classification – Carl Woese (1977)
11. Seven kingdom classification – Cavalier- Smith (1987)
12. Binomial nomenclature – Carolus Linnaeus
13. John Ray – Methodus Plantarum Nova and Historia Generalis Plantarum
14. Trinomial nomenclature – Huxley and Stricklandt
15. Charles Darwin – Origin of species
16. Ernst Haeckal – Cladogram

Question 34.
Distinguish between Monotypic genus and polytypic genus
Answer:

Monotypic genus:

• If a genus has only one species, it is called a monotypic genus.
• e.g. Ailurus fulgens (red panda)

Polytypic genus:

• If a genus has more than one species, it is called polytypic genus.
• e.g. Felis domestica Felis margarita

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Samacheer Kalvi 11th Commerce Solutions Chapter 1 Historical Background of Commerce in the Sub-Continent

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Samacheer Kalvi 11th Commerce Historical Background of Commerce in the Sub-Continent Textbook Exercise Questions and Answers

I. Choose the Correct Answer

11th Commerce Chapter 1 Question 1.
The place where the goods are sold is ……………
(a) Angadi
(b) Market
(c) Nalangadi
(d) Allangadi
Answer:
(a) Angadi

Trade And Commerce Was Common To Dynasty Question 2.
Hindrance of place is removed by ……………
(a) Transport
(b) Warehouse
(c) Salesman
(d) Insurance
Answer:
(a) Transport

11th Commerce 1st Chapter Question 3.
Who wrote “Arthasasthra”?
(a) Kautilya
(b) Chanakiya
(c) Thiruvalluvar
(d) Elangovadigal
Answer:
(a) Kautilya

Trade And Commerce Was Common To Dash Dynasty Question 4.
Trade and Commerce was common to …………… Dynasty.
(a) Pallava
(b) Chola
(c) Pandiya
(d) Chera
Answer:
(c) Pandiya

Historical Background Of Commerce Question 5.
…………… was first sultan who paved way in the dense forest and helped traders to move from one market place to others place for their commercial caravans.
(a) Balban
(b) Vasco da Gama
(c) Akbar
(d) Alauddin Khilji
Answer:
(a) Balban

II. Very Short Answer Questions

Samacheer Kalvi Guru 11th Commerce Question 1.
What is meant by Barter system?
Answer:
Goods were exchanged for goods prior to invention of money.

Question 2.
What is meant by Nalangadi?
Answer:
The place where the goods were sold was called ‘Angadi’ in the Pandiya period. Day market was called as Nalangadi.

Question 3.
What is meant by Allangadi?
Answer:
The night market was called as Allangadi according to Saint Poet Ilango in Silapathigaram, Madurai-Kanchi.

III. Short Answer Questions

Question 1.
Explain the meaning of the term “Vanigam”.
Answer:
The word ‘Vanigam’ has been widely used in Sangam literature like Purananuru and Thirukkural. The earliest trading population of India was Indus valley people, who used the word ‘trade’. The word vaniyam or vanipam would have had a Dravidian origin.

Question 2.
State the meaning of Maruvurapakkam and Pattinapakkam.
Answer:
Big cities like Poompuhar had the ‘Maruvurappakam’ (inland town) and ‘Pattinapakkam’ (coastal town), had market and bazaars where many merchants met one another for the purpose of selling or buying different kinds of commodities and food stuff.

Question 3.
What is the role of Sangam in trade development of ancient Tamil Nadu?
Answer:
Sangam work refers to great traders, their caravans, security force, markets, marts and guilds of such great traders.

Question 4.
What are the ports developed by Pandiya kingdom?
Answer:
Port towns like Tondi, Korkai, Puhar and Muziri were always seen as busy with marts and markets with activities related to imports and exports. In such a brisk trade, people of the coastal region, engaged themselves in coastal trade and developed their intercontinental trade contacts.

Question 5.
What was focused in Arthasasthra about creation of wealth?
Answer:
Arthasasthra focused on creation of wealth as the means to promote the well being of the state. Kautilya advocated maintenance of perfect balance between State management and people’s welfare through trading activities.

IV. Long Answer Questions

Question 1.
What are the hindrances of business?
Answer:
Hindrances of business:
1. Hindrance of Person: Manufacturers do not know the place and face of the consumers. It is the retailer who knows the taste, preference and location of the consumers. The chain of middlemen consisting of wholesalers, agents and retailers establish the link between the producers and consumers.

2. Hindrance of Place: Production takes place in one centre and consumers are spread throughout the country and world. Rail, air, sea and land transports bring the products to the place of consumer.

3. Hindrance of Time: Consumers want products whenever they have money, time and willingness to buy. Goods are produced in anticipation of such demands.

4. Hindrance of risk of deterioration in quality: Proper packaging and modern air conditioned storage houses ensure that there is no deterioration in quality of products.

5. Hindrance of risk of loss: Fire, theft, floods and accidents may bring huge loss to the business.

6. Hindrance of knowledge: Advertising and communication help in announcing the arrival of new products and their uses to the people.

7. Hindrance of exchange: Money functions as a medium of exchange and enables the buying and selling of any product or service by payment of the right price.

8. Hindrance of finance: Producers and traders may not have the required funds at the time of their need.

9. Hindrance of developing the exact product: Research and development helps in developing the exact product or service which can satisfy the specific wants of consumers and thus improve the standard of living of the people.

10. Hindrance of both selection and delivery at doorsteps: E – Commerce enables the consumer to select the product in the website, place online orders and make payment after receiving the product at the doorstep.

Question 2.
State the constraints in barter system.
Answer:
1. Lack of double coincidence of Wants: Unless two persons who have surplus have the demand for the goods possessed by each other, baiter could not materialize. If this “coincidence of wants” does not exist, Barter cannot take place.

2. Non – existence of common measure of value: Barter system could not determine the value of commodities to be exchanged as they lacked commonly acceptable measures to evaluate each and every commodity.

3. Lack of direct contact between producer and consumers: It was not possible for buyers and sellers to meet face to face in many contexts for exchanging the commodities for commodities.

4. Lack of surplus stock: Absence of surplus stock was one of the impediments in barter system. If the buyers and sellers do not have surplus then no barter was possible.

Question 3.
Explain the development of Commerce and Trade in North India.
Answer:
India was prosperous even during the medieval period from 12^th to 16^th centuries despite political upheavals. Balban was the first sultan who paved the way in the dense forest and helped traders and their commercial caravans to move from one market place to others. Allauddin Khilji brought the price to a very low ebb. He encouraged import of foreign goods from Persia and subsidised the goods.

Arabs were dominant players in India’s foreign trade. They never discouraged Indian traders like Tamils, Gujaratis, etc. The trade between the coastal ports were in the hands and Marwaris and Gujiratis, The overland trade with central and west Asia was in the hands of Multanis who were Hindus and Khurasanis who were Afghans, Iranians and so on.

Question 4.
Briefly explain the coastal trade in ancient Tamil Nadu.
Answer:
Big cities like Poompuhar had the ‘Maruvurappakam’ (inland town) and ‘Pattinapakkam’ (coastal town), had market and bazaars where many merchants met one another for the purpose of selling or buying different kinds of commodities and food stuff. Port towns like Tondi, Korkai, Puhar and Muziri were always seen as busy with marts and markets with activities related to imports and exports. In such a brisk trade, people of the coastal region, engaged themselves in coastal trade and developed their intercontinental trade contacts.

They were engaged in different kinds of fishing pearls, and conches and produced salts and built ships. Boats like ‘Padagu’, ‘Thimil’, ‘Thoni’, ‘Ambu’, ‘Odampunai’, etc., were used to cross rivers for domestic trade while ‘Kalam’, ‘Marakalam’, ‘Vangam’, ‘Navai’, etc., were used for crossing oceans for foreign trade.

Question 5.
What do you know about the overseas trading partners of ancient Tamil Nadu?
Answer:
Foreigners who transacted business were known as Yavanars. Arabs who traded with Tamil were called ‘Jonagar’. Pattinappalai praised Kaveripumpattinam as a city where various foreigners of high civilization speaking different languages assembled to transact business with the support of the then Kingdom.

Many ports were developed during the Sangam period. Kaveripumpattinam was the chief port of the Kingdom of Cholas while Nagapattinam, Marakannam, Arikamedu, etc., were other small ports on east coast. Similarly, Pandiyas developed Korkai, Saliyur, Kayal, Marungaurpattinam and Kumari for foreign trade. The State Govertments installed check posts to collect customs along the highways and the ports.

Samacheer Kalvi 12th Commerce Historical Background of Commerce in the Sub-Continent Additional Questions and Answers

I. Choose the Correct Answer:

Question 1.
…………….. is part and parcel of human life.
(a) Commerce
(b) Banking
(c) Insurance
(d) Warehousing
Answer:
(a) Commerce

Question 2.
…………….. means exchange of goods for goods.
(a) Insurance
(b) Money
(c) Barter System
(d) Transport
Answer:
(c) Barter System

Question 3.
Commerce activities are heading for a cashless system through ……………..
(a) e – commerce
(b) banking
(c) insurance
(d) warehousing
Answer:
(a) e – commerce

Question 4.
…………….. in Sangam period was both internal and external.
(a) Sales
(b) Education
(c) Industry
(d) Trade
Answer:
(d) Trade

Question 5.
Day market was called as ……………..
(a) Nalangadi
(b) Angadi
(c) Business
(d) Trade
Answer:
(a) Nalangadi

Question 6.
Night market was called as ……………..
(a) Allangadi
(b) Street store
(c) Shop
(d) Maligai
Answer:
(a) Allangadi

Question 7.
Which is called as sleepless city?
(a) Chennai
(a) Allangadi
(c) Tuticorin
(d) Salem
Answer:
(a) Allangadi

Question 8.
Boats like …………….. were used to cross rivers for domestic trade.
(a) Fishings
(b) Kalam
(c) Marakalam
(d) Thimil
Answer:
(d) Thimil

Question 9.
Boats like …………….. were used for crossing oceans for foreign trade.
(a) Vangam
(b) Thimil
(c) Ambu
(d) Thoni
Answer:
(a) Vangam

Question 10.
…………….. was the chief port of the Kingdom of Cholas.
(a) Marakannam
(b) Arikamedu
(c) Kayal
(d) Kaveripumpattinam
Answer:
(d) Kaveripumpattinam

Question 11.
…………….. Who advocated maintenance of perfect balance between State management and people’s welfare through trading activities.
(a) Kautilya
(b) Pandiya
(c) Hebrew
(d) Chola
Answer:
(a) Kautilya

Question 12.
The profit margin allowed for traders ranged from 5% for indigenous goods and ……………..
for imported goods.
(a) 10%
(b) 5%
(c) 15%
(d) 20%
Answer:
(a) 10%

Question 13.
Who encouraged import of foreign goods from Persia and subsidized the goods?
(a) Balban
(b) Sultan
(c) Alauddin Khilji
(d) Kautilya
Answer:
(c) Alauddin Khilji

Question 14.
Which one helps in announcing the arrival of new products and their uses to the people?
(a) Advertising
(b) Banking
(c) Trade
(d) Transport
Answer:
(a) Advertising

Question 15.
…………….. functions as a medium of exchange.
(a) Transport
(b) Banking
(c) Money
(d) Warehousing
Answer:
(c) Money

II. Very Short Answer Questions

Question 1.
What is cashless system?
Answer:
Commerce activities are heading for a cashless system through e-commerce which means business activities enabled through electronic modes like Online trading, Mobile banking and e – marketing.

Question 2.
What is ‘Angadi’?
Answer:
The place where the goods were sold was called ‘Angadi’ in Pandiya Dynasty.

Question 3.
Which city was called sleepless city?
Answer:
Madurai was called sleepless city due to round – the – clock business activities.

Question 4.
What type of boats were used to cross rivers for domestic trade?
Answer:
Boats like ‘Padgu’, ‘Thimil’, ‘Thom’, ‘Ambu’ ‘Odampunai’, etc., were used to cross rivers for domestic trade.

Question 5.
What type of boats were used to cross oceans for foreign trade?
Answer:
Boats like ‘Kalam’, ‘Marakalam’, ‘Vangam’, ‘Navai’, etc., were used for crossing oceans for foreign trade.

Question 6.
How were the foreigners called in Ancient Tamil Nadu?
Answer:
Foreigners who transacted business were known as ‘yavanars’.

Question 7.
What was the role of the state in trade?
Answer:
The role of the state in trade related to two aspects namely adequate infrastructure to sustain the trade and administrative machinery for taxation.

Question 8.
What was advocated by Kautilya?
Answer:
Kautilya advocated maintenance of perfect balance between State management and people’s welfare through trading activities.

Question 9.
With whom Cholas had a strong trading relationship?
Answer:
Cholas had a strong trading relationship with Chinese Song Dynasty.

Question 10.
What are the constraints of Barter System? (any two)
Answer:

1. Lack of double coincidence of wants
2. Non – existence of common measure of value

III. Short Answer Questions

Question 1.
How has the commerce activities emerged how?
Answer:
The whole of commerce activity has emerged from barter system into a multi – dimensional and multifaceted scientific system consisting of courses like Monetary system, Mail order business, Hire purchase system, Instalment purchase system and so on.

Question 2.
What is the course of commerce activities in today’s technology driven society?
Answer:
In a technology driven society today again the course of commerce activities is heading for a cashless system through e-commerce which means business activities enabled through electronic modes like Online trading, Mobile banking and e – marketing.

Question 3.
What were the constraints in barter system? (Any three)
Answer:
1. Lack of Double Coincidence of Wants:
Unless two persons who have surplus have the demand for the goods possessed by each other, barter could not materialize. If this “coincidence of wants” does not exist, barter cannot take place.

2. Non – existence of Common Measure of Value:
Barter system could not determine the value of commodities to be exchanged as they lacked commonly acceptable measures to evaluate each and every commodity.

3. Lack of Direct Contact between Producer and Consumers:
It was not possible for buyers and sellers to meet face to face in many contexts for exchanging the commodities for commodities.

Question 4.
Write any three hindrances of commerce.
Answer:
1. Hindrance of Person: Manufacturers do not know the place and face of the consumers. The chain of middlemen consisting of wholesalers, agents and retailers establish the link between the producers and consumers.

2. Hindrance of Place: Production takes place in one centre and consumers are spread throughout the country and world. Rail, air, sea and land transports bring the products to the place of consumer.

3. Hindrance of Time: Consumers want products whenever they have money, time and willingness to buy. Goods are produced in anticipation of such demands.

IV. Long Answer Questions

Question 1.
How did the ancient Tamil country trade with Rome, China and Europe?
Answer:
Roman and Greek traders frequented the ancient Tamil country and forged trade relationship with ancient Kings of Pandiya, Chola and Chera dynasties. Cholas had a strong trading relationship with Chinese Song Dynasty. The Cholas conquered the Sri Vijaya Empire of Indonesia and Malaysia to secure a sea trading route to China. During the 16^th and 18^th centuries, India’s overseas trade expanded due to trading with European companies. The discovery of new all sea routes from Europe to India via Cape of Good Hope by Vasco da Gama had far-reaching impact on the civilized world. The arrival of Portuguese in India was followed by the advent of other European communities. They gained a strong foothold in India’s maritime trade by virtue’ of their strong naval power.

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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 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 11th Chemistry Solutions Chapter 6 Gaseous State

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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

Question 15.
Helium diffuses more than air. Give reason.
Answer:
Take two balloons, one is filled with air and another with helium. After one day, the helium balloon was shrunk, because helium being lighter diffuses out faster than the air

Question 16.
Explain about the applications of Graham’s law of diffusion.
Answer:

• Graham’s law of diffusion is useful to determine the molecular mass of the gas if the rate of diffusion is known.
• Graham’s law forms the basis of the process of enriching the isotopes of U²³⁵ from other isotopes and also useful in isotopic separation of deuterium and protium.

Question 17.
What is compression factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT.
This is termed as compression factor.
Compression factor = Z = \(\frac {PV}{nRT}\)
For ideal gases Z = 1 at all temperature and pressures.

Question 18.

1. Define critical temperature.
2. What is the critical temperature of CO₂ gas?

Answer:

1. The temperature below which a gas can be liquefied by application of pressure is known as critical temperature.
2. The critical temperature of CO₂ gas is 303.98 K.

Question 19.

1. Define critical pressure.
2. What is the critical pressure of CO₂ gas?

Answer:

1. Critical temperature (P_C) of a gas is defined as the minimum pressure required to liquefy.
2. The critical pressure of CO₂ is 73 atm.

Question 20.
CO₂ gas cannot be liquefied at room temperature. Give reason.
Answer:
Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO₂ has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 3O₃ K. At room temperature, (critical temperature) even by applying large amount of pressure CO₂ cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO₂ remains as gas.

Question 21.
What is meant by Joule-Thomson effect?
Answer:
The phenomenon of lowering of temperature when a gas is made to expand adiabatically form a region of high pressure into a region of low pressure is known as Joule-Thomson effect.

Question 22.
Define inversion temperature.
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature (T_i).
Ti = \(\frac {2a}{Rb}\)

Question 23.
State and explain Boyle’s law. Represent the law graphically.
Answer:
It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.
P – \(\frac {1}{V}\)
PV = Constant (n and T are constant)
P₁v₁ = P₂V₂
Graphical Representation:

Question 24.
Give an expression for the van der Waals equation. Give the significance of the constants used in the equation. What are their units?
Answer:
\(\left(P+\frac{n^{2} a}{V^{2}}\right)\) (V- nb) = nRT
Where n is the number of moles present and ‘a’’ b’ are known as van der Waals constants.

Significance of Van der Waals constants:
Van der Waals constant ‘a’:
‘a’ is related to the magnitude of the attractive forces among the molecules of a particular gas. Greater the value of’a’, more will be the attractive forces.
Unit of’a’ = L² mol^-2

Van der Waals constant ‘b’:
‘b’ determines the volume occupied by the gas molecules which depends upon size of molecule.
Unit of ‘b’ = L mol^-1

Question 25.
What are ideal and real gases? Out of CO₂ and NH₃ gases, which is expected to show more deviation from the ideal gas behaviour?
Answer:
Ideal gas:
A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.

Real gases:
Gases which deviate from ideal gas behaviour are known as real gases. NH₃ is expected to show more deviation. Since NH₃ is polar in nature and it can be liquefied easily.

Samacheer Kalvi 11th Chemistry Gaseous State 3 – Mark Questions

Question 1.
Explain the graphical representation of 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 ∝ (T and n are fixed) .
If the pressure of the gas increases, volume will decrease and if the pressure of the gas decreases, the volume will increase. So PV = Constant.

Question 2.
What are the consequences of Boyle’s law?
Answer:
1. if the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
2. Boyle’s law also helps to relate pressure to density.
P₁V₁ = P₃V₃ (Boyle’s law)
\(P_{1} \frac{m}{d_{1}}=P_{2} \frac{m}{d_{2}}\)
Where ‘m’ is the mass, d₁ and d₂ are the densities of gases at pressure P₁ and P₂. The density of the gas is directly proportional to pressure.

Question 3.
Explain Charles’ law with an experimental illustration.
Answer:
Charles’ law states that for a fixed mass of a gas at constant pressure, the volume is directly proportional to temperature (K).
V ∝ T (or) \(\frac {V}{T}\) = Constant

Volume vs Temperature:
If a balloon is moved from an ice water bath to a boiling water bath, the gas molecules inside move faster due to increased temperature and hence the volume increases.

Question 4.
Explain the graphical representation of Charles’ law.
Answer:
1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3.  i.e. P₁<P₂ < P₃  <P₄ < P₅ . When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 5.
Explain graphicl representation of Gay Lussac’s law
Answer:
Gay Lussac’s law
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P – T (or) \(\frac {P}{T}\) = Constant
It can be graphically represented as shown here:
Lines in the pressure vs temperature graph are known as isochores (constant volume) of a gas.

Question 6.
Explain the graphical representation of Avogadro’s hypothesis.
Answer:
Avogadros hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure contain equaLnumbr of molecules.
V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant.
Where V₁ and are the volume and number 10- of moles of a gas and V₂ and n₂ are the different set of values of volume and number of moles of the same gas.

A better example to iflustrate Avogadro’s Number of Moles (n) hypothesis is to observe the effect of pumping more gas into a balloon. When more gas molecules (particularly CO₂) are passed, the volume of the balloon increases. The pressure and temperature stay constant as the balloon inflates, so the increase in volume is due to the increase in the quantity of gas inside the balloon.

Question 7.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant. Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.
Boyle’s law: V ∝ \(\frac{1}{P}\)
Charles’ law: V ∝ T
Avogadro’s law: V ∝ n
We can combine these equations into the following general equation that describes the physical
behaviour of all gases.
V ∝ \(\frac{nT}{P}\)
V = \(\frac{nRT}{P}\)
V = , where R = Proportionately constant.
The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where, R is also known as Universal gas constant.

Question 8.
Derive the various values of R, gas constant.
Answer:
1. For standard conditions in which P is 1 atm, volume 22.4 14 dm³ for 1 mole at 273.15 K.

= 0.08205 7 dm³ atm mol^-1 K^-1

2. Where P = 10⁵ Pascal, V = 22.71 x 10^-3 m³ for I mole of a gas at 273.15 K.

= 8.314 pa m³ K^-1 mol^-1
=8.314x 10^-2bar dm³ K^-1mol^-1
R = 8.314 J K^-1 mol^-1.

Question 9.
What is meant by Boyle temperature (or) Boyle point? How is it related with compression point?
Answer:
(1) Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
(2) The Boyle point varies with the nattirc of the gas.
(3) Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
(4) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.

Question 10.
Explain the different methods used for liquefaction of gases.
Answer:

• Linde’s method: Joule-Thomson effect is used to get liquid air or any other gas.
• Claude’s process: In addition to Joule-Thomson effect, the gas is allowed to perform mechanical work so that more cooling is produced.
• Adiabatic process: This method of cooling is produced by removing the magnetic property of magnetic material e.g. Gadolinium sulphate. By this method, a temperature of 10^-4 K i.e. as low as zero Kelvin can be achieved.

Samacheer Kalvi 11th Chemistry Gaseous State 5 – Mark Question

Question 1.
How CH₄ He and NH₃ are deviating from ideal behaviour? (or) Explain how real gases deviate from ideal behaviour.
Answer:
1. The gases which obeys gas equation PV = nRT are known as ideal gases. The gases which do not obey PV = nRT are known as real gases.

2. The gas laws and the kinetic theory are based on the assumption that molecules in the gas phase occupy negligible volume (assumption 1) and that they do not exert any force on one another either attractive or repulsive (assumption 2). Gases whose behaviour is consistent with these assumptions are said to exhibit ideal behaviour.

3. The following graph shows RT plotted against P for three real gases and an ideal gas at a given temperature.

4. According to ideal gas equation, PV/RT is equal to n. Plot PV/RT versus P for ‘n’ moles oigas at 0°C. For1 mole of an ideal gas PV/ RT is equal to 1 irrespective of the pressure of the gas.

5. For real gases, we observe various deviations from ideal behaviour at high pressure. At very low pressure, all gases exhibit ideal behaviour, ie. PV/RT values all converge to n as P approaches zero.

6. For real gases, this is true only at moderate low pressures. (≤ 5 atm) significant variation occurs as the pressure increases. Attractive forces operate among molecules at relatively short deviation.

7. At atmospheric pressure, the molecules in a gas are far apart and attractive forces arc negligible. At high pressure, the density of the gas increases and the molecules are much closer to one another. Intermolecular forces can be significant enough to affect the motion of the molecules and the gas will not behave ideally.

Question 2.
Derive Van der Waals equation of state.
Answer:
1. Consider the effect of intermolecular forces on the pressure exerted by a gas form the following explanation.

2. The speed of a molecule that is moving toward the wall of a container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is Q lower than the pressure the gas would exert, if it behave ideally.


Where ‘a’ is the proportionality constant and depends on the nature of the gas and n and V are the number of moles and volume of the container and respectively an²/ V² is the correction term.

3. The frequency of encounters increases with the square of the number of molecules per unit volume n²/ V². Therefore an²/ V² represents the intermolecular interaction that causes non-ideal behaviour.

4.  Another correction is concerned with the volume ¿ccupied by the gas molecules. ‘V’ represents the volume of the container. As every individual molecule of a real gas occupies certain volume, the effective volume V- nb which is the actually available for the gas, n is the number of moles and b is a constant of gas.

5.  Hence Van der Waals equation of state for real gases are given as \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT Where a and b are Van der Waals constants.

Question 3.
Explain about Andrew’s experimental isotherms of CO₂ gas.
Answer:
1. Andrew plotted isotherms of carbon dioxide at different temperatures. it is then proved that many real gases behave in a similar manner like CO₂.

2. At a temperature of 303.98 K, CO₂ remains as a gas. Below this temperature, CO₂, turns into liquid CO₂ at 7 3atm. It is called the critical temperature of CO₂.

3. At 303.98 K and 73 atm pressure, CO₃,, becomes a liquid but remains a gas at higher temperature.

4. Below the critical temperature, the behaviour of CO₂ is different. For example, consider an isotherm of CO₂ at 294.5 K, it is a gas until the point B, is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist. At C, the gas is completely condensed.

5. If the pressure is higher than at C, only the liquid is compressed so, a steep rise in pressure is observed. Thus, there exist a continuity of state.

6.  A gas below the critical temperature can be liquefied by applying pressures.

Activity – 1
The table below contains the values of pressure measured at different temperatures for 1 moie of an ideal gas. Plot the values in a graph and verify the Gay Lussac’s law. [Lines in the pressure vs temperature graph are known as iso chores (constant volume) of a gas]

Solution:
Gay Lussac’s law at constant volume = \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

If temperature increases, pressure also increases.
So \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3

11th Maths Exercise 1.3 Answers Question 1.
Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as “x related toy if the student x belongs to the section y”. Is this relation a function? What can you say about the inverse relation? Explain your answer.
Solution:
(i) A = {set of students in 11^th standard}
B = {set of sections in 11sup>th standard}
R : A ➝ B ⇒ x related to y
⇒ Every students in eleventh Standard must in one section of the eleventh standard.
⇒ It is a function.
Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard.

Exercise 1.3 Class 11 Maths State Board Question 2.
Write the values of f at – 4, 1, -2, 7, 0 if

Solution:
f(-4) = -(-4) + 4 = 8
f(1) = 1 – 1² = 0
f(-2) = (-2)² – (-2) = 4 + 2 = 6
f(7) = 0
f(0) = 0

11th Maths Exercise 1.3 Question 3.
Write the values of f at -3, 5, 2, -1, 0 if

Solution:
f(-3) = (-3)² – 3 – 5 = 9 – 8 = 1
f(5) = (5)² + 3(5) – 2 = 25 + 15 – 2 = 38
f(2) = 4 – 3 = 1
f(-1) = (-1)² + (-1) – 5 = 1 – 6 = -5
f(0) = 0 – 3 = -3

11th Maths Exercise 1.3 Solutions Question 4.
State whether the following relations are functions or not. If it is a function check for one-to-oneness and ontoness. If it is not a function, state why?
(i) If A = {a, b, c] and/= {(a, c), (b, c), (c, b)};(f: A ➝ A).
(ii) If X = {x, y, z} and/= {(x, y), (x, z), (z, x)}; (f: X ➝ X).
Solution:
(i) f : A ➝ A

It is a function but it is not 1 – 1 and not onto function.

(ii) f : X ➝ X

x ∈ X (Domain) has two images in the co-domain x. It is not a function.

11th Maths Exercise 1.3 Answers Samacheer Question 5.
Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A ➝ B for each of the following:
(i) neither one-to-one nor onto.
(ii) not one-to-one but onto.
(iii) one-to-one but not onto.
(iv) one-to-one and onto.
Solution:
A = {1, 2, 3, 4}
B = {a, b, c, d}.

R = {(1, b) (2, b) (3, c) (4, d)} is not 1-1 and not onto

(iii) Not possible

(iv)

11 Maths Exercise 1.3 Question 6.
Find the domain of \(\frac{1}{1-2 \sin x}\)
Solution:

11th Maths Exercise 1.3 In Tamil Question 7.
Find the largest possible domain of the real valued function f(x) = \(\frac{\sqrt{4-x^{2}}}{\sqrt{x^{2}-9}}\)
Solution:

∴ No largest possible domain
The domain is null set

Samacheer Kalvi 11 Maths Solutions Question 8.
Find the range of the function \(\frac{1}{2 \cos x-1}\)
Solution:
The range of cos x is – 1 to 1

Exercise 1.3 Class 11 Maths Question 9.
Show that the relation xy = -2 is a function for a suitable domain. Find the domain and the range of the function.
Solution:
xy = – 2 ⇒ y = -2/x
which is a function
The domain is (-∞, 0) ∪ (0, ∞) and range is R – {0}

Samacheer Kalvi Guru 11th Maths Question 10.
If f, g : R ➝ R are defined by f(x) = |x| + x and g(x) = |x| – x, find gof and fog.
Solution:

Samacheer Kalvi 11th Maths Question 11.
If f, g, h are real valued functions defined on R, then prove that
(f + g)oh = foh + goh. What can you say about fo(g + h) ? Justify your answer.
Solution:
Let f + g = k

= (f + g((h(x))
= f[h(x)] + g [h(x)]
= foh + goh
(i.e.,)(f + g)(o)h = foh + goh
fo(g + h) is also a function

Samacheer Kalvi Guru 11 Maths Question 12.
If f: R ➝ R is defined by f(x) = 3x – 5, prove that f is a bijection and find its inverse.
Solution:
P(x) = 3x – 5
Let y = 3x – 5 ⇒ 3x = y + 5

Samacheer Kalvi 11th Maths Book Back Answers Question 13.
The weight of the muscles of a man is a function of his body weight x and can be expressed as W(x) = 0.35x. Determine the domain of this function.
Solution:
W(x) = 0.35x
Since body weight x is positive and if it increases then W(x) also increase.
Domain is (0, ∞) i.e.,x > 0

Samacheerkalvi.Guru 11th Maths Question 14.
The distance of an object falling is a function of time t and can be expressed as s(t) = -16t². Graph the function and determine if it is one-to-one.
Solution:
s(t) = -16t²
Suppose S(t₁) = S(t₂)

since time cannot be negative, we to take t₁ = t₂
Hence it is one-one.

Samacheer Kalvi.Guru 11th Maths Question 15.
The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.
Solution:
C – base cost,
S = fuel surcharge,
m = mileage
C(m) = 0.4 m + 50
S(m) = 0.03 m
Total cost = C(m) + S(m)
= 0.4 m + 50 + 0.03 m
= 0.43 m + 50
for 1600 miles
T(c) = 0.43 (1600) + 50 = 688 + 50 = ₹ 738

Samacheer Kalvi Class 11 Maths Question 16.
A salesperson whose annual earnings can be represented by the function A(x) = 30, 000 + 0.04x, where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total family income if they each sell Rupees 1,50,00,000 worth of merchandise.
Solution:
A(x) = 30, 000 + 0.04x, where x is merchandise rupee value
S(x) = 25000 + 0.05 x
(A + S) (x) = A(x) + S(x)
= 30000 + 0.04x + 25000 + 0.05 x
= 55000 + 0.09x
(A + S) (x) = 55000+ 0.09x
They each sell x = 1,50,00,000 worth of merchandise
(A + S) x = 55000 + 0.09 (1,50,00,000)
= 55000 + 13,50,000
∴ Total income of family = ₹ 14,05,000

Samacheer Kalvi Class 11 Maths Solutions Question 17.
The function for exchanging American dollars for Singapore Dollar on a given day is f(x) = 1.23x, where x represents the number of American dollars. On the same day the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.
Solution:
f(x) = 1. 23x where x is number of American dollars.
g(y) = 50.50y where y is number of Singapore dollars.
gof(x) = g(f(x))
= g(1. 23x)
= 50.50 (1.23x)
= 62.115 x

Question 18.
The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D(x) = 200 – x. Express his day revenue, total cost and profit on this meal as functions of x.
Solution:
cost of one meal = ₹ 100
Total cost = ₹ 100 (200 – x)
Number of customers = 200 – x
Day revenue = ₹ (200 – x) x
Total profit = day revenue – total cost
= (200 – x) x – (100) (200 – x)

Question 19.
The formula for converting from Fahrenheit to Celsius temperatures is \(y=\frac{5 x}{9}-\frac{160}{9}\)
Find the inverse of this function and determine whether the inverse is also a function.
Solution:

Question 20.
A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines).
Solution:
f(x) = 3x – 4
Let y = 3x – 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3 Additional Questions

Question 1.
Find the domain and range of the function \(f(x)=\frac{1}{\sqrt{x-5}}\)
Solution:
Given that : f(x) \(f(x)=\frac{1}{\sqrt{x-5}}\)
Here, it is clear that / (x) is real when x – 5 > 0 ⇒ x > 5
Hence, the domain = (5, ∞)
Now to find the range put

For x ∈ (5, ∞), y ∈ R⁺.
Hence, the range of f = R⁺.

Question 2.
If \(f(x)=\frac{x-1}{x+1}\), then show that

Solution:

Question 3.
Find the domain of each of the following functions given by:
\(f(x)=\frac{x^{3}-x+3}{x^{2}-1}\)
Solution:

Here, f(x) is not defined if x² – 1 ≠ 0
(x – 1) (x + 1) ≠ 0
x ≠ 1, x ≠ -1
Hence, the domain of f = R – {-1, 1}

Question 4.
Find the range of the following functions given by
f(x) = 1 + 3 cos 2x
Solution:
Given that: f(x) = 1 + 3 cos 2x
We know that -1 ≤ cos 2x ≤ 1
⇒ -3 ≤ 3 cos 2x ≤ 3 ⇒ -3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1
⇒ -2 ≤ 1 + 3 cos 2x ≤ 4 ⇒ -2 ≤ f(x) ≤ 4
Hence the range of f = [-2, 4]

Question 5.
Find the domain and range of the function \(f(x)=\frac{x^{2}-9}{x-3}\)
Solution:

Domain off: Clearly f(x) is not defined for x – 3 = 0 i.e. x = 3.
Therefore, Domain (f) = R – {3}
Range off: Let f(x) = y. Then,

It follows from the above relation that y takes all real values except 6 when x takes values in the set R – {3}. Therefore, Range (f) = R {6}.

Question 6.
Find the range of the following functions given by f(x) = \(\frac{1}{2-\sin 3 x}\)
Solution:

Students can Download Computer Science Chapter 9 Introduction to C++ Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 9 Introduction to C++

Samacheer Kalvi 11th Computer Science Introduction to C++ Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 9 Book Back Answers Question 1.
Who developed C++?
(a) Charles Babbage
(b) Bjame Stroustrup
(c) Bill Gates
(d) Sundar Pichi
Answer:
(b) Bjame Stroustrup

11th Computer Science Evaluate Yourself Answers Question 2.
What was the original name given to C++?
(a) CPP
(b) Advanced C
(c) C with Classes
(d) Class with C
Answer:
(c) C with Classes

Samacheer Kalvi Guru 11th Computer Science Question 3.
Who coined C++?
(a) Rick Mascitti
(b) Rick Bjame
(c) Bill Gates
(d) Dennis Ritchie
Answer:
(a) Rick Mascitti

Samacheer Kalvi Computer Science Question 4.
The smallest individual unit in a program is:
(a) Program
(b) Algorithm
(c) Flowchart
(d) Tokens
Answer:
(d) Tokens

Computer Science Samacheer Kalvi Question 5.
Which of the following operator is extraction operator of C++?
(a) >>
(b) <<
(c) <>
(d) ^^
Answer:
(a) >>

Samacheer Kalvi 11th Computer Science Question 6.
Which of the following statements is not true?
(a) Keywords are the reserved words convey specific meaning to the C++ compiler.
(b) Reserved words or keywords can be used as an identifier name.
(c) An integer constant must have at least one digit without a decimal point.
(d) Exponent form of real constants consists of two parts
Answer:
(b) Reserved words or keywords can be used as an identifier name.

11th Samacheer Kalvi Computer Science Question 7.
Which of the following is a valid string literal?
(a) ‘A’
(b) ‘Welcome’
(c) 1232
(d) “1232”
Answer:
(d) “1232”

11th Computer Science Samacheer Kalvi Question 8.
A program written in high level language is called as ……………….
(a) Object code
(b) Source code
(c) Executable code
(d) All the above
Answer:
(b) Source code

Samacheer Kalvi Computer Science Book Question 9.
Assume a = 5, b = 6; what will be result of a & b?
(a) 4
(b) 5
(c) 1
(d) 0
Answer:
(a) 4

11 Computer Science Samacheer Kalvi Question 10.
Which of the following is called as compile time operators?
(a) sizeof
(b) pointer
(c) virtual
(d) this
Answer:
(a) sizeof

PART – 2
II. Answer to all the questions

Samacheer Kalvi.Guru 11th Computer Science Question 1.
What is meant by a token? Name the token available in C++.
Answer:
C++ program statements are constructed by many different small elelments such as commands, variables, constants and many more symbols called as operators and punctuators. Individual elements are collectively called as Lexical units or Lexical elements or Tokens. C++ has the following tokens:

1. Keywords
2. Identifiers
3. Literals
4. Operators
5. Punctuators

Samacheer Kalvi 11th Computer Science Book Back Answers Question 2.
What are keywords? Can keywords be used as identifiers?
Answer:
Keywords are the reserved words which convey specific meaning to the C++ compiler. They are the essential elements to construct C++ programs. Most of the keywords are common to C, C++ and Java. Keywords are reserved and cannot be used as identifiers.

Question 3.
The following constants are of which type?

1. 39
2. 032
3. OXCAFE
4. 04.14

Answer:

1. 39 – Decimal
2. 032 – Octal
3. OXCAFE – Hexadecimal
4. 04.14 – Decimal

Question 4.
Write the following real constants into the exponent form:

1. 23.197 00
2. 7.214
3. 0.00005
4. 0.319

Answer:

1. 23.197 = 0.23197 x 10² = 0.23197E2
2. 00 7.214 = 0.7214 x 10¹ = 0.7214E1
3. 0.00005 = 0.5 x 10^-4 = 0.5E – 4
4. 0.319 = 3.19 x 10^-1 = 3.19E – 1

Question 5.
Assume n = 10; what will be result of n>>2;?
Answer:

Question 6.
Match the following:

Answer:

PART – 3
III. Answer to all the questions

Question 1.
Describe the differences between keywords and identifiers?
Answer:
Keywords:

• Keywords are the reserved words which convey specific meaning to the C++ compiler.
• They are essential elements to construct C++ programs.
• Most of the keywords are common to C, C++ and Java.

Identifiers:

• Identifiers are the user-defined names given to different parts of the C++ program.
• They are the fundamental building blocks of a program.
• Every language has specific rules for naming the identifiers.

Question 2.
Is C++ case sensitive? What is meant by the term “case sensitive”?
Answer:
C++ is a case sensitive programming language so, all the keywords must be in lowercase. Case sensitive means that the uppercase and lowercase letters are considered differently.

Question 3.
Differentiate “=” and “==”.
Answer:

Question 4.
Assume a = 10, b = 15; What will be the value of a^b?
Answer:

Question 5.
What is the difference between “Run time error” and “Syntax error”?
Answer:
Run – time Error:

• A run time error occurs during the execution of a program. It occurs because of some illegal operation that takes place.
• For example, if a program tries to open a file which does not exist, it results in a run – time error.

Syntax Error:

• Syntax errors occur when grammatical rules of C++ are violated.
• Example: if you type as follows, C++ will throw an error.
  cout << “Welcome to Programming in C++”

Question 6.
What are the differences between “Logical error” and “Syntax error”?
Answer:
Logical Error : Logical errors occur when there is an incorrect usage of variable / operator / order of execution etc. It is also called as Semantic Error.

Syntax Error : Syntax errors occur when grammatical rules of C++ are violated.

Question 7.
What is the use of a header file?
Answer:
Header files contain definitions of Functions and Variables, which is imported or used into any C++ program by using the pre – processor #include statement. Header files have an extension “.h” which contains C++ function declaration and macro definition.
Example: #include

Question 8.
Why is main function special?
Answer:
C++ program is a collection of functions. Every C++ program must have a main function. The main() function is the starting point where all C++ programs begin their execution. Therefore, the executable statements should be inside the main() function.

Question 9.
Write two advantages of using include compiler directive.
Answer:

1. The program is broken down into modules, thus making it more simplified.
2. More library functions can be used, at the same time size of the program is retained.

Question 10.
Write the following in real constants.

1. 15.223
2. 211.05
3. 0.00025

Answer:

1. 15.223 = 0.15223E2
2. 211.05 = 0.21105E3
3. 0.00025 = 0.25E – 3

PART – 4
IV. Answer all the questions

Question 1.
Write about Binary operators used in C++.
Answer:
Binary Operators require two operands:
Arithmetic operators that perform simple arithmetic operations like addition, subtraction, multiplication, division (+, -, *, %, /) etc. are binary operators which requires minimum of two operands.

Relational operators are used to determine the relationship between its operands. The relational operators (<, >, >=, <=, ==, !=) are applied on two operands, hence they are binary operators. AND, OR (logical operator) both are binary operators. Assignment operator is also a binary operator (+=, – =, *=, /=, %=).

Question 2.
What are the types of Errors?
Answer:

Question 3.
Assume a = 15, b = 20; What will be the result of the following operations?
(a) a&b
(b) a|b
(c) a^b
(d) a>>3
(e) (~b)
Answer:

PART – 1
I. Choose the correct answer

Question 1.
How many categories of data types available in C++?
(a) 5
(b) 4
(c) 3
(d) 2
Answer:
(c) 3

Question 2.
Which of the following data types is not a fundamental type?
(a) signed
(b) int
(c) float
(d) char
Answer:
(a) signed

Question 3.
What will be the result of following statement?
char ch= ‘B’;
cout<< (int) ch;
(a) B
(b) b
(c) 65
(d) 66
Answer:
(d) 66

Question 4.
Which of the character is used as suffix to indicate a floating point value?
(a) F
(b) C
(c) L
(d) D
Answer:
(a) F

Question 5.
How many bytes of memory allocates for the following variable declaration if you are using
Dev C++? short int x;
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(a) 2

Question 6.
What is the output of the following snippet?
charch =‘A’;
ch = ch + 1;
(a) B
(b) A1
(c) F
(d) 1A
Answer:
(a) B

Question 7.
Which of the following is not a data type modifier?
(a) signed
(b) int
(c) long
(d) short
Answer:
(b) int

Question 8.
Which of the following operator returns the size of the data type?
(a) sizeof()
(b) int ()
(c) long ()
(d) double ()
Answer:
(a) sizeof()

Question 9.
Which operator is used to access reference of a variable?
(a) $
(b) #
(c) &
(d) !
Answer:
(c) &

Question 10.
This can be used as alternate to endl command:
(a) \t
(b) \b
(a) \0
(d) \n
Answer:
(d) \n

PART – 2
II. Answer to all the questions

Question 1.
Write a short note const keyword with an example.
Answer:
const is the keyword used to declare a constant, const keyword modifies / restricts the accessibility of a variable. So, it is known as Access modifier.

Question 2.
What is the use of setw( ) format manipulator?
Answer:
setw manipulator sets the width of the field assigned for the output. The field width determines the minimum number of characters to be written in output.
Syntax:
setw(number of characters)

Question 3.
Why is char often treated as integer data type?
Answer:
Character data type accepts and returns all valid ASCII characters. Character data type is often said to be an integer type, since all the characters are represented in memory by their associated ASCII Codes. If a variable is declared as char, C++ allows storing either a character or an integer value.

Question 4.
What is a reference variable? What is its use?
Answer:
Reference variable in C++ is alias for existing variable. They store nothing but the address of the variable used at the time of its declaration. It is important to assign the reference variable at the time of declaration, else it will show an error.

Question 5.
Consider the following C++ statement. Are they equivalent?
char ch = 67;
charch = ‘C’;
Answer:
Both the statements are equivalent as they declare ‘ch’ to be char and initialize it with the value of 67. Since this is the ASCII code for ‘C’, the character constant also can be used to initialize ‘ch’ to 67.

Question 6.
What is the difference between 56L and 56?
Answer:
56L – The suffix L forces the constant to be represented as long, which occupies 4 bytes.

56 – This will be represented as int type constant which occupies 2 bytes as per Turbo C++.

Question 7.
Determine which of the following are valid constant? And specify their type.

1. 0 0.5
2. ‘Name’
3. ‘\t’
4. 27,822

Answer:

1. 0.5 – is a valid constant. It is a decimal.
2. ‘Name’ – Invalid constant as single quote is not allowed.
3. ‘\t’ – Escape sequence (or) non – graphical character (horizontal tab).
4. 27,822 – Invalid constant. Comma is not allowed.

Question 8.
Suppose x and y are two double type variable that you want add as integer and assign to an integer variable. Construct a C++ statement for the doing so.
Answer:
double x = 10.5, y = 4.5; int a;
a = int (x) + int (y);

Question 9.
What will be the result of following if num=6 initially.
(a) cout << num;
(b) cout << (num==5);
Answer:
(a) 6 (b) False

Question 10.
Which of the following two statements are valid? Why? Also w rite their result, int a; a = 3,014; a=(3,014);
Answer:
It is invalid as comma is not allowed in an integer constant. It is valid. Comma in bracket is allowed.

PART – 3
III. Answer to all the questions

Question 1.
What are arithmetic operators in C++? Differentiate unary and binary arithmetic operators. Give example for each of them.
Answer:
Arithmetic operators : perform simple arithmetic operations like addition, subtraction, multiplication, division etc.
Unary Operators : Require only one operand . Example: +, -, *, /, %, >, <, <=, AND, OR

Binary Operators:

1. Require two operands
2. Example: ++ (Plus, Plus) Increment operator, – – (Minus, Minus) Decrement operator, NOT, ~,

Question 2.
Evaluate x + = x + + + x; Let x = 5;
Answer:
x + = x + + + x (x = 5) x + = x + + + 5 (x becomes 6)
x + = 6 + 6
x + = 12
x = 6 + 12
x = 18

Question 3.
How relational operators and logical operators are related to one another?
Answer:
Relational operators are used to determine the relationship between its operands. When the relational operators are applied on two operands, the result will be a boolean value 1 or 0 which represents True or False respectively which represents logical operator.

Question 4.
Evaluate the following C++ expressions where x, y, z are integers and m, n are floating point numbers. The value of x = 5, y = 4 and m = 2.5;
Answer:

1. n = x + y / x;
2. z = m * x + y;
3. z = (x++) * m + x;

Answer:
1. n = x + y / x;
= 5 + 4/5
= 5 + 0 (both x and y are int type. Therefore only integer part of quotient is considered)
=5

2. z = m * x + y;
= 2.5 * 5 + 4 (m is float type, so x value is promoted to float [implicit conversion])
= 12.5 + 4 ‘
= 16 (2 is int type. So ‘.2’, the fractional part is discarded)

3. z = (x++) * m + x;
= 5*2.5 + x
= 12.5 + 5
= 18 (z is int type, therefore the fractional part is removed, x is incremented after the addition)

Samacheer kalvi 11th Computer Science Introduction to C++ Additional Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
The latest standard version published in December 2017 as ISO/IEC …………….. which is informally known as C++ 17.
(a) 14882 : 1998
(b) 14883 : 2017
(c) 14882 : 2017
(d) 14882 : 2000
Answer:
(c) 14882 : 2017

Question 2.
The smallest individual unit in a program is known as ……………..
(a) token
(b) lexical unit
(c) lexical element
(d) all the above
Answer:
(d) all the above

Question 3.
Integer constant is also called as ……………..
(a) fixed point constant
(b) floating point constant
(c) real constants
(d) boolean literals
Answer:
(a) fixed point constant

Question 4.
Exponent form of real constants consists of parts.
(a) 3
(b) 2
(c) 5
(d) 4
Answer:
(b) 2

Question 5.
…………….. relational operators are binary operators.
(a) 7
(b) 8
(c) 6
(d) 2
Answer:
(c) 6

Question 6.
Match the following

(a) 1 – (i) 2 – (ii) 3 – (iii) 4 – (iv)
(b) 1 – (iv) 2 – (ii) 3 – (iii) 4 – (i)
(c) 1 – (i) 2 – (iii) 3 – (ii) 4 – (iv)
(d) 1 – (i) 2 – (ii) 3 – (iv) 4 – (iii)
Answer:
(b) 1 – (iv) 2 – (ii) 3 – (iii) 4 – (i)

Question 7
…………….. used to label a statement.
(a) colon
(b) comma
(c) semi – colon
(d) parenthesis
Answer:
(a) colon

Question 8.
Match the following

(a) 1 – (i) 2 – (ii) 3 – (iii) 4 – (iv)
(b) 1 – (ii) 2 – (iv) 3 – (i) 4 – (iii)
(c) 1 – (i) 2 – (iii) 3 – (ii) 4 – (iv)
(d) 1 – (i) 2 – (ii) 3 – (iv) 4 – (iii)
Answer:
(d) 1 – (i) 2 – (ii) 3 – (iv) 4 – (iii)

Question 9.
IDE stands for ……………..
(a) Integrated Development Environment
(b) International Development Environment
(c) Integrated Digital Environment
(d) None of the above
Answer:
(a) Integrated Development Environment

Question 10.
In programming language …………….. are referred as variables and the values are referred to as data.
(a) constant
(b) integer
(c) fields
(d) files
Answer:
(c) fields

Question 11.
…………….. data type signed more precision fractional value.
(a) char
(b) short
(c) long double
(d) signed doubles
Answer:
(c) long double

Question 12.
Syntax for reference is ……………..
(a) =
(b) = <& reference>
(c) <&reference_variable>=
(d) None of these
Answer:
(c) <&reference_variable>=

Question 13.
…………….. manipulator is the member of iomanip header file.
(a) setw
(b) setfill
(c) setf
(d) all the above
Answer:
(d) all the above

Question 14.
…………….. is used to set the number of decimal places to be displayed.
(a) Set precision
(b) Garbage
(c) Constant
(d) All the above
Answer:
(a) Set precision

Question 15.
Match the following

(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)
(b) 1 – (i) 2 – (ii) 3 – (iii) 4 – (iv)
(c) 1 – (iv) 2 – (i) 3 – (i) 4 – (iii)
(d) 1 – (iv) 2 – (iii) 3 – (i) 4 – (ii)
Answer:
(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)

PART – 2
II. Short Answers

Question 1.
Mention any two benefits of C++.
Answer:

1. C++ is a highly portable language and is often the language of choice for multi – device, multi – platform app development.
2. C++ is an object – oriented programming language and includes classes, inheritance, polymorphism, data abstraction and encapsulation.

Question 2.
What is the use of Boolean literals?
Answer:
Boolean literals are used to represent one of the Boolean values (True or false). Internally true has value 1 and false has value 0.

Question 3.
What are the types of C++ operators based on the number of operands?
Answer:
The types of C++ operators based on the number of operands are:

1. Unary Operators – Require only one operand
2. Binary Operators – Require two operands
3. Ternary Operators – Require three operands

Question 4.
What are the types of bitwise operators?
Answer:
In C++, there are three kinds of bitwise operators, which are:

1. Logical bitwise operators
2. Bitwise shift operators
3. One’s compliment operators

Question 5.
What is stream extraction operator?
Answer:
C++ provides the operator >> to get input. It extracts the value through the keyboard and assigns it to the variable on its right; hence, it is called as “Stream extraction” or “get from” operator.

Question 6.
Expand IDE and GNU.
Answer:
IDE stands for Integrated Development Environment and GNU stands for General Public License.

Question 7.
What are the main types of C++ datatypes?
Answer:
In C++, the data types are classified as three main categories

1. Fundamental data types
2. User – defined data types and
3. Derived data types.

Question 8.
Write C++ program to accept any character and display its next character.
Answer:
C++ Program to accept any character and display its next character

The output produced by the program will be
Enter a character: A
The Next character: B

Question 9.
Write C++ program to find the area of circle.
Answer:
C++Program to find the area of circle

Output:
Enter Radius: 6.5
The Area of the circle is 132.665

Question 10.
What is Junk (or) Garbage values?
Answer:
If you declare a variable without any initial value, the memory space allocated to that variable will be occupied with some unknown value. These unknown values are called as “Junk” or “Garbage” values.

PART – 3
III. Explain in Brief

Question 1.
What are the uses of Set precision?
Answer:
setprecision () prints the values from left to right. For the above code, first, it will take 4 digits and then prints one digit from fractional portion. setprecision can also be used to set the number of decimal places to be displayed. In order to do this task, you will have to set an ios flag within setf() manipulator. This may be used in two forms: fixed and (i) fixed (ii) scientific These two forms are used when the keywords fixed or scientific are appropriately used before the setprecision manipulator.

Question 2.
Define – Expression?
Answer:
An expression is a combination of operators, constants and variables arranged as per the rules of C++. It may also include function calls which return values.

Question 3.
What is Automatic conversion and Type promotion?
Answer:
An Implicit type conversion is a conversion performed by the compiler automatically. So, implicit conversion is also called as “Automatic conversion”. This type of conversion is applied usually whenever different data types are intermixed in an expression. If the type of the operands differ, the compiler converts one of them to match with the other, using the rule that the “smaller” type is converted to the “wider” type, which is called as “Type Promotion”.

Question 4.
What is type casting. Write its syntax?
Answer:
C++ allows explicit conversion of variables or expressions from one data type to another specific data type by the programmer. It is called as “type casting”.
Syntax:
(type – name) expression;
Where type – name is a valid C++ data type to which the conversion is to be performed.

Question 5.
What kind of constants are the following?

1. 0X568
2. – 27
3. – 27

Answer:

1. Hexadecimal
2. Decimal
3. Octal

Question 6.
What are the types of C++operators?
Answer:
C++ Operators are classified as:

1. Arithmetic Operators
2. Relational Operators
3. Logical Operators
4. Bitwise Operators
5. Assignment Operators
6. Conditional Operator
7. Other Operators

Question 7.
Convert the following real constants into exponent form:

1. 12.0005
2. 0.00000009
3. 7.9283

Answer:
0 12.0005 = 0.120005 x 10² = 0.120005E2
0.00000009 = 0.9 x 10^-7 = 0.9E-7
7.9283 = 0.79283 x 10¹ = 0.79283E1

Question 8.
If a = 15; what is the result of a<<3 and a >> 2?
If a = 15; equivalent binary value of a is 00001111
Answer:

Question 9.
If b = 15 what is the result of ~b?
If b = 15; equivalent binary value of b is 00001111
Answer:

Question 10.
If a = 65, b = 15, what is the result of a|b?
Equivalent Binary values of 65 = 01000001; 15 = 00001111
Answer:

PART – 4
IV. Explain in Detail

Question 1.
Explain steps involved in creating and execution of C++ program.
Answer:
For creating and executing a C++ program, one must follow four important steps.
1. Creating Source code : Creating includes typing and editing the valid C++ code as per the rules followed by the C++ Compiler.

2. Saving source code with extension .cpp
After typing, the source code should be saved with the extension .cpp

3. Compilation
This is an important step in constructing a program. In compilation, compiler links the library files with the source code and verifies each and every line of code. If any mistake or error is found, it will inform you to make corrections. If there are no errors, it translates the source code into machine readable object file with an extension .obj

4. Execution
Execution is the final step of construction of a C++ Program. In this stage, the object file becomes an executable file with extension .exe. Once the program becomes an executable file, the program has an independent existence. This means, you can run your application without the help of any compiler or IDE.

Question 2.
Explain working with Dev C++.
Answer:
Dev C++ is an open source, cross platform (alpha version available for Linux), full featured Integrated Development Environment (IDE) distributed with the GNU General Public License for programming in C and C++. It is written in Delphi. It can be downloaded from
http://www.bloodshed.net/dev/devcpp.html

After installation Dev C++ icon is available on the desktop. Double click to open IDE. Dev C++ IDE appears.
To create a source file, Select File → New → Source file or Press Ctrl + N.
On the screen that appears, type your C++ program, and save the file by clicking File → Save or Pressing Ctrl + S. It will add .cpp by default at the end of your source code file. No need to type .cpp along with your file name.

After save, Click Execute → Compile and Run or press F11 key.
After successful compilation, output will appear in output console, as follows

Question 3.
Draw memory allocation of a variable and int type variable.
Answer:

Question 4.
Write the C++ program to find the curved surface area of the cylinder.
C++ Program to find the Curved Surface Area of a cylinder (CSA) (CSA = 2 nr * h)
Answer:
#include
using namespace std;
int main()
{

float pi = 3.14, radius, height, CSA;
cout << “\n Curved Surface Area of a cylinder”;
cout << “\n Enter Radius (in cm):”; cin >> radius;
cout << “\n Enter Height (in cm):”; cin >> height;
CSA = (2*pi*radius)*height; system (“els”);
cout << “\n Radius:” << radius « “cm”;
cout << “\n Height:” << height« “cm”;
cout << “\n Curved Surface Area of a Cylinder is” << CSA «“sq. cm.”;

}

Output:
Curved Surface Area of a cylinder
Enter Radius (in cm) : 7
Enter Height (in cm) : 20
Radius : 7cm
Height : 20cm
Curved Surface Area of a Cylinder is 879.2 sq. cm.

Question 5.
Write the C++ program to calculate Net salary.
Program to Calculate Net Salary
Answer:
#include
#include using namespace std;
int main()
{

float basic, da, hra, gpf, tax, gross, np; char name[30];
cout << “\n Enter Basic Pay:”; cin >> basic;
cout<< “\n Enter D.A cin >> da;
cout << “\n Enter H.R.A:”; cin >> hra;
gross = basic+da+hra; // sum of basic, da nad hra
gpf = (basic+da) * 0.10; //10% of basic and da
tax = gross * 0.10; //10% of gross pay
np = gross – (gpf+tax); //netpay = earnings – deductions
cout << setw(25) << “Basic Pay:” << setw(10) << basic << endl;
cout << setw(25) << “Dearness Allowance:”<< setw(10) << da << endl;
cout << setw(25) << “House Rent Allowance:” << setw(10) << hra << endl;
cout << setw(25) << “Gross Pay:” << setw(10) << gross << endl;
cout << setw(25) << “G.P.F:” << setw(10) << gpf << endl;
cout << setw(25) << “Income Tax:” << setw(10) << tax << endl;
cout << setw(25) << “Net Pay:” << setw(10) << np << endl;

}
The output will be,
Enter Basic Pay : 12000
Enter D.A : 1250
Enter H.R.A : 1450
Basic Pay : 12000
Dearness Allowance : 1250
House Rent Allowance : 1450
Gross Pay : 14700
G.P.F : 1325
Income Tax : 1470
Net Pay : 11905

Question 6.
If a = 17, b = 24, what is the result of the following?

1. a & b
2. a|b
3. a^b
4. a>>3
5. ~ b

Answer:

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Tamilnadu Samacheer Kalvi 11th English Solutions Prose Chapter 4 Tight Corners

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Check out the topics covered in Chapter 4 Tight Corners Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Chapter 4 Tight Corners Questons and Answers. This helps to improve your communication skills.

Tight Corners Summary Warm Up

Many of us have unused, old but valuable items at home. If we wish to get rid of them, we can sell them at an auction. Items like paintings, jewels, household articles, vehicles, even houses can be auctioned.

The flowchart below will help you understand how an auction is conducted.

Samacheer Kalvi 11th English Tight Corners Textual Questions

Tight Corner Questions And Answers Question (a)
Describe the activity that was going on in the sale – room at King Street.
Answer:
Christie Auction house was full. The auction house was selling Barbizon pictures and getting tremendous sums for each. Some were sold for two thousand and some for three thousand pounds. It was surprising to observe that all the sold items were little bit of things like forest scenes, pools at evening, shepherdesses and the regular subjects.

Tight Corners Pdf Question (b)
What can you say about the author’s attitude when he high-handedly participated in the auction?
Answer:
The author behaved like a gambler. He risked high banking on serendipity alone. In reality, he should have a minimum balance of 500 pounds in his bank account to be eligible to be a bidder in the auction. But he had only sixty three pounds. He didn’t have rich acquaintances or relatives who could bail him out-of a financial crisis of such a serious nature at a short notice. So, the author’s participation in the auction in a high handed manner demonstrates his audacity combined with absurdity.

Tight Corners Question (c)
Why was the author sure he would not be caught?
Answer:
The author decided to bid safely by just raising the stake a little bit and leave it for real millionaires to go ahead. Thus he was sure that he would not be caught.

Tight Corners Paragraph Question (d)
What made the author ignore his friend’s warning?
Answer:
The author was confident that he couldn’t run any risks by a playful participation in the auction. So, he ignored his friend’s warning.

Tight Corners Lesson Question (e)
How had the author managed the auction without getting involved in the deal?
Answer:
Although many bids ended up in four figures, they were started with a modest price of fifty to hundred guineas only. He ventured till the figures reached.only upto three digits. Thus he managed the auction without getting involved in the deal.

Summary Of The Lesson Tight Corners Question (f)
What came as a shock to the author?
Answer:
Unlike previous lots, one painting’s launch price was four thousand pounds. When, the millionaires were too stunned to react, the author had sheepishly said “and fifty”. The auctioneer clinched the deal in the author’s favour. This was a rude shock to the author because he did not wish to buy any painting.” .

Tight Corners Book Back Answers Question (g)
What did the falling of the hammer indicate?
Answer:
The falling of the hammer indicated “closure of the bid” and it mandated the highest bidder to pay and collect his purchase.

Tight Corners Theme Question (h)
What made the friend laugh heartily?
Answer:
The author’s friend realized that he had got into a pickle and there was no possible escape. Looking.at the crisis from outside made him laugh heartily.

Tight Corners Summary In English Question (i)
What kind of excuses did the narrator think he could make?
Answer:
The author speculated on the possibility of confessing his poverty to one of Christie ’s staff and request to put up the picture for sale once again.

Tight Corners Summary In Tamil Question (j)
Why did the friend desert the narrator, a second time?
Answer:
The author’s friend was so much tickled by the comedy of the situation that he deserted the author for a second time to have a hearty laugh alone.

Tight Corners Moral Question (k)
How does the narrator describe the man who approached him?
Answer:
The man who approached the author wore a green baize apron and spoke in a husky cockney tones. He had come to find out if he would accept the offer of fifty guineas for his expression of interest for Daubigny.

Question (l)
How does the Narrator show presence of mind in the sudden turn of events?
Answer:
The author should have been grateful for the stranger’s offer to bail him out of potential insolvency. He could have embraced him and even accepted fifty farthings for restoring him from a mental agony. But he asked the mediator if that was the most he could offer. This was . nothing but wordly guile. The man offered to find out saying there was no harm trying for a bit more. The author gave his ultimatum that he would take a hundred. He got a cheque for hundred guineas.

Question (m)
The narrator would not forget two things about his friend. What are they?
Answer:
The author’s friend only persuaded him to go to Christie’s auction. Secondly, he was the only witness to the author ’s mental agony in trying to get out of the crisis.

1. Choose the most appropriate answer for the following questions:

Question (a)
‘Tight Comer’ means a
(i) difficult situation
(ii) crowded comer
(iii) tragic incident
(iv) fierce fight
Answer:
(i) difficult situation

Question (b)
Barbizon refers to a
(i) kind of paint
(ii) type of architecture
(iii) a region in Britain
(iv) a French school of painting
Answer:
(iv) a French school of painting

Question (c)
The narrator visited the sale-room as he
(i) wished to see an auction
(ii) had a painting to sell
(iii) was persuaded by his friend
(iv) wanted to buy a painting
Answer:
(iii) was persuaded by his friend

Question (d)
The narrator had been a safe contributor at the auction, as
(i) there were bidders quoting higher prices
(ii) he had a sound financial background
(iii) his friend had lent him money
(iv) he did not make any bidding
Answer:
(i) there were bidders quoting higher prices

Question (e)
‘And I got it.’ Here‘it’refers to the’
(i) the picture
(ii) the price
(iii) an award
(iv) the card
Answer:
(iv) the card

2. Answer the following questions:

Question (a)
What is a tight corner? What happens when one finds oneself in a tight corner?
Answer:
Tight comer is a difficult situation. When one finds oneself in a tight comer, one worries and thinks seriously about the ways of getting out of it.

Question (b)
What is the difference between a physical and mental tight corner?
Answer:
Physical tight comers are those situations which threaten the lives of an individual. Mental tight comers are worries for which no solution is in sight. It upsets the individuals and confounds them. .

Question (c)
Why did the narrator visit Christie’s?
Answer:
The narrator visited Christie’s to watch the process of auctioning.

Question (d)
The narrator heard his own voice saying, ‘and fifty’. What does this suggest?
Answer:
The narrator heard his voice saying “and fifty”. This suggested that he offered to buy the painting by paying four thousand fifty guineas.

Question (e)
What was the narrator’s financial condition?
Answer:
The narrator had just sixty three pounds in his bank account. The tragic fact was that he did not even have 500 pounds which was the security deposit to be eligible to bid for the paintings.

Question (h)
The narrator could not pretend to have made a mistake in bidding. Why?
Answer:
The narrator had made bids for many paintings. Now he could not confess his poverty. So, he could not pretend tahave made a genuine mistake.

Question (g)
What could have been the best way for the narrator, to get himself out of the tight corner?
Answer:
The author could have confessed his poverty and
requested the auctioned picture to be put up again for “sale” again to get himself freed from the auction.

Question (h)
Why did the narrator feel he could have welcomed a firing party?
Answer:
The author had made many bids in a low margin and got escaped. But he got trapped by saying “and fifty” when a picture was put up for sale with a starting price of 4000 guineas. No one else raised the stake. The auctioner rang the bell and the author realized with alarm how on earth he could ever raise that much money. He thought he could find a firing party to shoot him down. Death is better than public disgrace.

Question (i)
What was the bidder’s offer to the narrator?
Answer:
The bidder offered 50 guineas to the narrator to give up his claim of the painting.

Question (j)
How did the narrator take advantage of the situation?
Answer:
The author asked the mediator of the bidder if that was the most he could offer. The mediator said that there was no harm in asking for a little more. Then the narrator gave his ultimatum that he would take hundred. This showed how guile the narrator was though the stranger was inadvertently rescuing him from a tight comer.

3. Form a meaningful summary of the lesson by rewriting the numbers in the correct sequence:

(a) The narrator had only 63 pounds with him and did not know how to manage the situation. [ ]
(b) The narrator thought of ail his relations from whom he could borrow. [ ]
(c) Unfortunately he had made the highest bid. [ ]
(d) The narrator entered Christie’s as his friend persuaded him to visit the sale room. [ ]
(e) Every time someone else made a higher bid and hence the narrator was not caught. [ ]
(f) The narrator on a sudden impulse added 50 more guineas, to the amount offered. [ ]
(g) His friend joined him then but left immediately unable to control his laughter. [ ]
(h) He even thought of borrowing from money lenders and considered the possibility of confessing the truth to the staff at Christie’s. [ ]
(i) The picture was declared sold to the narrator. [ ]
(j) After sometime a picture was put up and a bid for 4000 guineas was raised. [ ]
(k) A sudden stroke of luck befell the narrator when he heard that the gentleman who had made the bid of 4000 guineas would offer him the additional 50 guineas and buy the picture. [ ]
(l) The narrator kept bidding just for fun. [ ]
(m) The picture was given away to the other bidder and the narrator was saved from humiliation. [ ]
(n) His friend had left the place roaring with laughter at the narrator’s predicament. [ ]
(o) The narrator was too happy at the offer but demanded 100 guineas instead of the 50. Now there was no need for him to make any payment. [ ]

Answer:

1. The narrator entered Christie’s as his friend persuaded him to visit the sale room.
2. The narrator kept bidding just for fun.
3. Every time someone else made a higher bid and hence the narrator was not caught.
4. After sometime a picture was put up and a bid for 4000 guineas was raised.
5. The narrator on a sudden impulse added 50 more guineas, to the amount offered.
6. Unfortunately he had made the highest bid.
7. The picture was declared sold to the narrator.
8. The narrator had only 63 pounds with him and did not know how to manage the situation.
9. The narrator thought of all his relations from whom he could borrow.
10. He even thought of borrowing from moneylenders and considered the possibility of confessing the truth to the staff at Christie’s,
11. His friend had left the place roaring with laughter at the narrator’s predicament. .
12. A sudden stroke of luck befell the narrator when he heard that the gent who had made the bid of 4000 guineas would offer him the additional 50 guineas and buy the picture.
13. The narrator was too happy at the offer but demanded 100 guineas instead of the 50. Now there was no need for him to make any payment.
14. The picture was given away to the other bidder and the narrator was saved from humiliation.
15. His friend joined him then but left immediately unable to control his laughter.

4. Answer the following questions in a paragraph of about 100 – 150 words.

Question (a)
Narrate the circumstances that led to the narrator getting into a tight corner, by his own folly.
Answer:
Lucas learnt that an auction was in progress. His friend suggested that they peeped in, to watch the fun. Despite the caution from his friend, he started bidding at moderate rates. He had only 63 pounds in his account. A bidder was supposed to have a minimum of 500 pounds to take part in the bid. As bidding for most of the paintings were started with two or three digits in guineas, the author sailed through raising the stakes of many paintings and staying behind watching millionaires bid with higher prices. But one painting viz big Daubigny was launched at an offer price of 4000 Guineas.

Only one bidder showed interest. The rest were in silence. The author heard himself say “and fifty”. After seconds of uncomfortable silence, the dealer banged the hammer indicating the acceptance of the narrator’s offer of 4050 guineas for the painting. It was only then the narrator realized that he was in a tight comer. He wished a firing squad would be welcomed to eliminate him and put an end to his mental agony. He had no friend or relative or even money lenders who could extend him a loan to raise the money. He had got into a mess of his own choice.

“Auction houses run a rigged game. They know exactly how many people will be bidding on a work and exactly who they are. In a gallery, works of art just needs on to pay. ”

Question (b)
Trace the thoughts that went on in the mind of the narrator when picture after picture was put up and sold at the auction.
Answer:
The author was enthusiastically participating in the bid at Christie with very little money on him. He sailed smooth for a long time raising the stakes on many paintings and carefully staying behind other competitors. It was fun watching till he got trapped in a net, set by his own tongue. When one particular painting was offered for 4000 guineas, the bidders maintained an uncomfortable silence when the author heard himself foolishly saying “and fifty”. The auctioner banged the hammer finalizing the deal in the narrator’s favour.

It was then the narrator realized with alarm that he had no money on him. Suddenly he lost interest in fun bidding. He started thinking fast for a way out of the tight comer he had created for himself. Many small and big paintings were offered and sold out fast. The Barbizon pictures were selling fast like hot cakes for 2000 to 3000 guineas. The author was running over the names of friends, relatives and even money lenders who might bail him out of the tight comer. He even speculated on the possibility of confessing his poverty to the staff of Christie and request them to put up the picture again for sale. Such a genuine mistake could have been rectified at the early stages of auction. As he had enthusiastically participated in the bid for many paintings,

the auctioners wouldn’t buy his justification for the “genuine mistake”. As bidders stood in a queue to hand in their cheques/cash to collect their paintings, the narrator stood deliberately at the end. He never felt such a fool or had colder feet all his life.

“People do not wish to appear foolish; to avoid the appearance of foolishness, they are will to
remain actually fools but wait in patience for the right time

Question (c)
Explain how the narrator got out of the tight corner that he was in.
Answer:
When the author was perplexed beyond measure and was even ready to welcome a firing squad to bail him out of the current crisis, a divine chance presented itself to the narrator. The narrator had stupidly given an open bid to buy “big Daubigny” for 4050 guineas when he had only 63 pounds in his bank account. However hard he tried, he could not recall a name of an “uncle” or a friend who could extend him a loan to cover the price of the painting. To delay disgrace, he was standing at the end of the queue of the successful bidders. Like a providential intervention, a mediator from the starting bidder who was ready to take the same painting for 4000 guineas enquired the narrator in a husky cockney tone if he was the gentleman who had bought, “big Daubigny”.

The narrator admitted it. To the narrator’s great relief, the mediator said the first bidder wanted to know if he would take 50 guineas for his interest. The author should have embraced him and wept for joy for bailing him out of a potential disgrace. But he made the best use of the opportunity exhibiting his guile, by asking him if that was the most he could offer. The mediator said that there was no harm in asking for some more. The narrator said he would take hundred guineas. When the man left to find out the possibility both the author and his friend laughed.

But when the author saw the cheque for hundred guineas, he became serious. He said with joy and shock, “of all the luck! well, I’m hanged”. Thus the narrator had a narrow escape from a tight comer. One could even say that the narrator escaped by the skin of his teeth.

“Call it a narrow escape, maybe it’s your lucky day. ”

Question (d)
As the narrator, make a diary entry about the tight corner you faced at Christie’s and how you were saved from the dire situation.
Answer:
Thursday, 17th Nov. 20xx
I was lunching at a club in King James’s street. While passing along Kingstreet later, my friend suggested that we peeped in at Christie’s where an auction of Barbizon pictures was going on. The prices of the paintings were pertaining to forest scenes, pools at evening, shepherdesses, and the regular subjects were tremendous for each ranging from two to three thousand guineas each. The remarkable thing was that nothing was sold at three figures. After watching the auction for fun for a while I found myself bidding.

I had exactly sixty three pounds in my account in the bank. I knew that any bidder must have a minimum of five hundred pounds in the bank to stand as security to bid for the artistic works. I enthusiastically participated in many bids as the starting price for each paining was a modest fifty to hundred guineas. Things went on well for me for quite sometime. But a cruel fate awaited me. A short red-faced man electrified the room by fixing the starting price at 4000 guineas. There was a rustle of excitement followed by terrible silence. But I found myself saying “and fifty”. The dealer looked at the opener and at the company. To my surprise and horror, the dealer shot his bolt.

My heart stopped and my blood congealed. I was in possession of the picture I did not want to buy. I was the top purchaser in the auction with just 63 pounds in the bank account. I turned to my friend for some moral support but he had deserted me to have a hearty laugh at a distance. With great alarm, I saw many other Barbizon pictures being put up and sold. The auction came to an end. The bidders stood in a queue to pay the price and collect the pictures. I stayed behind at the end of the queue as I could not recall the name of any uncle, aunt or even a relative who could offer me 4050 guineas to buy the painting. I wished that a firing squad could give me a welcome relief by shooting me down. I preferred death to public disgrace. But something divine turned my tragedy into a comedy.

Just then one gentleman enquired if I was the gentlemen who bought “big Daubigny”. I admitted. The mediator asked if 1 could take 50 guineas for my interest and give up my claim. I would have hugged him and wept for joy of relief from the tight comer. But I had the guile/presence of mind to ask, “Is that the most he would offer?” The mediator said that there was no harm in trying for a bit more. I said, “Tell him I will take hundred” myself and my friend started laughing. But when I saw the cheque for hundred guineas, I became grave, My friend said to me that it was he who brought me to Christie. I admitted, “I shall never forget it. It is indelibly branded in letters of fire on my heart”

Additional Questions

Question (e)
Describe the activity that was going on in the sale-room at King street. What can you say about the author’s attitude when he high handedly participated in the auction?
Answer:
The sale-room at Christie’s was full. Billionaire’s and millionaires who wished to showcase their taste in art had assembled there. The dealer was offering Barbizon pictures on sale. The paintings had mainly forest scenes, pools at evening, shepherdesses and the regular subjects. But each of the little paintings was sold at tremendous prices ranging from two to three thousand guineas. Nothing was sold at three digits. The narrator had just 63 pounds in his bank. The auctioneer had made it mandatory that each bidder should have a minimum of 500 pounds in the bank to stand as security for the bid.

Being aware of this condition, the narrator’s friend cautioned him against participating in the auction for fun. But the narrator ignored his friend’s caution and started bidding for many paintings. He promised his friend that he was not going to run any risk. As most of the paintings were offered with the initial modest price of either 50 or 100 guineas, he carefully hiked the price a little and derived great pleasure in the hot bidding that ensued among millionaires. His over confidence and high handed participation did land him in trouble.

“Life is not a competition. Each one is on their own journey. Live according to your choices,capacity, values and principles. ”

Vocabulary

(i) Auction House Puzzler
You have come across many terms associated with auction, in the lesson. Now solve the crossword puzzle with words from the lesson. Make use of the clues given

Across
1. conducts
2. a protective garment
3. strip with numbers
4. offer

1. painter
2. school of painting
3. auction houses
4. painting Answers

Answers:

Across
1. Auctioneer
2. Apron
3. Card
4. Bid

Down
1. Daubigny
2. Barbizon
3. Christie
4. Art

A. Here are some more idioms taken from the lesson. Find out their meanings and use them in sentences of your own.

Question 1.
Tight corners – difficult situations.
Answer:
Those who act on impulse find themselves in tight comers.

Question 2.
Shoot one’s bolt – exhaust one’s effort
Answer:
Mithali Raj had shot her bolt and made 54 runs to help India win the World Cup Cricket title.