CA Foundation Business and Commercial Knowledge Study Material Chapter 2 Business Environment – Test Questions

CA Foundation Business & Commercial Knowledge Study Material Chapter 2 Business Environment – Test Questions

CA Foundation Business and Commercial Knowledge Study Material Chapter 2 Business Environment – Test Questions

1. Which of the following is a characteristic of business environment?
(a) Aggregative
(b) Dynamic
(c) Uncertain
(d) All of them.

2. Which of the following is an element of micro environment:
(a) Customers
(b) Competitors
(c) Suppliers
(d) All of them.

3. Which of following relates to population?
(a) Demographic environment
(b) Social environment
(c) Cultural environment
(d) Natural environment

4. Understanding of environment enables a business enterprise to
(a) focus on customers
(b) gain the first mover advantage
(c) become aware of impending threat
(d) All of them.

5. Demonetization and GST are examples of changes in
(a) Political environment
(b) Social environment
(c) Technological environment
(d) Global environment

6. Merger of associate banks of SBI into SBI is an example of changes in
(a) Economic environment
(b) Technological environment
(c) Social environment
(d) Demographic environment

7. Tick (✓) the correct alternative.
Demographic trends are a part of:
(a) economic environment III
(b) social environment
(c) political environment
(d) legal environment

8. State whether the following statements are True or False:
Wait and watch is a response of least resistance.
Big and powerful firms adopt the response of gaining command over the environment.
Innovative approach requires no feedback system.
Adaptation response involves anticipation of changes in business environment.

9. State whether the following statements are true or false.
Privatisation and globalisation are components of economic liberalisation.
Pressures for structural adjustments are a reason for globalisation.
Denationalisation is a form of privatisation.
Closure of small scale firms is a positive effect of economic liberalisation.

10. Fill in the blanks:

  1. Business environment is the totality of …………… forces
  2. Different elements of business environment are ……………
  3. When business environment changes rapidly and suddenly …………… increases.
  4. Industrial policy, monetary policy and fiscal policy are elements of …………… environment of business.
  5. Business gets …………… from the environment and supplies …………… to the environment.

11. Match the items in column A with those in column B

ca-foundation-business-commercial-knowledge-study-material-meaning-and-elements-of-macro-environment-2

12. Age, family size, sex composition and other people related elements are part of
(a) political environment
(b) economic environment
(c) demographic environment
(d) natural environment.

CA Foundation Business & Commercial Knowledge Study Material – Meaning and Elements of Macro Environment

CA Foundation Business & Commercial Knowledge Study Material Chapter 2 Business Environment – Meaning and Elements of Macro Environment

Meaning and Elements of Macro Environment

Macro environment refers to the general environment or remote environment within which a business firm and forces in its micro environment operate. A company does not directly or regularly interact with the macro environment. Therefore, macro environment is also known as Indirect Action Environment. Forces in the macro environment, however, create opportunities for and pose threats to the company. The macro environment forces are less controllable than the micro forces. Therefore, success of an enterprise depends on its ability to adapt to the macro environment. For example, when there is a substantial increase in the cost of imported raw materials due to depreciation of the Rupee, production of such materials within the country may become necessary.

Macro environment consists of the following components:

  1. Demographic environment
  2. Political and legal environment
  3. Social and cultural environment
  4. Economic environment
  5. Technological environment
  6. Natural environment
  7. Global environment.

1. Demographic Environment: Demographic environment means various dimensions of country’s population. The demographic environment is important to business because people constitute the market for a business. Moreover, business management involves management of people and the efficiency of business depends largely on the competence and motivation of its people. Business firms often use demographic factors (e.g., age, sex, family size, occupation, family life cycle, education, social class, income distribution) as the basis of market segmentation. The demographic environment differs from country to country and from one place to another within a country. The demographic factors which have very significant implications for business are as follows:

  • Size and growth rate of population,
  • Age and sex composition of population,
  • Life expectancy,
  • Rate of employment,
  • Density of population,
  • Rural urban distribution,
  • Family size,
  • Ethnic composition,
  • Literacy levels, and
  • Income levels.

2. Economic Environment – The economic environment comprises all those economic forces which influence the functioning of business enterprises, e.g., the nature and structure of the economy, the stage of economic development, economic resources, the level of income, economic policies, distribution of income, etc. The main components of economic environment are as follows:

  • The nature of economic system-capitalist, socialist or mixed economy.
  • Economic structure-occupational distribution of labour force, structure of national output, capital formation, investment pattern, composition of trade, balance/imbalance between different sectors, five year plans.
  • Economic policies-industrial policy, export-import policy, monetary policy, fiscal policy, foreign investment and technology policy.
  • Organisation and development of the capital market-banking system, securities markets, etc.
  • Economic indices-gross national product, per capita income, rate of savings and investment, price level, balance of payments position, interest rates, etc.
  • Economic infrastructure and stage of development of the economy.
  • Product markets and factor markets-degree of competition, market size, etc.

3. Political and Legal Environment – Political environment comprises the elements relating to Government affairs. It serves as the regulatory framework of business. The main constituents of a country’s political and legal environment are as follows:

  • The constitution of the country.
  • Political organisation-organisation and philosophy of political parties, ideology of the Government, nature and extent of bureaucracy, influence of primary groups, business donations to political parties, political consciousness, etc.
  • Political stability-structure of military and police force, election system, law and order situation, President’s Rule, foreign infiltrations, secessionist activities, etc.
  • Image of the country and its leaders.
  • Foreign policy-alignment or non-alignment, relations with neighbouring countries.
  • Defence and military policy.
  • Laws governing business, and legal system.
  • Flexibility and adaptability of laws-constitutional amendments and direction of public policies.
  • The judicial system-implementation and effectiveness of laws.

4. Social and Cultural Environment – Social environment refers to the characteristics of the society in which a business firm exists. Social and cultural environment consists of the following:

  • Social institutions and groups.
  • Caste structure and family organisation.
  • Educational system and literacy rates.
  • Customs, attitudes, beliefs, values and life styles.
  • Tastes, preferences of people, and their buying behaviour.
  • Religions, etc.

Family, marriage, education, religion, attitudes to work and wealth and ethics are some examples of socio-cultural factors.

ca-foundation-business-commercial-knowledge-study-material-meaning-and-elements-of-macro-environment-1
Fig: Elements of Macro Environment

5. Technological and Physical Environment – The main elements of technological and physical environment are the following:

  • Sources and types of technology.
  • Rate of technological change.
  • Approaches to production of goods and services.
  • New processes and equipment.
  • Research and Development (R&D) systems.

6. Natural Environment – The main natural forces are as follows:

  • Climatic and geographical conditions.
  • Agricultural, commercial and other natural resources.
  • Ecological system.
  • Levels of pollution.

7. Global Environment – International agencies (World Bank, IMF, WTO, EEC, etc.), international conventions, treaties and agreements, economic and business conditions in other countries, etc. Certain developments such as a hike in the crude oil price have global impact. Developments in information and communication technologies facilitate rapid spread of culture across countries. Economic conditions abroad affect Indian firms. For example, exports increase when markets expand abroad. International political factors can also affect business. For example, improvements in relations between India and Pakistan has led to higher trade between the two countries. WTO regulations have far reaching impact on business in India. Import and investment liberalisation by WTO has led to greater competition in India. The main determinants of international environment are as follows:

  • The state of the world economy and distribution of world output.
  • International economic cooperation.
  • International market structure and competition.
  • Barriers to international trade and investment.
  • National economic policies of different countries.
  • Role of multilateral economic institutions.
  • International economic laws, treaties, agreements, codes and practices.
  • Political system and conditions in different countries.
  • Cultural factors in different countries.
  • Growth and transfer of technology.
  • Growth and spread of multinationals.

 

CA Foundation Business & Commercial Knowledge Study Material – Meaning and Elements of Micro Environment

CA Foundation Business & Commercial Knowledge Study Material Chapter 2 Business Environment – Meaning and Elements of Micro Environment

Meaning And Elements of Micro Environment

Micro environment or task environment refers to those individuals, groups and agencies with which the organisations comes into direct and frequent contact in the course of its functioning. In the words of Philip Kotler, “micro environment consists of the actors in company’s immediate environment that affect the performance of the company.” Micro environmental factors exercise a direct and intimate influence on the operations of the enterprise. Therefore, it is also known as Direct Action Environment or specific forces or Stakeholders. Micro environment consists of the groups in the company’s immediate operating environment which have a stake in the company. However, the micro forces may not influence all the firms in a particular industry in the same manner. For example, one firm’s supplier environment may be entirely different from that of another firm which has inhouse supplies. Even when all the competing firms in an industry have similar micro environment, their relative success depends on how effectively they face the micro forces.

Micro environment consists of the following elements:

1. Customers –
The people who buy a firm’s products and services are its customers. A business exists to create and satisfy customers. A firm may have different types of customers like individuals, households, Government departments, commercial establishments, etc. For example, the customers of a paper company may include students, teachers, educational institutions, business firms and other users of stationery.

ca-foundation-business-commercial-knowledge-study-material-meaning-and-elements-of-micro-environment-1
Fig: Elements of Micro Environment

In order to be successful a company must understand and meet the needs and expectations of its customers. A firm can select the target customer group or market segment on the basis of factors like profitability, elasticity of demand, dependability, degree of competition and growth prospects. It is generally risky to depend upon a single customer group. The customer environment is becoming global due to increasing globalisation and
liberalisation of the economy. With the opening up of Indian market and foreign markets, the customer is becoming more global in the matter of shopping.

2. Competitors –
A company may have both direct and indirect competitors. Direct competitors are the other firms which offer the same or similar products and services. For example, Sony TV faces direct competition from other brands like LG, Samsung, Onida, Videocon, BPL, etc. Indirect competition comes from firms vying for discretionary income. For example, a cinema house, faces indirect competition from Casino, and other firms marketing entertainment. Due to economic liberalisation and globalisation, Indian companies are now facing competition from both domestic firms and multinational corporations. In order to understand the full range of its competition, a company must look at from buyers viewpoint.

3. Suppliers –
Suppliers refer to the people and groups who supply raw materials and components to the company. Reliable sources of supply enable the company to carry on uninterrupted operations and to minimise inventory carrying costs. Suppliers also influence quality levels and costs of manufacturing. It is very risky to depend on a single supplier. A strike*-or any other production problem of the supplier may cause interruptions in manufacturing. Therefore, it is advisable to develop and sustain multiple sources of supply. Some companies like Maruti Suzuki undertake vendor development to ensure timely and regular supply of materials and parts. The relationship between the suppliers and the firm reflects a power equation which is based on the extent to which each of them is dependent on the other.

4. Marketing Intermediaries –
Several marketing intermediaries help a company in promoting, selling and distributing its products to consumers. Middlemen like agents, wholesalers, and retailers serve as a link between the company and its customers. Transportation firms and warehouses assist in the physical distribution of products. Advertising agencies, marketing research agencies and insurance companies are other types of marketing intermediaries. Countrywide retail distribution network has contributed significantly to the success of companies like Hindustan Unilever and Dabur India.

5. Financiers –
The shareholders, financial institutions, debenture holders and banks provide finance to a company. Financial capacity, policies and attitudes of financiers are important factors for the company. For example, the company cannot raise funds through shares if the financiers are not risk taking.

6. Publics –
Publics include all those groups who have an actual or potential, interest in the company or who influence the company’s ability to achieve its objectives. Media groups, environmentalists, non-government organisations (NGOs), consumer associations and local community are examples of publics. These publics can have both positive and negative impact on a business firm. For example, media groups can be used to disseminate useful information. A company can cooperate with the local people to improve its image as well as to provide some benefit to the people. On the negative side, local community concerned with public health can force a company to suspend operations or to take pollution control measures. Non- government organisations often organise protests against firms suspected of being guilty for child labour, cruelty against animals and damage to nature. For example, one of the leading companies in India was attacked by the media for writing advertisements on rocks near a famous hill station. Such activities of publics can tarnish the image of business.

7. Workers and Trade Union –
Workers and their union are an important component of micro environment. A firm’s relations with its workers and trade union have a significant impact on its functioning and performance. Company’s work environment and industrial relations system must be conducive to efficient functioning.

According to Philip Kotler, “companies must put their primary energy into effectively managing their relationships with their customers, distributors and suppliers. Their overall success will be affected by how other publics in the society view their activity. Companies would be wise to spend time monitoring all their public, understanding their needs and opinions and dealing with them constructively.”

CA Foundation Business & Commercial Knowledge Study Material – Meaning of Business Environment

CA Foundation Business & Commercial Knowledge Study Material Chapter 2 Business Environment – Meaning of Business Environment

Business enterprises do not function in isolation. They operate within charging environment. Various elements of this environment and changes in them exercise a significant influence on the working and performance of business firms.

Meaning of Business Environment

The term business environment means “the aggregate of all the forces, factors and institutions which are external to and beyond the control of an individual business enterprise but which exercise a significant influence on the functioning and growth of individual enterprises.” Keith Davis defines business environment as “the aggregate of all conditions, events and influences that surround and affect business.”

According to Bayord O. Wheeler, business environment refers to “the total of all things external to firms and industries which affect their organisation and operation”.

In the words of Arthtur M. Weimer, “business environment encompasses the climate or set of conditions, economic, social, political, or institutional in which business operations are conducted”.

Thus, business environment means all those internal and external factors that have an impact on business.

Nature (Characteristics) of Business Environment

Business environment is characterised by the following features:

  1. Aggregative –
    Business environment is the totality of all the internal and external forces which influence the working and decision-making of an enterprise.
  2. Inter-related –
    Different elements of business environment are closely inter-related and interdependent. A change in one element affects the other elements. Economic environment influences the non-economic environment which in turn affects the economic conditions. For example, economic liberalisation in India since 1991 has opened up new opportunities for private sector and foreign entrepreneurs. Similarly, social pressures against pollution led to the enactment of anti-pollution laws. Therefore, managers should not consider environmental factors in isolation from one another. A holistic approach is necessary for proper understanding of business environment.
  3. Dynamic –
    Business environment is dynamic in nature as it keeps on changing from time to time.
  4. General and Specific Forces –
    Business environment consists of both general and specific forces. General forces such as economic, social, political, legal, natural and technological conditions influence all business enterprises. Specific forces such as investors, customers, competitors, suppliers, etc. affect individual enterprises directly.
  5. Relative –
    Business environment is a relative concept. It differs from country to country and even region to region. Capitalist economies like those of USA and UK have a different kind of environment than communist economies. The nature of economic system in a country affects the environment of business.
  6. Inter-temporal –
    Business environment is also an inter-temporal concept as it changes over time. For example, business environment in India today is much different from that prevailing before 1991. In the short run business environment may remain static. But in the long run, it does change.
  7. Uncertain –
    Business environment is largely uncertain because it is very difficult to forecast the future environment. When the environment is volatile, ie., changes very fast, uncertainty increases.
  8. Contextual –
    Business environment provides the macro framework within which the business firm (a micro unit) operates. The environmental forces are largely those given within which an individual enterprise and its management must function.
    Business environment exercises tremendous influence on the working and success of business firms. Different elements of business environment have different types and degrees of influence on business. A factor that has a favourable impact on one firm may adversely effect another firm. Therefore, management of a business enterprise must have a deep understanding and appreciation of the environment. The changes taking place in environment must be continuously monitored to judge their impact on business. Appropriate and timely steps must be taken to face the environmental changes.

Significance of Business Environment

The survival and success of any enterprise depends upon its inherent capabilities (physical, financial, human and other resources) and its ability to adapt to the changing environment.

It is very important for business firms to understand their environment and changes occurring in it. Business enterprises which know their environment and are ready to adapt to environmental changes would be successful. On the other hand, firms which fail to adapt to their environment are unlikely to survive in the long run. For example, some Indian firms suffered considerably because they failed to appreciate the tightening regulations against environmental pollution. Knowledge of environmental changes is very helpful in the formulation and implementation of business plans. A business can obtain this knowledge through environmental scanning. Environmental scanning is the process by which organisations monitor their relevant environment to identify opportunities and threats affecting their business. With the help of environmental scanning, an enterprise can consider the impact of different events, trends, issues and expectations on its business operations. Firms which systematically analyse and diagnose the environment are more effective than those which do not.

Some of the direct benefits of understanding the environment are given below:

  1. First Mover Advantage – Awareness of environment helps an enterprise to take advantage of early opportunities instead of losing them to competitors. For example, Maruti Suzuki became the leader in small car market because it was the first to recognise the need for small car on account of rising petroleum prices and a large middle class.
  2. Early Warning Signal – Environmental awareness serves as an early warning signal. It makes a firm aware of the impending threat or crisis so that the firm can take timely action to minimise the adverse effects, if any. For example, ‘when new firms entered in the mid segment cars (threat), Maruti Suzuki increased the production of its Esteem threefold. Increase in production enabled the company to make faster delivery. As a result the company captured a substantial share of the market and became a leader in this segment.
  3. Customer Focus – Environmental understanding makes the management sensitive to the changing needs and expectations of consumers. For example, Hindustan Unilever and several other FMCG companies launched small sachets of shampoo and other products realising the wishes of customers. This move helped the firms to increase sales.
  4. Strategy Formulation – Environmental monitoring provides relevant information about the business environment. Such information serves as the basis for strategy making. For example, ITC realised that there is a vast scope for growth in the travel and tourism industry in India and the Government is keen to promote this industry because of its employment potential. With the help of this knowledge, ITC planned new hotels both in India and abroad. Study of environment enables an organisation to analyse its competitors’ strategies and thereby formulate effective counter strategies. All strategic decisions such as what business to do, whether to expand or reduce a business, and so on require a thorough understanding of the internal and external environment of the organisation.
  5. Change Agent – Business leaders act as agents of change. They create a drive for change at the gross root level. In order to decide the direction and nature of change, the leaders need to understand the aspirations of people and other environmental forces through environmental scanning. For example, contemporary environment requires prompt decision-making and power to people. Therefore, business leaders are increasingly delegating authority to empower their staff and to eliminate procedural delays.
  6. Public Image – A business firm can improve its image by showing that it is sensitive to its environment and responsive to the aspirations of public. Leading firms like Reliance Industries, ICICI Bank and others have built good image by being sensitive and responsive to environmental forces. Environmental understanding enables business to be responsive to their environment.
  7. Continuous Learning – Environmental analysis serves as broad based and ongoing education for business executives. It keeps them in touch with the changing scenario so that they are never caught unaware. With the help of environmental learning managers can react in an appropriate manner and thereby increase the success of their organisations. Knowledge of changing environment can keep the organisation dynamic in its approach.

There are two major components of business environment-micro and macro.

CA Foundation Business & Commercial Knowledge Study Material Chapter 1 Introduction to Business – Test Questions

CA Foundation Business & Commercial Knowledge Study Material Chapter 1 Introduction to Business – Test Questions

1. Classify the following into economic and non-economic activities.
(a) A mother cooking food for her children
(b) A teacher teaching in a coaching centre.
(c) A young boy helps an old man in crossing the road.
(d) The Chaat seller outside a school.
(e) A lady officer in Delhi Police

2. The income of a profession is called
(a) wage
(b) salary
(c) fee
(d) profit

3. In employment a person receives
(a) profit
(b) fee
(c) salary
(d) dividend

4. Income of a businessman is called
(a) wage
(b) salary
(c) fee
(d) profit

5. Which of the following is not a business activity:
(a) A housewife selling old newspapers
(b) A person selling vegetables in your locality
(c) A teacher teaching his daughter at home
(d) A painter painting houses of others

6. In which of the following occupations services of specialised nature are provided for fee
(a) Employment
(b) Business
(c) Profession
(d) Farming

7. Which of the following occupations involve maximum risk
(a) Employment
(b) Business
(c) Profession
(d) Teaching

8. Which of the following statements best describe economic activities.
(a) These activities are undertaken to earn money.
(b) These activities provide pleasure.
(c) These activities involve no risk
(d) These activities involve working outside the house.

9. The head of the joint Hindu family business is called
(a) Partner
(b) Manager
(c) Karta
(d) Member

10. The maximum number of partners in a partner-ship firm can be :
(a) 10
(b) 20
(c) 50
(d) 100

11. State whether the following statements are True or False:

(a) Taking photographs of mountains as a hobby is an economic activity.
(b) Teaching in a school is a non-economic activity.
(c) Service or employment is an economic activity.
(d) Business involves buying and selling of goods and services.
(e) Running a hair-cutting saloon is a business.
(f) Occupations of doctors, lawyers and char¬tered accountants are called professions.
(g) Economic activities are undertaken for earning a living.

12. Answer the following in Yes or No:

(a) A business activity can be carried out without sale or exchange.
(b) Business involves the creation of utilities.
(c) A business activity may be without a profit motive.
(d) Risk is an essential element of every business activity.
(e) Earning profits is the sole objective of business.
(f) Business is an economic activity.
(g) Profession involves greater risk than busi¬ness.

13. Name the following:

(a) The industry in which useful products are obtained from the earth and the sea.
(b) The trade in which goods are imported and exported to some other country.
(c) The branch of commerce which removes the hindrance of time.
(d) Buying and selling of goods in small quan-tities.

14. Fill in the blanks:

(a) Business = Industry + ………….
(b) …………. means the production of goods and services.
(c) Commerce includes ‘trade and ………….’
(d) Conversion of raw materials and component parts into finished products is known as …………… industry.
(e) Agriculture is an …………… industry.
(f) Transport removes the hindrance of ……………

15. State whether the following statements are True or False:

(a) Industry and commerce are interchangeable terms.
(b) Commerce is a wider term than trade.
(c) Trade between two countries is known as internal trade.
(d) Conversion of cotton into cloth is an example of genetic industry.

16. Match the following:

ca-foundation-business-commercial-knowledge-study-material-chapter-1-introduction-to-business-test-questions-1

17. Answer in Yes or No:

(a) The sole proprietor is the exclusive master of his business.
(b) The liability of a sole proprietor is limited.
(c) A sole proprietorship has no legal existence apart from its owner.
(d) Sole proprietorship is the oldest form of business organisation.
(e) It is difficult to set up a sole proprietorship.

18. Fill in the blanks:

(a) Sole proprietorship is the …………… form of business enterprises, (latest/oldest)
(b) Sole proprietorship is most suitable form …………… for business, (small/large)
(c) Liability of a sole proprietor is …………… (limited/ unlimited)
(d) Sole proprietorship …………… a separate legal entity, (has/does not have)
(e) It is to set up a sole proprietorship firm, (easy/difficult)

19. Fill in the blanks:

(a) The membership of a joint Hindu family business is acquired by ……………
(b) The liability of the karta in a joint Hindu family business is ……………
(c) A member of the joint Hindu family business demand division/of …………… property.
(d) All the members of a joint Hindu family business …………… part in its man agement.
(e) A member of the j oint Hindu family business …………… call for account of past profits while leaving the business.

20. State whether the following statements are True or False:

(a) A joint Hindu family business is the result of a contract between the members of the family.
(b) Every member of a joint Hindu family business is an agent of the firm.
(c) There is no limit on the number of members in a joint Hindu family business.
(d) Female members of a family do not have a share in the joint Hindu family business.
(e) A joint Hindu family business is not dissolved on the death of a member.

21. Match the following:

ca-foundation-business-commercial-knowledge-study-material-chapter-1-introduction-to-business-test-questions-2

22. Name the following:

(a) A partnership in which the liability of one or more partners is limited.
(b) A partnership which is set up for an indefinite period of time.
(c) The document containing the terms and conditions of a partnership.
(d) The partner who contributes capital but does not take active part in the management
(e) Person who lends his name and goodwill for the benefit of a partnership firm.

23. State whether the following statements are True or False:

(a) Registration of a partnership is legally compulsory.
(b) Limited partnership is not allowed in India.
(c) A partnership can take as many partners as it likes.
(d) A minor can become a partner.
(e) Partners may come and go but the part-nership goes on forever.
(f) A person becomes a partner on his birth in the family.

24. Name the following:

(a) The person who promotes a business.
(b) The document containing the bylaws of a company.
(c) The documents inviting subscriptions for shares and debentures.
(d) The amount of money which must be raised before allotment of share.

25. Mark the following statements True or False:

(a) A private company must file a statement in lieu of prospectus.
(b) A company which does not want to prepare its own Articles of Association can adopt Table A.
(c) A company wanting to raise capital must issue a prospectus.
(cl) A private company can start its business immediately after incorporation.
(e) Capital clause is a part of the Articles of Association
(f) The Memorandum of Association defines the relationship of a company with or out¬siders.
(g) Every company must issue a prospectus to raise share capital.

26. Fill in the blanks:

(a) A company is legally …………… from its members.
(b) Liability of every member in a company is ……………
(c) Registration of a company is ……………
(d) A company is managed by a ……………
(e) In a private company, there must be at least …………… members.
(f) A company has …………… succession.

27. Name the following:
(a) A company in which the number of members cannot exceed 50:
(b) A company having at least seven members:

28. Match the following:

ca-foundation-business-commercial-knowledge-study-material-chapter-1-introduction-to-business-test-questions-3

29. The capital of a company is divided into number of parts, each one of which is called
(a) Dividend
(b) Profit
(c) Interest
(d) Share

30. The form of business organisation in which there is separation of ownership and management is called
(a) Sole proprietorship
(b) Partnership
(c) Company
(d) All these.

31. State whether the following statements are True or False:

(a) A joint stock company is separate legal entity.
(b) A company survives even if all its members die.
(c) A joint stock company is a voluntary association of persons.
(d) A joint stock company is less stable than a partnership.
(e) Registration of a joint stock company is optional.
(f) There can be no conflict of interest in a joint stock company.
(g) A company is an artificial legal person.

32. Fill in the blanks:

(a) The East India Company was a …………… company
(b) The Reserve Bank of India is a …………… company.
(c) Reliance Industries Limited is a …………… company.
(d) A does not have share capital.
(e) The liability of members of a company is unlimited.
(f) A company which holds 51% more of the shares of another company is called ……………
(g) Steel Authority of India Limited is a company.
(h) Coca Cola Corporation is a company.
(i) Minimum share capitalin a private company is Rs. ……………

33. State whether the following statements are True or False.

(a) There is no chartered company in India.
(h) LIC is a statutory company.
(c) At least five persons are needed to form a public company.
(d) The minimum paid up capital of a private company must be Rupees five lakh.
(e) A private company can invite public deposits.
(f) The shares of a public company are transferable.
(g) A subsidiary of a public company is also a public company.
(h) The concept of private company deemed to be public no longer exists in India.

CA Foundation Business & Commercial Knowledge Study Material – Forms of Business Organizations

CA Foundation Business & Commercial Knowledge Study Material Chapter 1 Introduction to Business – Forms of Business Organizations

FORMS OF BUSINESS ORGANIZATIONS

The main forms of ownership in private sector are as follows:

  1. Sole Proprietorship
  2. Joint Hindu Family Business
  3. Partnership
  4. Joint Stock Company
  5. Cooperative Society

Sole Proprietorship

Sole trader is a person who carries on business exclusively for himself He alone establishes the business, arranges its finances, manages its affairs and bears all its risk. He acts both as the owner and manager of his business. He alone, is responsible for the profits and losses of his business. He may borrow funds and employ people to help him but the ultimate authority and responsibility lie in his hands. Sole trader business is thus a one-man show.

Some popular definitions of sole trader are given below:

  • The individual proprietorship is the form of business organisation at the head of which stands an individual as one who is responsible, who directs its operations and who alone runs the risk of failure. – L.H. Haney
  • A sole proprietor is a person who carries on business exclusively by and for himself. He is not only the owner of the capital of the undertaking, but is usually the organiser and manager and takes all the profits or responsibility for losses. – Janies Stephenson
  • A sole trader business is a type of business unit where one person is solely responsible for providing the capital, for bearing the risk of the enterprise and for the management of business. – J. L. Hansen
  • Under the sole proprietorship form of ownership, a single individual organises and operates the business in his own name. He is not only responsible for its management but also for its risks. – J. M, Shubin
  • Sole proprietorship is a form of business where the individual proprietorship is the supreme judge of all matters pertaining to his business. – Kimball and Kimball
  • Sole proprietorship is an informal type of business owned by one person. – J. L. Lundy
  • The sole proprietorship is that form of business ownership which is owned and controlled by a single individual. He receives all the profits and risks all of his property in the success or failure of the enterprise. – B. O. Wheeler

The distinguishing characteristics of sole proprietorship are as follows:

  1. Single ownership – A sole proprietorship is wholly owned by one individual. The individual supplies the total capital from his own wealth or from borrowed funds.
  2. One-man control – The proprietor alone takes all the decisions pertaining to the business. He is not required to consult anybody. Ownership and management are vested in the same person. Some persons may be employed to help the owner but ultimate control lies with him.
  3. No legal entity – A sole proprietorship has no legal identity separate from that of its owner. The law makes no distinction between the proprietor and his business. The business and the owner exist together. If the owner dies or becomes insolvent the business is dissolved. Business and the proprietor are one and the same.
  4. Unlimited liability – The proprietor is personally liable for all the debts of the business. In case the assets are insufficient to meet its debts, the personal property of the proprietor can be attached.
  5. No profit-sharing – The sole proprietor alone is entitled to all the profits and losses of business. He bears the complete risk and there is nobody to share the profits or losses.
  6. Small size – The scale of operations carried on by a sole proprietorship is generally small. A sole trader can arrange limited funds and managerial ability. Therefore, the area of operations is limited.
  7. No legal formalities – No legal formalities are required to start, manage and close.

Advantages

  1. Ease of formation
  2. Incentive to work
  3. Independence in control
  4. Prompt decisions
  5. Business secrecy
  6. Personal touch
  7. Flexibility of operations
  8. Inexpensive management
  9. Minimum Government regulations
  10. Easy dissolution
  11. Social advantage.

Disadvantages

  1. Limited capital
  2. Lack of specialisation
  3. Lack of stability
  4. Unlimited liability
  5. No scope for expansion and growth.

Suitability:

Thus, sole proprietorship has several advantages and disadvantages. According to William R. Basset, “the one man control is the best in the world if that man is big enough to manage everything.” But, one man can rarely manage and control everything. Therefore, sole proprietorship is a suitable form of organisation in the following cases:

  • Where the market is local, e.g., small-scale retailers;
  • Where personal attention to the needs and preferences of customers is essential, e.g., tailoring, beauty parlours, etc.
  • Where fashions change very frequently, e.g., artistic jewellery;
  • Where small amount of capital is required but personal skills are more important, e.g., health clinic;
  • Where quick decision and prompt action are necessary, e.g., stock brokers; and
  • Where risk involved is negligible, e.g., doctors, lawyers, chartered accountants.

Joint Hindu Family Business

The joint Hindu family business refers to a business which is owned and managed by the members of a joint Hindu family. It is also known as Hindu Undivided Family Business: It is governed by the Hindu Succession Act. This form of business is created by the law of succession. The joint Hindu family form of business is one in which the family possesses some inherited property. The share of ancestral property is inherited by a male member from his father grandfather and great grandfather. Thus, three successive generations can simultaneously inherit the ancestral property. All the male members having a share in family property are known as coparceners and the oldest male member is called the karta.

Features: The main characteristics of Joint Hindu Family Business are as follows:

  1. Membership – A person becomes a member in the family business by virtue of his birth in the family. No formal agreement is necessary between the family members. The membership is restricted to three successive generations. Only male members can be coparceners. A female relative of a deceased male coparcener will have a share after the death of the coparceners. Minors are also full-fledged members of the family business. There is no limit on the number of members.
  2. Management – The management of joint Hindu family business is rested in the karta. The karta may, however, associate other members to assist him in the management of family business.
  3. Liability – The liability of the karta is unlimited. The liability of other members is limited to the extent of their share in the property of the family business.
  4. Right to Accounts – Coparceners are not entitled to inspect the accounts of the business. However, a coparcener who is leaving the family business can demand accounts from the karta.
  5. Dissolution – Joint Hindu family business is not dissolved on the death of a coparcener. It comes to an end when all the members notify that they are not members of the joint Hindu family.

JOINT HINDU FAMILY BUSINESS AT A GLANCE

Merits

  1. Ease of formation
  2. Freedom of action
  3. Personal contact
  4. Utmost secrecy
  5. Limited liability
  6. Stability
  7. Incentive to work hard
  8. Quick decisions
  9. Economy of operations
  10. Flexibility of operations

Demerits

  1. Limited financial resources
  2. Limited managerial ability
  3. Unlimited liability of karta
  4. Hasty decisions
  5. Source of Conflicts
  6. Misuse of authority by Karta

Partnership

The partnership form of business organisation grew out from the limitations of sole proprietorship. When the business expands, one man is unable to arrange the financial resources and bear the risks. He cannot supervise and manage all the functions of business personally. Therefore two or more persons join hands and combine their capital and skill to start and run a business. Partnership is thus an extension of sole proprietorship.

A partnership is a voluntary association of two or more persons who agree to carry on some business jointly and share its profits and losses. They combine their funds and skills to carry on business together. Some popular definitions of partnership are as follows:

L.H. HANNEY: Partnership is the relations existing between persons competent to make contact, who agree to carry on lawful business in connection with a new to private gain.

THE PARTNERSHIP ACT : “Partnership is the relation between persons who have agreed to share profits of a business carried on by all or any one of them acting for all”.

A Partnership is a form of business organisation in which two or more persons upto a maximum twenty join together to undertake some form of business activity. – J.L. Hanson

Two or more individuals may form a partnership by making a written or oral agreement that they will jointly assume full responsibility for the conduct of business. – John Shubin

The persons who enter into partnership with the one another are individually called ‘partners’ and collectively a ‘firm’. The name under which they carry on business is called the ‘firm name’.

The essential characteristics of partnership are as follows :

  1. Two or more persons –
    There must be at least two persons to form a partnership. A person cannot enter into partnership with himself. The maximum number of persons in a partnership should not exceed 50. If the number of partners exceeds the prescribed maximum, it would become an illegal association of persons. A firm cannot become a partner of another firm though its partners can join any other firm as partners.
  2. Agreement –
    Partnership is the outcome of an agreement between persons. The relation of partnership arises from the formation of a contract and not from status or birth. If a proprietor gives hare in profits to his employee it will not be called a partnership unless there is an agreement of partnership between the two. The agreement may be oral or in writing but it must satisfy all the essentials of a valid contract.
  3. Lawful business –
    A partnership can be formed only for the purpose of carrying on a business. An association of persons who jointly own a house without carrying on a business is not partnership. Moreover, the business carried on by the partners must be lawful. Illegal acts such as theft, dacoity, smuggling, etc., cannot be called partnership.
  4. Sharing of profits –
    The agreement between the partners must be to share the profits of business. There can be no partnership without the intention of mutual gain. The profits must be distributed among the partners in an agreed ratio. Similarly, losses should be shared among the partners. However, sharing of profits is not a conclusive proof of partnership. For example, a manager may be given a share in profits of the firm.
  5. Mutual agency –
    Partnership business can be carried on by all the partners or by any of them acting on behalf of the others. In other words, every partner is an implied agent of the other partners and of the firm. Each partner is liable for acts performed by other partners on behalf of the firm.
    The above mentioned features are the real tests of partnership. In addition, partnership has the following characteristics:
  6. Utmost good faith –
    The relations between partners are based upon mutual trust and confidence. Every partner is expected to act in the best interests of other partners and of the firm as a whole. He must observe utmost good faith in all the dealings with his co-partners. He must render true accounts and make no secret profits from the business.
  7. Unlimited liability –
    Every partner is jointly and severally liable to an unlimited extent for the debts of the partnership firm. In case the assets of the firm are insufficient to pay the debts in full, the personal property of each partner can be attached to pay the creditors of the firm.
  8. Restriction on transfer of interest –
    No partner can transfer his share in the partnership without the prior consent of all other partners.
  9. Joint ownership and control –
    A partnership is owned and controlled jointly by all the partners.

DISTINCTION BETWEEN PROPRIETORSHIP AND PARTNERSHIP

S.No. Point of Distinction Sole Proprietorship Partnership
1. Members One-man show Minimum: 2 Maximum: 50
2. Agreement Not required Essential
3. Capital Contributed by the owner Contributed by the’partners
4. Registration No provision for registration Desirable
5. Management Lies with the owner Lies with the partners
6. Secrecy Easy to maintain Difficult to maintain
7. Risk Borne entirely by the owner Shared by the partners
8. Continuity Depends on the life of the owner Depends on mutual trust and unity among the partners
9. Scale of operations Small scale Medium scale

MERITS AND DEMERITS OF PARTNERSHIP

Merits

  1. Ease of formation
  2. Larger financial resources
  3. Combined judgment
  4. Direct motivation
  5. Close supervision
  6. Flexibility of operations
  7. Secrecy
  8. Protection of minority interest
  9. Mutual Cooperation
  10. Scope for expansion

Demerits

  1. Limited resources
  2. Unlimited liability
  3. Uncertain life
  4. Conflicts
  5. Risk of implied agency
  6. Restriction on transfer of interest
  7. Lesser Public confidence

Limited Liability Partnership (LLP) – Limited partnership is now allowed in India under the Limited.
Liability Partnership Act, 2008. The chief characteristics of a limited liability partnership are as follows:

  • A limited liability partnership must be registered under the Act with a minimum of two partners. There is no limit on the maximum number of partners.
  • An LLP is a body corporate having a separate legal entity and perpetual succession.
  • In an LLP the liability of partners is limited to their agreed contributions to the LLP. No partner would be liable on account of any unauthorised or independent actions of other partners.
  • An LLP must maintain annual accounts reflecting the true and fair view of its state of affairs.
  • The liability of partners and the firm would become unlimited in case the firm or its partners carry out any act with the intention to defraud the creditors or any other person or for any fraudulent purpose.
  • Thus, LLP is a hybrid form of business organisation combining features of both partnership firm and joint stock company.

Advantages: Limited liability partnership offers the following benefits:

  • An LLP is a separate legal entity independent of the partners. It is capable of owning and holding property in its own name.
  • It is much more stable than a general partnership because it is not dissolved by the retirement, insolvency, death, etc. of a partner. It enjoys perpetual existence.
  • The Liability of partners in LLP is limited, they have not to take unlimited risk.
  • As there is no limit on the number of partners, an LLP can raise huge funds for expansion and growth of business.

Disadvantages: An LLP suffers from the following disadvantages:

  • It has to be registered under the Act. It has to spend time and money in the documents and formalities of incorporation.
  • There is less secrecy of business affairs as it has to fulfill legal requirements.
  • Credit standing of an LLP is reduced due to the limited liability of partners.

DISTINCTION BETWEEN COMPANY AND PARTNERSHIP

S.No Basis of Distinction Company Partnership
1. Formation By incorporation-legal formalities By agreement- No legal formalities
2. Registration Compulsory Optional
3. Legal Status Separate from members No separate entity
4. Number of members public company. Maximum 200 in private company and no limit in public company Minimum 2, maximum 50
5. Liability Limited Unlimited
6. Transferability of interest Freely transferable except in case of private company Transferable with the consent of all partners
7. Implied agency Members not agents of the company Partners are agents of the firm
8. Management By elected representatives of members By partners themselves
9. Statutory control Legal formalities concerning accounts audit, directors, etc. No legal formalities
10. Durability Life independent of members Life dependent on members
11. Dissolution Through legal process of winding up By agreement between partners
12. Audit Compulsory in all cases Not always compulsory
13. Scale of operations Large scale Medium size

Joint Stock Company

The company form of business organization came into existence to overcome the limitations of sole proprietorship and partnership.

According to James Stephenson, “a company is an association of many persons who contribute money or money’s worth to a common stock and employ it in some trade business, and who share the profit and loss arising therefrom”. Under the Companies Act a company having share capital may be is defined as “a company limited by shares having a permanent paid up or nominal capital of fixed amount divided into shares, also of fixed amount held and transferable as stock and formed on the principles of having its members only the holders of those shares or stock and no other persons”.

According to Prof. Haney, “Joint stock company is a voluntary association of individuals for profit, having a capital divided into transferable shares, the ownership of which is the condition of membership”. In the words of Justice Lindley, “By a company is meant an association of many persons who contribute money or money’s worth to a common stock and employ it for a common purpose. The common stock so contributed is denoted in money, and is the capital of the company. The persons who contribute it are its members”.

The distinctive features of the company form of organisation are as follows:

  1. Separate legal existence –
    A company has a distinct legal entity independent of its members. It can own property, make contracts and file suits in its own name. Shareholders are not the joint owners of the company’s property. A shareholder cannot be held liable for the acts of the company. Similarly, members of the company are not its agents. There can be contracts between a company and its members. A creditor of the company is not a creditor of its members.
  2. Perpetual succession –
    Perpetual succession means continued existence. A company is a creation of the law and only the law can bring an end to its existence. Its life does not depend on the life of its members. The death, insolvency or lunacy of members does not affect the life of a company. It continues to exits even if all its members die. Members may come and go but the company goes on until it is wound up.
  3. Limited liability –
    As a company has a separate legal entity, its members cannot be held liable for the debts of the company. The liability of every member is limited to the nominal value of the shares bought by him or to the amount of guarantee given by him. For instance, if a member has 50 shares of Rs. 10 each, his liability is limited to Rs. 500. Even if the assets of the company are insufficient to satisfy fully the claims of the creditors, no member can be called to pay anything more than what is due from him. However, if the members of the company so desire, they may form a company with unlimited liability.
  4. Transferability of shares –
    The capital of a company is divided into parts. Each part is called a share. These shares are generally transferable. A shareholder is free to withdraw his membership from the company by transferring his shares. However, in actual practice some restrictions are placed on the transfer of shares.
  5. Common seal –
    Being an artificial entity, a company cannot act and sign itself. Therefore, it acts through human beings. All the acts of the company are authorised by its common seal. The name of the company is engraved on its common seal. The common seal is affixed on all important documents as a token of the Company’s approval. The common seal is the official signature of the company. Any document which does not bear the common seal of the company is not binding on the company.
  6. Separation of ownership and control –
    Members have no right to participate directly in the day- to-day management of a company. They elect their representatives, called directors, who manage the company’s affairs on behalf of the members. Thus, the ownership of a company is distributed among the shareholders while management is vested in the board of directors. The management of a company is delegated and centralised.
  7. Voluntary association –
    A joint stock company is a voluntary association of certain persons formed to carry out a particular purpose in common. Members of a company can join it and leave it at their own freewill.
  8. Artificial legal person –
    A company is an artificial person created by law. It exists only in contemplation of law. It is competent to enter into contracts and to own property in its own name. But it docs not take birth like a natural person and it has no physical body of a natural human being.
  9. Corporate finance –
    The share capital of a company is generally divided into a large number of shares of small value. These shares are purchased by a large number of people from different walks of life.
  10. Statutory regulation and control –
    Government exercises control through company law over the management of joint stock companies. A company is required to comply with several legal formalities and to file several documents with the Registrar of Companies.

JOINT STOCK COMPANY AT A GLANCE

Advantages

  1. Large capital resources
  2. Limited liability
  3. Stability
  4. Efficient management
  5. Transferability of shares
  6. Economies of scale
  7. Democratic management
  8. Public goodwill
  9. Social utility

Disadvantages

  1. Legal formalities
  2. Lack of motivation
  3. Delay in decisions
  4. Economic oligarchy
  5. Corrupt management
  6. Excessive Government control
  7. Unhealthy speculation
  8. Conflict of interests
  9. Social evils

Private Company – It means a company which has a minimum paid-up capital of one lakh rupees or such higher paid-up capital as may be prescribed, and which by its Articles of Association:

  • restricts the right of its members to transfer shares, if any;
  • except in case of one person company limits the number of its members to 200, excluding members who are or were in the employment of the company;
  • prohibits any invitation to the public to subscribe for any securities of, the company;
  • prohibits any invitation or acceptance of deposits from persons other than its members directors or their relatives.

The minimum number of members required to form a private company is two. Such a company must use the word ‘private’ in its name. A private company enjoys special privileges and exemptions under the Companies Act. It is generally a family affair.

A private company enjoys several privileges under the Companies Act, 2013.

The main privileges available to a private company are as follows—

  1. A private company can be started with just two members whereas a public company ‘requires at least seven members.’
  2. A private company can start its business immediately after incorporation. Unlike a public company, it is not required to obtain the certificate of commencement of business.
  3. A private company is not required to issue or file a prospectus or statement in lieu of prospectus with the Registrar of Companies.
  4. A private company is neither required to hold statutory meeting nor to file statutory report with the Registrar of Companies.
  5. A private company can directly allot shares. There is no restriction of minimum subscription.
  6. It can have only two directors whereas a public company must have at least three directors.
  7. It is not required to offer new issue of shares to the existing shareholders on a pro rata basis. It can issue shares with disproportionate voting rights.
  8. Its directors need not retire by rotation. Consent to act as directors or to take up the qualification shares need not be filed with the Registrar. A private company need not take the consent of the Government to grant loans to its directors. Its directors can vote on a contract in which they are interested. Persons can be appointed to office of profit.
  9. A private company is exempted from restrictions concerning remuneration to managerial personnel, appointment of persons to office of profit, inter company investments and publication of accounts.
  10. Two persons personally present is sufficient quorum for general meeting of a private company unless the Articles of Association provide otherwise.
  11. A private company is exempted from preparing an index to its register of members.

Public Company – It means a company which—
(a) is not a private company;
(b) has a minimum paid-up capital of five lakh rupees or such higher paid-up capital, as may be prescribed;
(c) is a private company which is a subsidiary of a company which is not a private company.

Thus, a public company is one which:

  • does not restrict the transfer of its shares,
  • does not limit the number of members,
  • can invite the general public to subscribe to its share; and debentures, and
  • can invite or accept deposits from the public.

At least seven persons are required to form a public company.

One Person Company (OPC) – The Companies Act, 2013 allows the formation of one person company. As the name suggests, a one person company has only one member. The company’s name will carry a suffix ‘OPC’. The process of setting upon OPC is the same as that for a private limited company. Since the company is owned by a single person, he must nominate someone to take charge of it in case of his death or disability. The nominee must give his consent in writing which has to be filed with the Registrar of companies.

An OPC is exempt from certain procedural formalities, such as conducting annual general meetings, general meetings and extraordinary general meetings. No provisions have been prescribed on holding board meetings if there is only one director, but two meetings need to be organised every year if there is more than one director. Any resolution passed by the sole member must be communicated to the company and entered in the minutes book. There is, however, no relief from the provisions on audits, financial statements and accounts, which are applicable to private companies.

Benefits & drawbacks of an OPC – The biggest advantage of a one person company is that its identity is distinct from that of its owner. Therefore, if the firm is embroiled in a legal controversy, the owner will not be sued, only the company will. Another advantage is limited liability. Since the company is distinct from that of its owner, the personal assets of the shareholders and directors remain protected in case of a credit default.

On the other hand, an OPC is not easy to set up. It requires a lot of paperwork and is a time-consuming and costly process.

Despite the advantages that a one person company offers, it may not be a viable option for everyone. In contrast to a company, a proprietorship is easy to set up. The paperwork involved is minimal. Of course, the risk in a proprietorship is higher as the owner is personally responsible for the business.

COMPARISON BETWEEN PROPRIETORSHIP AND OPC

S.No. Basis of Comparison Proprietorship OPC
1. Legal status The firm and the owner are one It is a separate legal entity
2. Registration Registration is not compulsory It must be registered
3. Liability of the owner Liability of the owner is unlimited Liability of the owner is limited
4. Setting up It is easy to set up, minimum paper is involved It is difficult to set up, involves more paperwork. It is a time consuming process
5. Formalities No legal formalities involved Formalities concerning board meetings, audit etc. make it difficult to run.

DISTINCTION BETWEEN PARTNERSHIP AND PRIVATE COMPANY

S.No. Basis of Distinction Partnership Private Company
1. Number of members Maximum: 2 Minimum: 2
Maximum: 50 Maximum: 200
2. Registration Operational Compulsory
3. Legal status No separate legal entity Separate legal entity
4. Minimum paid up capital Not prescribed One lakh
5. S. Liability of members Unlimited Limited
6. Directors Not required Minimum two
7. None Not required Must be “Private Limited” after its name
8. Regulating show The Partnership Act, 1932 The Companies Act, 2013

DISTINCTION BETWEEN PRIVATE AND PUBLIC COMPANIES

S.No. Point of Distinction Private Company Public Company
1. Number of Members Minimum-2, Maximum-200 Minimum-7, Maximum- No limit
2. Name The name must include the words “Private Limited”. The name must include the word “Limited”.
3. Prospectus Need not issue and file prospectus Must issue and file a prospectus or a statement in lieu of prospectus.
4. Allotment of shares No restrictions on allotment of shares. No binding on further issue of shares Cannot allot shares without raising minimum subscription and without complying with other legal formalities. Further issues of shares must in the first instance, be offered to the existing members.
5. Share Warrants Cannot issue share warrants Can issue share warrants per bearer
6. Minimum Capita One lakh Five lakh
7. Listing on stock exchange Cannot be listed Can be listed
8. Managerial

Remuneration

No restrictions on director’s remuneration Total annual remuneration must not exceed 11 per cent of the net profits. In case of insufficiency of profits the maximum limit is t 50,000 per annum
9. Filing Need not send the list of Directors and their consent to the Registrar Must sent list of directors, their consent to the Registrar
10. Quorum at annual general meeting Two members personally present At least five members personally present

Cooperative Society

Cooperative organisation developed as a means of protecting the interests of the weaker sections of society against exploitation and oppression by the economically strong and powerful sections. It is a form of organisation wherein persons associate together voluntarily and on equal basis to further their common interests. For example, consumers may join hands to provide goods at cheaper rates by establishing direct contacts with manufacturers and thereby eliminating the profits of middlemen. Similarly, people belonging to the working class may form a cooperative society to provide houses at low costs to the members. A cooperative society is based on the principles of self help and mutual help and its primary motive is to render service to the members.

Some popular definitions of cooperative organisation are given below:

  • Cooperative organisation is “a society which has its objectives for the promotion of economic interests of its members in accordance with cooperative principles”. —The Indian Cooperative Societies Act, 1912
  • A cooperative society is “a form of organisation wherein persons voluntarily associate together as human beings on the basis of equality for the promotion of the economic interests of themselves”. —H.C. Calvert
  • Cooperative is a joint enterprise of those who are not financially strong and, therefore, come together not with a view to get profits but to overcome disability arising out of the want of adequate financial resources. —H.N. Kernzen
  • Cooperative is an association of individuals to secure a common economic goal by honest means. —Sir Horace Plumkett

The distinctive features of cooperative organisation are as follows :

  1. Voluntary association – A cooperative society is essentially a voluntary association of persons desirous of improving their economic status through joint action. Everyone having a common interest is free to join a cooperative society. He can also leave the society after giving proper notice. He can withdraw his capital but cannot transfer his share to another person. No body is forced to become a member or to continue as a member.
  2. Religious and political neutrality – The membership of a cooperative society is open to all irrespective of caste, creed, religion or political affiliation. New members are always welcome to join the society. Cooperative societies represent universal brotherhood.
  3. Separate legal entity – After registration a cooperative society becomes a distinct body independent of its members. It can own property and make contracts in its own name. It becomes an autonomous and self-governing organisation.
  4. One-man one vote – Every member has one vote irrespective of the number of shares held by him. Rich persons holding more shares cannot dictate terms. The organisation of a cooperative society is democratic and all members have an equal voice in its management.
  5. Service motive – The primary aim of a cooperative society is to provide service to its members. Its motto is ‘each for all and all for each’. However, a cooperative society may earn some profits for the benefit of its members.
  6. Disposal of surplus – The surplus of a cooperative society is not distributed among members in proportion to their capital. According to law, at least one fourth of the profits must be transferred to general reserve. A portion of the profits, not exceeding ten per cent, may be utilised for the welfare of the locality in which the society is functioning. Thus, profits of a cooperative society are utilised for the benefit of its members and the local community.
  7. Limited return on capital – A dividend not exceeding 10 per cent can be paid to members on the capital. The members of a cooperative society are thus assured of a fixed return on their investment.
  8. Cash Trading – Cooperative societies sell goods on cash basis. Cash trading provides protection against bad debts and maintains working capital. However, an exception, may be made in the case of members.
  9. Perpetual existence – Once registered a cooperative society enjoys perpetual existence. It is created by law and law alone can dissolve it.
  10. State control – Every cooperative society must be registered under the Cooperative Societies Act, 1912 or respective State cooperative laws. It is required to observe the prescribed rules and regulations. Government exercises supervision and control over cooperative societies to ensure their proper functioning.

COOPERATIVE ORGANISATION AT A GLANCE

Advantages

  1. Easy to form
  2. Open membership
  3. Limited liability
  4. Continuity
  5. Democratic control
  6. Low operating costs
  7. State patronage
  8. Internal financing
  9. Cheaper and better supplies
  10. Social utility

Disadvantages

  • Limited finance
  • Lack of expertise
  • Lack of incentive
  • Non-transferability of interest
  • Lack of secrecy
  • Legal formalities
  • Disputes among members

DISTINCTION BETWEEN COMPANY AND COOPERATIVE

S.No. Point of Difference Company Cooperative
1. Formation Many legal formalities Few legal formalities
2. Governing Law Under the Companies Act Under the Cooperative Societies Act
3. Number of members Minimum two in private company and seven in public company. Maximum 200 in private company and no limit in public company. Membership open to all. Minimum ten, maximum no limit. Membership restricted to a particular locality or group
4. Management By Board of Directors By Managing Committee
5. Basic Object Earning profits Service to members
6. Voting rights One share one vote One member one vote
7. Transferability of shares Freely transferable Not transferable, but returnable to the society
8. Capital Subscription No limit on individual holding of shares. New shares first offered to existing members. Subscription list closed. Individual subscription to capital may be limited. New shares issued to admit new members. Membership open.
9. Distribution of profits Dividend in proportion to share in capital. Limited dividend, rest on equitable basis.
10. Privileges No special exemptions Special exemptions relating to stamp duty, income tax, etc.
11. Return of capital No member can demand back his capital except at the time of winding up A member can demand his capital during life time of the society.
12. Scale of operations Large scale Small scale

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2

More Exercises

Question 1.
Draw an equilateral triangle of side 4 cm. Draw its circumcircle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q1.1
Solution:
Steps of Construction :
(i) Draw a line segment BC = 4 cm
(ii) With centres B and C, draw two arcs of radius 4 cm
which intersect each other at A.
(iii) Join AB and AC. ∆ ABC is an equilateral triangle.
(iv) Draw the right bisector of BC and AC intersecting each other at O.
(v) Join OA, OB and OC.
(vi) With centre O, and radius equal to OB or OC or OA,
draw a circle which will pass through A, B and C.
This is the required circumcircle of ∆ ABC.

Question 2.
Using a ruler and a pair of compasses only, construct: (i) a triangle ABC given AB = 4cm, BC = 6 cm and ∠ABC = 90°.
(ii) a circle which passes through the points A, B and C and mark its centre as O. (2008)
Solution:
Steps of Construction:
(i) Draw a line segment AB = 4cm
(ii) At B, draw a ray BX making an angle of 90°
and cut off BC = 6 cm.
(iii) Join AC.
(iv) Draw the perpendicular bisectors of sides
AB and AC intersecting each other at O.
(v) With centre O, and radius equal to OB or OA or OC,
draw a circle which passes through A, B and C.
This is the required circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q2.1

Question 3.
Construct a triangle with sides 3 cm, 4 cm and 5 cm. Draw its circumcircle and measure its radius.
Solution:
Steps of Construction
(i) Draw a line segment BC = 4 cm.
(ii) With centre B and radius 3 cm and with centre C
and radius 5 cm draw two arcs which intersect each other at A.
(iii) Join AB and AC.
(iv) Draw the perpendicular bisector of sides BC and AC
which intersect each other at O.
(v) Join OB.
(vi) With centre O and radius OB, draw a circle
which will pass through A, B and C.
(vii) On measuring the radius OB = 2.5cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q3.1

Question 4.
Using ruler and compasses only :
(i) Construe a triangle ABC with the following data: Base AB = 6 cm, AC = 5.2 cm and ∠CAB = 60°.
(ii) In the same diagram, draw a circle which passes through the points A, B and C. and mark its centre O.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6 cm.
(ii) At A, draw a ray making an angle of 60°.
(iii) With centre B and radius 5-2 cm.
draw an arc which intersects the ray at C.
(iv) Join BC
(v) Draw the perpendicular bisector of AB and BC
intersecting each other at O.
(vi) With O as a centre and OA as a radius
draw a circle which touches the ∆ABC at A, B and C.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q4.1

Question 5.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm and draw its inscribed circle. Measure the radius of the circle.
Solution:
Steps of Construction :
(i). Draw a line segment BC = 5 cm
(ii) With centre B and C and radius 5 cm,
draw two arcs intersecting each other at A.
(iii) Join AB and AC.
(iv) Draw the angle bisectors of ∠B and ∠C intersecting each other at I.
(v) From I, draw a perpendicular ID on BC.
(vi) With centre I and radius ID, draw a circle
which touches the sides of the triangle internally.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q5.1
This is the required in circle.
Measure the radius ID which is 1.5 cm (approx)

Question 6.
(i) Conduct a triangle ABC with BC = 6.4 cm, CA = 5.8 cm and ∠ ABC = 60°. Draw its incircle. Measure and record the radius of the incircle.
(ii) Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. (2014)
Solution:
Steps of Construction :
(i) Draw a line segment BC = 6.4 cm
(ii) Construct ∠ DBC = 60° at B.
(iii) With C as centre and radius CA = 5.8 cm.
Draw an arc cutting BD at A.
(iv) Join AC. Then ABC is the required triangle.
(v) Draw the angle bisectors of ∠B and ∠C which intersect each other at O.
(vi) Draw OE ⊥ BC, intersecting BC in E.
(vii) With O as centre and OE as radius draw the required incircle.
Measure the radius OE which is = 1.5cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q6.1
Steps of construction:
1. Draw a line BC = 6.5 cm.
2. With centre B and C draw arcs AB = 5.5 cm and AC = 5 cm
3. Join AB and AC, ABC is the required triangle.
4. Draw the angle bisetors of B and C. Let these bisectors meet at O.
5. Taking O as centre. Draw a incircle which touches all the sides of the ∆ABC.
6. From O draw a perpendicular to side BC which cut at N.
7. Measure ON which is required radius of the incircle. ON = 1.5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q6.2

Question 7.
The bisectors of angles A and B of a scalene triangle ABC meet at O.
(i) What is the point O called?
(ii) OR and OQ are drawn a perpendicular to AB and CA respectively. What is the relation between OR and OQ ?
(iii) What is the relation between ∠ACO and ∠BCO?
Solution:
(i) The point O where the angle bisectors meet is called the incentre of the triangle.
(ii) The perpendiculars drawn from O to AB and CA are equal i.e. OR and OQ.
(iii) ∠ACO = ∠BCO
OC will bisect the ∠C
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q7.1

Question 8.
Using ruler and compasses only, construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and the perpendicular from A on BC is 2.5 cm. Draw the circumcircle of triangle ABC and measure its radius.
Solution:
Steps of Construction
(i) Draw a line segment BC = 4 cm.
(ii) At B, draw a perpendicular and cut off BE = 2.5 cm.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q8.1
(iii) From E, draw a line EF parallel to BC.
(iv) From C, draw a ray making an angle of 45° which intersects EF at A.
(v) Join AB.
(vi) Draw the perpendicular bisectors of
sides BC and AC intersecting each other at O.
(vii) Join OB, OC and OA.
(viii) With centre 0 and radius OB or OC or OA draw a circle
which will pass through A, B and C.
This circle is the circumcircle of ∆ABC. On measuring its radius OB = 2 cm.

Question 9.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q9.1
Solution:
Steps of Construction
(i) Draw a line segment BC = 4 cm.
(ii) At A and B draw rays making on angle of 120° each and cut off AF = BC = 4cm.
(iii) At F and C, draw rays making angle of 120° each and cut off EF = CD = 4cm.
(iv) Join ED.
ABCDEF is the required hexagon.
(v) Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
(vi) With centre O and radius equal OA or OB draw a circle
which passes through the vertices of the hexagon.
This is the required circumcircle of hexagon ABCDEF.

Question 10.
Draw a regular hexagon of side 4 cm and construct its incircie.
Solution:
Steps of constructions :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 Q10.1
(i) Draw a regular hexagon ABCDEF of side 4 cm.
(ii) Draw the angle bisectors of ∠A and ∠B
which intersect each other at O.
(iii) Draw OL ⊥ AB.
(iv) With centre O and radius OB, draw a circle
which touches the sides of the hexagon

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 More Exercises

Question 1. Use a ruler and compass only in this question. (i) Draw a circle, centre O and radius 4 cm. (ii) Mark a point P such that OP = 7 cm. Construct the two tangents to the circle from P. Measure and record the length of one of the tangents. Solution: Steps of Construction:

  1. Draw a circle with centre O and radius 4 cm.
  2. Take a point P such that OP = 7 cm.
  3. Bisect OB at M.
  4. With centre M and diameter OP, draw another circle intersecting the given circle at A and B.
  5. Join PA and PB. PA and PB is a pair of tangents to the circle.
  6. On measuring PA, it is equal to 5.5 cm. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 Q1.1

Question 2. Draw a line AB = 6 cm. Construct a circle with AB as diameter. Mark a point P at a distance of 5 cm from the mid-point of AB. Construct two tangents from P to the circle with AB as diameter. Measure the length of each tangent Solution: Steps of Construction:

  1. Take a line segment AB = 6 cm.
  2. Draw its perpendicular bisector bisecting it at O.
  3. With centre O and radius OB, draw a circle.
  4. Produce AB to P such that OP = 5 cm.
  5. Draw its perpendicular bisector intersecting it at M.
  6. With centre M and radius OM, draw a circle which intersects the given circle at T and S.
  7. Join OT, OS, TP and SP. PT and PS are the required tangents to the given circle.
  8. On measuring, each tangent is 4 cm long i.e. PT = PS = 4 cm. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 Q2.1

Question 3. Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation. Solution: Steps of construction:

  1. Take a point O.
  2. With centre O and radii 4 cm and 6 cm, draw two concentric circles.
  3. Join OA and take its mid-point M.
  4. With centre M and radius MA, draw another circle which intersects the first circle at P and Q.
  5. Join AP and AQ. AP and AQ are the required tangents to the first circle from point A. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 Q3.1

Question 4. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. Solution: Steps of construction:

  1. Take a point O and with centre O, and radius 3 cm, draw a circle.
  2. Produce its diameter both sides and cut off OP = OQ = 7 cm.
  3. Take mid-points of OP and OQ as M and N respectively.
  4. With centres M and N and OP and OQ as diameters, draw circles which intersect the given circle at A, B and C and D respectively.
  5. Join PA, PB, QC and QD. PA, PB and QC and QD are the required tangents. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 Q4.1

Question 5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Solution: Steps of construction:

  1. Draw a line segment AB = 8 cm.
  2. With centre A and radius 4 cm and with centre B and radius 3 cm, draw circles.
  3. Draw a third circle AB as diameter which intersects the given two circles at C and D and P and Q respectively.
  4. Join AC and AD, BP and BQ.
  5. AC and AD, BP and BQ are the required tangents. ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 Q5.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Constructions Ex 16.1 are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Electrochemistry Class 12 Important Extra Questions Chemistry Chapter 3

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 3 Electrochemistry. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 3 Important Extra Questions Electrochemistry

Electrochemistry Important Extra Questions Very Short Answer Type

Question 1.
Write the name of the cell which Is generally used in hearing aids. Write
the reactions taking place at the anode and the cathode of this cell. (CBSE AI 2017)
Answer:
Mercury cell
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 1
At Cathode: HgO + H2O + 2e → Hg (I) + 2OH

Question 2.
Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell. (CBSE Al 2017)
Answer:
Dry cell
Anode: Zn (s) → Zn2+ + 2e
Cathode: MnO2 + NH4+ + e → MnO(OH) + NH3

Question 3.
Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell. (CBSE Al 2017)
Answer:
Lead storage battery
Anode: Pb (s) + SO42-(aq) → PbSO4(s) + 2e
Cathode: PbO2 + SO42-(aq) + 4H+ + 2e → PbSO4(s) + 2H2O(l)

Question 4.
From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life? (CBSE Delhi 2016)
Answer:
(i) Mercury cell
(ii) Fuel cell
(iii) Lead storage cell
(iv) Dry cell

Question 5.
What happens if external potential applied becomes greater than E°el, of electrochemical cell? (CBSE Al 2016)
Answer:
If E°cell (external) is greater than E°cell, the cell starts acting as an electrolytic cell. In this case, electrical energy is used to carry out non-spontaneous chemical reaction.

Question 6.
Using the E° values of A and B, predict which one is better for coating the surface of iron [E°(Fe2+/Fe) = – 0.44V] to prevent corrosion and why?
Given: E°(A2+|A) = -2.37 V and E°(B2+|B) = – 0.14 V (CBSE Al 2016)
Answer:
‘A’ will prevent iron from rusting. So, we can coat the iron surface with metal A because it has more negative value.

How to calculate emf of a cell Calculate the emf of the cell in which the following reaction takes place.

Question 7.
Given that the standard electrode potentials (E°) of metals are:
K+/K = – 2.93V, Ag+/Ag = 0.80V, Cu2+/Cu = 0.34V, Mg2+/Mg = – 2.37V, Cr3+/Cr = – 0.74V and Fe2+/Fe = – 0.44V. (CBSE Al 2010)
Arrange the metals in the increasing order of their reducing power.
Answer:
Ag+/Ag < Cu2+/Cu < Fe2+/Fe < Cr3+/Cr < Mg2+/Mg < K+/K.

Question 8.
What is change in free energy for
(a) galvanic cell and
(b) electrolytic cell?
Answer:
(a) For a galvanic cell, free energy decreases, i.e. ΔG < 0.
(b) For electrolytic cell, free energy increases, i.e. ΔG > 0.

Question 9.
What is role of ZnCl2 in a dry cell?
Answer:
ZnCl2 combines with NH3 produced to form the complex [Zn(NH3)2Cl2], otherwise the pressure developed due to NH3 would crack the seal of the cell.

Question 10.
When the silver electrode having reduction potential 0.80 V is attached to NHE, will it act as anode or cathode?
Answer:
It will act as cathode.

Question 11.
What is the effect of carbon dioxide in water on corrosion?
Answer:
The presence of carbon dioxide in water increases rusting of iron. Water containing CO2 acts as an electrolyte and increases the flow of electrons from one place to another.

Question 12.
Why is it not possible to measure the voltage of an isolated half reaction?
Answer:
It is not possible to measure the voltage of an isolated half reaction because neither the oxidation nor the reduction can occur by itself. Therefore, we can only calculate the relative electrode potential by connecting it to some standard electrode.

Question 13.
Why does a dry cell become dead after a long time, even if it has not been used?
Answer:
A dry cell becomes dead after a long time because the acidic NH4Cl corrodes the zinc container.

Question 14.
Why does the cell potential of mercury cell remain constant throughout its life? (CBSE AI 2015)
Answer:
This is because the overall cell reaction does not involve any ion in the solution whose concentration changes during its life time.

Question 15.
How can you increase the reduction potential of an electrode?
Answer:
By increasing the concentration of the ions

Question 16.
The E° values of MnO4, Ce4+ and Cl2 are 1.507, 1.61 and 1.358 V respectively. Arrange these in order of increasing strength as oxidising agent.
Answer:
Cl2 < MnO4 < Ce4+.

Question 17.
E° values for Fe3+/Fe2+ and Ag+/Ag are respectively 0.771 V and 0.800 V. Is the reaction:
Fe3+ + Ag → Fe2+ + Ag+ spontaneous or not?
Answer:
E° for the reaction is 0.771 – 0.800 = – 0.029 V.
Therefore, the reaction is not spontaneous.

Question 18.
What is the use of platinum foil in the hydrogen electrode?
Answer:
It is used for inflow and outflow of electrons.

Question 19.
What is meant by limiting molar conductivity? (CBSE 2010)
Answer:
The molar conductivity at infinite dilution or when concentration approaches zero is called limiting molar conductivity.

Question 20.
Express the relation among the conductivity of solution in the cell, the cell constant and the resistance of solution in the cell. (CBSE Delhi 2011)
Answer:
The conductivity (K), cell constant (G°) and resistance (R) of the solution are related as:
K = G° × 1/R
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 3

Question 21.
Explain the relation between conductivity and molar conductivity of a solution held in a ceil. (CBSE Delhi 2011)
Answer:
Conductivity (κ) and molar conductivity are related as:
m = \(\frac{\kappa \times 1000}{M}\)
where M is the molarity of the solution.

Question 22.
Is it safe to stir 1 M AgNO2 solution with a copper spoon?
Given E° (Ag+/Ag) = 0.80 V, E° (Cu2+/Cu) = 0.34 V. Explain.
Answer:
No, copper spoon will dissolve as Cu2+ ions because copper has more tendency to get oxidised than silver.

Question 23.
The e.m.f. of the cell:
Zn | Zn2+ (1M) || H+ (1M) | H2 (1 atm),
Pt is 0.76 V. What is the electrode potential of Zn2+/Zn electrode?
Answer:
cell = E°(H+/H2) – E°(Zn2+/Zn)
0. 76 = 0 – E°(Zn2+/Zn)
∴ E°(Zn2+/Zn) = – 0.76 V.

Question 24.
Write relationship between
(i) standard free energy change and e.m.f. of a cell.
(ii) standard free energy change and equilibrium constant.
Answer:
(i) ∆G° = – nFE°cell
(ii) ∆G° = – RT In Kc

Question 25.
Give the units of specific conductance and molar conductance.
Answer:
Specific conductance: ohm-1 cm-1
Molar conductance: ohm-1 cm2 mol-1

Question 26.
Give one example each of primary cell and secondary cell.
Answer:
Primary cell: Dry cell
Secondary cell: Lead storage battery

Question 27.
How are cell constant and specific conductance related to one another?
Answer:
Specific conductance = Cell constant × Conductance.

Question 28.
Why is the equilibrium constant K related to only E°cell and not Ecell?
Answer:
This is because at equilibrium Ecell = 0.

Question 29.
What is the effect of decreasing concentration on the molar conductivity of weak electrolytes?
Answer:
With the decrease in concentration of weak electrolytes, the molar conductivity increases.

Question 30.
Why is it not possible to determine the potential of a single electrode?
Answer:
Oxidation and reduction cannot occur alone. Therefore, it is not possible to measure single electrode. Moreover, it is a relative tendency and can be measured with respect to a reference electrode only.

Question 31.
Suggest a metal which can be used for cathodic protection of iron?
Answer:
Zinc.

Question 32.
Write the overall cell reaction for lead storage battery.
Answer:
The overall cell reaction is:
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO42- (aq) → 2PbSO4 (s) + 2H2O.

Electrochemistry Important Extra Questions Short Answer Type

Question 1.
The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1? (CBSE AI 2008; CBSE Delhi 2007, 2008, 2012)
Ans.
Conductivity, K = 0.146 × 10-3 S cm-1
Resistance, R = 1500 ohm
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 3
= Conductivity (κ) × Resistance (R)
∴ Cell constant = 0.146 × 10-3 ohm-1 cm-1 × 1500 ohm = 0.219 cm-1.

Question 2.
The conductivity of 0.20 M KCl solution at 298 K is 0.025 S cm-1. Calculate its molar conductivity.
Answer:
Molar conductivity
m = \(\frac{\kappa \times 1000}{\mathrm{C}}\)
k = 0.025 cm-1, C = 0.20 M
∴ ∧m = \(\frac{0.025 \times 1000}{0.20}\) = 125.0 S cm2 mol-1

Question 3.
Resistance of a conductivity cell filled wIth 0.1 M KCl solution is 100. If the resistance of the same cell when filled with 0.02 M KCl solution is 520Ω, calculate the conductivity and molar conductivity of 0.02 M KCl solution. (The conductivity of 0.1 M KCl solution is 1.29 S m-1.) (CBSE AI 2006, CBSE Delhi 2014)
Answer:
Step 1. Let us first calculate the cell constant.
Cell constant, G* = Conductivity (κ) × Resistance (R)
Resistance of 0.1 M KCl solution = 100 Ω
Conductivity of 0.1 M KCl solution = 1.29 S m-1
∴ Cell constant = 1.29 (S m-1) × 100Ω = 129 m-1
or = 1.29 cm-1

Step II. Calculation of conductivity of 0.02 M KCl soLution.
Resistance of solution = 520 Ω
Cell constant (G*) = 1.29 cm-1
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 4

Step III. Calculation of molar conductivity.
m = \(\frac{1000 \times \kappa}{C}\)
C = 0.02 M, κ = 0.248 × 10-2 S cm-1
∴ ∧m = \(\frac{1000 \times 0.248 \times 10^{-2}}{0.02}\) – 124 S cm2 mol-1

Question 4.
The electrical resistance of a column of 0.05 M NaOH+ solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its
(i) resistivity
(ii) conductivity, and
(iii) molar conductivity. (CBSE AI 2012)
Answer:
Cell constant, G* = \(\frac{1}{a}\)
l = 50 cm, diameter = 1 cm
∴ radius = 0.5 cm
Area of cross-section, a = πr2
= 3.14 × (0.5)2 = 0.785 cm2
∴ G* = \(\frac{50}{0.785}\) = 63.694 cm-1

(i) Resistivity,
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 5

(ii) Conductivity,
κ = \(\frac{1}{\rho}=\frac{1}{87.135}\) = 1.48 × 10-2 cm-1

(iii) Molar conductivity,
m = \(\frac{\kappa \times 1000}{\mathrm{C}}\)
c = 0.05 M
∴ ∧m = \(\frac{1.148 \times 10^{-2} \times 1000}{0.05}\) = 229.6 s cm2 mol-1

Question 5.
The molar conductivities at infinite dilution for sodium acetate, hydrochloric acid and sodium chloride are 91.0, 425.9 and 126.4 S cm2 mol-1 respectively at 298 K. Calculate the molar conductivity of acetic acid at infinite dilution. (CBSE Delhi 2010)
Answer:
Molar conductivity at infinite dilution for acetic acid can be calculated as:
∧° (CH3COOH ) = λ°H+ + ∧°CH3COO
= λCH3COO + λ°Na+ + λ°H+ + λ°cl – λ°Na+ – λ°cl
∧°(CH3COOH+) = ∧°(CH3COONa) + ∧° (H+Cl) – ∧° (NaCl)
∧°(CH3COONa) = 91.0 S cm2 mol-1
∧°(H+Cl) = 425.9 S cm2 mol-1
∧°(NaCl) = 126.4 S cm2 mol-1
∴ ∧°(CH3COOH+) = 91.0 + 425.9 – 126.4 = 390.5 S cm2 mol-1.

Question 6.
(a) Write the reaction that occurs at anode on electrolysis of concentrated H2SO4 using platinum electrodes.
(b) What is the effect of temperature on ionic conductance? (CBSE Al 2019)
Answer:
(a) At anode:
2SO42-(aq) → S2O82-(aq) + 2e
(b) Ionic conductance will increase with increase in temperature.

Question 7.
Write anode and cathode reactions that occur in dry cell. How does a dry cell differ from a mercury cell? (CBSE Al 2019)
Answer:
Anode: Zn → Zn2+ + 2e
Cathode:
NH4+(aq) + MnO2(s) + e → MnO(OH) + NH3
The potential of a dry cell decreases slowly but continuously, while the potential of mercury cell remains constant throughout its life.

Question 8.
Calculate the limiting molar conductivity of CaS04 if limiting molar conductivities of calcium and sulphate ions are 119.0 and 106.0 S cm2 mol-1 respectively. (CBSE Sample Paper 2012)
Answer:
∧°m (CaSO4) = λm (Ca2+) + λ°m (SO42-)
= 119.0 + 106.0
= 225.0 S cm2- mol-1.

Question 9.
Arrange the following solutions in the decreasing order of specific conductance.
(i) 0.01M NaCl (ii) 0.05M NaCl (iii) 0.1M NaCl (iv) 0.5M NaCl
Ans.
(iv) 0.5M NaCl > (iii) 0.1M NaCl > (ii) 0.05M NaCl > (i) 0.01M NaCl

Question 10.
Write the Nernst equation and calculate the e.m.f. of the following cell at 298 K:
Cu(s) | Cu2+ (0.130 M) | | Ag+ (1.0 × 10-4 M)) | Ag (s)
Given: E(Cu2+/Cu) = + 0.34V and E(Ag2+/Ag) = + 0.80V (CBSE Al 2004)
Answer:
The electrode reactions and cell reaction are:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 7

Question 11.
Calculate Ecell, for the following reaction at 298K:
2Cr(s) + 3Fe2+ (0.01 M) → 2Cr3+ (0.01M) + 3Fe(s)
(Given: Ecell = 0.261v)                (CBSE AI 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 8
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 9

Question 12.
For the reaction:
2AgCl (s) + H2 (g) (1atm) → 2Ag(s) + 2H+ (0.1M) + 2Cl (0.1M)
ΔG° = – 43600 J at 25°C
Calculate the e.m.f of the cell. (CBSE AI 2018)
Answer:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 10

Question 13.
Write the name of two fuels other than hydrogen used in fuel cell. Write two advantages of fuel cell over an ordinary cell. (CBSE Al 2019)
Answer:
Methane, oxygen
Advantages of fuel cells:
(i) Fuel cells have greater efficiency than ordinary cells.
(ii) These do not produce any harmful by-product and therefore, do not cause any pollution.

Question 14.
Define conductivity and molar conductivity for the solution of an electrolyte. Why does the conductivity of solution decrease with dilution? (CBSE 2019C)
Answer:
Conductivity of a solution at any given concentration is the conductance of one unit volume of solution kept between two platinum electrodes with unit area of cross section and at a distance of unit length.

Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore,
m = \(\frac{\mathrm{κA}}{\mathrm{l}}\)
Since l = 1 and A = V (Volume containing 1 mole of electrolyte).
m = κV
Conductivity always decreases with decrease in concentration both for weak and strong electrolytes. It is because the number of ions per unit volume that carry the current in solution decreases on dilution.

Question 15.
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? (CBSE AI 2012)
Answer:
The conductivity (κ), cell constant (G*) and resistance (R) of solution are related as:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 11
Molar conductivity (∧m) is related to conductivity (κ) as:
m = \(\frac{\kappa \times 1000}{M}\)

Question 16.
The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1. Calculate the conductivity of this solution. (CBSE AI 2012)
Answer:
m = 138.9 S cm2 mol-1, M = 1.5 M
138.9 = \(\frac{\kappa \times 1000}{1.5}\)
∴ κ = \(\frac{138.9 \times 1.5}{1000}\) = 0.208 ohm-1 cm-1.

Electrochemistry Important Extra Questions Long Answer Type

Question 1.
(a) Following reaction takes place in the cell:
Zn (s) + Ag2O (s) + H2O (l) → Zn2+ (aq) + 2Ag (s) + 20H (aq)
Calculate ΔrG° of the reaction.
[Given: E°(Zn2+/Zn) = – 0.76 V,
(Ag2+/Ag) = 0.80 V, 1 F = 96,500 C mol-1]
Answer:
cell = E°(Ag2+/Ag) – E°(Zn2+/Zn)
= 0.80 – (-0.76) = 1.56V
ΔG° = -nFE°cell          (Here n = 2)
= – 2 × 96500 × 1.56
= – 301080 J mol-1
or = – 301.080 kJ mol-1

(b) How can you determine limiting molar conductivity (∧m°,) for strong electrolyte and weak electrolyte? (CBSE Al 2019)
Answer:
For strong electrolytes ∧m° can be obtained as intercept from the plot of Am vs C-1/2 graph. For weak electrolytes ∧m° can be obtained from Kohlrausch law.

Question 2.
(i) Explain the following:
(a) CO2 is always present in natural water. Explain its effect (increases, stops or no effect) on rusting of iron.
(b) Rusting of iron is quicker in saline water than in ordinary water. Explain.
Answer:
(i) (a) Presence of CO2 in natural water increases rusting of iron. It dissolves in water to form H2CO3 which gives H+ ions. The H+ ions accelerate the process of corrosion.
In rusting of iron, Fe oxidises to Fe2+ ions
Fe (s) → Fe2+ (aq) + 2e (anode)
The released electrons go to the cathode and reduce oxygen in the presence of H+ ions (obtained from H2CO3). The reaction occurs at cathode. Thus, CO2 increases rusting.
O2 (g) + 4H+(aq) + 4e → 2H2O (l)

(b) Rusting of iron is quicker in saline water (salt water) than in ordinary water. This is mainly due to the fact that saline water increases the electrical conduction of electrolyte solution formed on the metal surface. Therefore, rusting becomes more serious problem where salt water is present.

(ii) Discuss electrical protection for preventing rusting of iron pipes in underground water.
Answer:
In this method iron articles which are in contact with water such as underground water pipes are protected from rusting. The article of iron is connected with more active metals like magnesium or zinc. This prevents its tendency to lose electrons and therefore, corrosion is prevented. The cathodes of magnesium of zinc can be fixed to the surface of iron or burned in sub-soil water near by the pipes.

Question 3.
Give the construction and working of hydrogen standard electrode potential?
Answer:
The standard hydrogen electrode consists of platinum wire sealed in a glass tube and has a platinum foil attached to it. The foil is coated with finely divided platinum and acts as platinum electrode. It is dipped into an acid solution containing H+ ions in 1 M concentration (1 M HCl). Pure hydrogen gas at 1 atmosphere pressure is constantly bubbled into the solution at constant temperature of 298 K. The surface of the foil acts as a site for the reaction. This is shown in figure.
The following reactions occur in this half cell depending upon whether it acts as an anode or as a cathode:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 12
If S.H.E. acts as anode:
H2(g) → 2H+ + 2e
If S.H.E. acts as cathode:
2H+ + 2e → H2(g)
The electrode potential of an electrode can be determined by connecting this half cell with a standard hydrogen electrode. The electrode potential of the standard hydrogen electrode is taken as zero.

Measurement of the standard electrode potential (E°): The standard electrode potential of a metal electrode is measured with respect to a standard hydrogen electrode. A cell is prepared in which the metal electrode constitutes one half cell and the S.H.E. as the other half cell (anode). The electrons released by the metal in the oxidation half cell are accepted by the H+ ions of the acid in the reduction half cell.
The cell may be represented as:
Pt(s) | H2(g, 1 atm) | H+ (aq, 1 M) || Mn+ (aq, 1M) | M
Now, e.m.f. of cell
e.m.f. = ER – EL
Since the potential of S.H.E. has been fixed to be zero, i.e. EL = 0 so that
e.m.f. = ER – 0
or ER = e.m.f.
From the knowledge of e.m.f. of the cell, the electrode potential of the electrode can be calculated. For example, if we wish to determine the electrode potential of zinc electrode in 1 M solution of ZnSO4, it is combined with S.H.E. The e.m.f. of the cell is found to be – 0.76 V so that
e.m.f. = EL – EL
– 0.76 = ER – 0
or ER = – 0.76 V.

Question 4.
What are fuel cells? Discuss briefly hydrogen-oxygen fuel cell?
Answer:
Fuel cells. These are voltaic cells in which the reactants are fed continuously to the electrodes. These are designed to convert the energy from the combustion of fuel such as H2, CO, CH4, etc. directly into electrical energy. The common example is hydrogen-oxygen fuel cell as described below:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 13
Fig. A simple H2 – O2 fuel cell.

In this cell, hydrogen and oxygen are bubbled through a porous carbon electrode into concentrated aqueous sodium hydroxide as shown in figure. The diffusion rates of the gases into the cell are carefully regulated to get maximum efficiency. In the anode compartment hydrogen is oxidised while oxygen in the cathode compartment is reduced. The net reaction is the same as burning of hydrogen and oxygen to form water.
The reactions are given below:
At anode:
2 [H2 (g) + 2OH(aq) → 2H2O (l) + 2e]

At cathode:
O2 (g) + 2H2O (l) + 4e → 4OH(aq)

Overall reaction:
2H2 (g) + O2(g) → 2H2O (l)
The catalysts (Pt, Ag or Co) are also added.
This cell runs continuously as long as the reactants are fed. These fuel cells are more efficient than conventional methods of generating electricity on a large scale by burning hydrogen, carbon fuels because these fuel cells convert the energy of the fuel directly into electricity.
The cell has been used for electric power in the Apollo space programme.

Question 5.
What is corrosion? What are the factors which affect corrosion?
Answer:
Corrosion. It is a process of eating away of metals when exposed to the atmosphere surrounding it. Many metals when exposed to the atmosphere, react with air or water in the environment to form undesirable compound on their surface. In case of iron, the corrosion is called rusting. The red or orange coating that forms on the surface of iron when exposed to air and moisture is called rust. Chemically, rust is a hydrated form of ferric oxide, Fe2O3 . xH2O.
Factors which affect corrosion. The main factors which affect corrosion are:
1. Position of metals in e.m.f. series: The reactivity of a metal depends upon its position in the electrochemical series. More the reactivity of the metal, more will be the possibility of the metal getting corroded.

2. Presence of impurities in metals: The impurities help in setting up a voltaic cell, which increases the speed of corrosion.

3. Presence of electrolytes: Presence of electrolytes in water also increases the rate of corrosion. For example, corrosion of iron in sea water takes place to larger extent than in distilled water.

4. Presence of CO2 in water. Presence of CO2 in natural water increases rusting of iron. Water containing CO2 acts as an electrolyte and increases the flow of electrons from one place to another.

5. Presence of protective coatings. When the iron surface is coated with layers of metals more active than iron, then the rate of corrosion is retarded. For example, coating of zinc on iron prevents rustings.

Question 6.
What type of battery is lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery. (CBSE 2011)
Answer:
It is a secondary cell.
Anode reaction:
Pb (s) + SO42- (aq) → PbSO4 (s) + 2e

Cathode reaction:
PbO2 (s) + 4H+ (aq) + SO42- (aq) + 2e → PbSO4 (s) + 2H2O (l)

Overall reaction:
Pb (s) + PbO2 (s) + 2H2S04 → 2PbSO4 (s) + 2H2O (l)

Question 7.
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. (CBSE 2011)
Answer:
The chemistry of corrosion of iron is an electrochemical theory which involves oxidation and reduction reactions. According to this theory it is believed that non-uniform surface of metal or impurities present in iron behave like small electric cells (called corrosion couples) in the presence of water containing dissolved oxygen or carbon dioxide. A film of moisture with dissolved CO2 acts as electrolytic solution covering the metal surface at various places. This is shown in Fig. In these small electrolytic cells, pure iron acts as anode while cathodes are impure portions. The overall rusting involves the following steps:
Oxidation occurs at the anode of each electrochemical cell. Therefore, at each anode iron is oxidised to Fe2+ ions.

At anode:
Fe (s) → Fe2+ (aq) + 2e …… (i)
Thus, the metal atoms in the lattice pass into the solution as ions, leaving electrons on the metal itself. These electrons move towards the cathode region through the metal.

At the cathode of each cell, the electrons are taken up by hydrogen ions (reduction takes place). The H+ ions are obtained either from water or from acidic substance in water:
H2O ⇌ H+ + OH …… (ii)
or CO2 + H2O → H+ + H+CO3 …… (iii)

At cathode:
H+ + e → H+ …… (iV)
Thus, hydrogen atoms on the iron surface reduce dissolved oxygen.
4H + O2 → 2H2O ……..(v)
Therefore, the overall reaction at cathode of different electrochemical cells may be written as:
4H+(aq) + O2 (g) + 4e → 2H2O (l) ……. (vi)
The overall redox reaction may be written by multiplying reaction at anode Eq. (i) by 2 and adding reaction at cathode Eq. (iv) to equalise number of electrons lost and gained, i.e. Oxidation half reaction:
Fe (s) → Fe2+ (aq) + 2e × 2

Reduction half reaction:
4H+(aq) + O2 (g) + 4e → 2H2O (l)

Overall cell reaction:
2Fe (s) + 4H+(aq) + O2 (g) → 2Fe2+(aq) + 2H2O (l)
The ferrous ions are oxidised further by atmospheric oxygen to Fe3+ (as Fe2O3) and form rust
4Fe2+ + O2 (g) + 4H2O → 2Fe2O3 + 8H+
and Fe2O3 + xH2O → Fe2O3. xH2O
The H+ ions produced above are also used for reaction (iv).
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 14

Mechanism of rusting of air.

Question 8.
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? (CBSE 2014)
Answer:
Kohlrausch law of independent migration of ions states that at infinite dilution when the dissociation is complete, each ion makes a definite contribution towards molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated. Kohlrausch law also means that the limiting molar conductivity of an electrolyte is sum of the individual contributions of the ions of the electrolyte.

Conductivity (κ) of the electrolyte solution decreases with dilution because the number of ions per unit volume furnished by an electrolyte decreases with dilution.

Question 9.
(i) Following reactions occur at cathode during the electrolysis of aqueous silver chloride
solution:
Ag+(aq) + e → Ag(s)        E° = + 0.80 V
H+(aq) + e → \(\frac{1}{2}\)H2(g)         E ° = 0.00 V
On the basis of their standard electrode potential (E°) values which reaction is feasible at the cathode and why?
(ii) Define limiting molar conductivity. Why does conductivity of an electrolyte solution decrease with the decrease in concentration? (CBSE Delhi 2015)
Answer:
(i) The reaction, Ag+(aq) + e → Ag(s) (E° = + 0.80 V) is feasible at cathode because its reduction potential is higher than other reaction (H+ + e → \(\frac{1}{2}\) H2; E° = 0.0 V)

(ii) The molar conductivity of the solution when the concentration approaches zero (infinite dilution) is called limiting molar conductivity. Conductivity is the conductance of one centimeter cube of the solution. Upon diluting the solution, the concentration of ions per centimeter cube decreases and hence conductivity decreases with dilution.

Question 10.
(i) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride
solution:
Na+(aq) + e → Na(s)        E° = -2.71V
H+(aq) + e → H2(g)         E ° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
Answer:
(i) The reaction: H(aq) + e → H2(g) (E° = 0.00 V) is feasible at cathode because its reduction
potential is higher than the other electrode reaction.

(ii) Why does the cell potential of mercury cell remain constant throughout its life? (CBSE 2015)
Answer:
The cell potential of mercury cell remains constant because the overall reaction does not involve any ion in the solution whose concentration changes during its life time.

Question 11.
Zinc rod is dipped in 0.01 M solution of zinc sulphate when temperature is 298 K. Calculate the electrode potential of zinc.
(Given: E°zn2+/Zn = – 0.76 V; log 10 = 1) (CBSE 2019 C)
Answer:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 15

Question 12.
Write Nernst equation and calculate e.m.f. of the following cells at 298 K:
(i) Mg (s) | Mg2+ (0.001 M) | | Cu2+ (0.0001 M) | Cu (s)
Given: EMg2+/Mg = -2.37 V, ECu2+/Cu = 0.34V
Answer:
(i) The electrode reactions and cell reactions are:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 16
Since the reaction involves 2 moles of electrons and therefore, n = 2 and the Nernst equation for the cell at 298 K is:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 17
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 18

(ii) Fe (s) | Fe2+ (0.001 M) | | H+ (1M) | H2 (1 atm) | Pt
Given: EFe2+/Fe = -0.44V (CBSE Delhi 2013)
Answer:
Fes | Fe2+ (0.001 M) || H+ (1 M) | H2 (1 atm) | Pt
The electrode reactions and overall cell reactions are:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 20
Since the reaction involves 2 moles of electrons, therefore, n = 2 and the Nernst equation at 298 K
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 19

(iii) Sn (s) | Sn2+ (0.050 M) | | H+ (0.020 M) | H2 (1 atm) | Pt
Given: ESn2+/Cn = – 0.14V (CBSE Al 2018)
Answer:
The cell is:
Sn(s) | Sn2+(0.050M) | | H+(0.020 M) | H2(1 atm) | Pt
The electrode reactions and cell reactions are:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 21
The reaction involves 2 moles of electrons, therefore, n = 2 and the Nernst equation is:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 22

Question 13.
(a) The e.m.f. of the following cell at 298 K is 0.1745 V:
Fe (s) | Fe2+ (0.1 M) | | H+ (x M) | H2 (g) (1 bar) | Pt (s)
Given: E°Fe2+Fe = – 0.44 V
Calculate the H+ ions concentration of the solution at the electrode where hydrogen is being produced.
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 23

(b) Aqueous solutions of copper sulphate and silver nitrate are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.
Answer:
The mass of copper and silver deposited at the cathode will be different.
Faraday’s second law of electrolysis: It states that when same quantity of electricity is passed through different electrolytic solutions connected in series, the weights of the substances produced at the electrodes are directly proportional to their chemical equivalent weights.

For example, when same current is passed through two electrolytic solutions, containing copper sulphate (CuSO4) and silver nitrate (AgNO3) connected in series, the weights of copper and silver deposited are:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 24

OR

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10-5 S cm-1.
Given λ°H+ = 349.6 S cm2 mol-1; λ°CH3COO = 40.9 S 2 mol-1
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 25
Electrolyte B is a strong electrolyte.

(b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer. (CBSE Sample Paper 2019)
Answer:
Limiting molar conductivity increases only to a smaller extent for a strong electrolyte, as on dilution the interionic interactions are overcome. Limiting molar conductivity increases to a larger extent for a weak electrolyte, as on dilution the degree of dissociation increases, therefore the number of ions in total volume of solution increases. Therefore, ‘B’ is a strong electrolyte.

Question 14.
(i) Calculate E°cell for the following reaction at 298 K:
2Cr(s) + 3Fe2+ (0.01 M) → 2Cr3+ (0.01 M) + 3Fe(s)
Given: Ecell = 0.261 V
Answer:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 26

(ii) Using the E° values of A and B, predict which one is better for coating the surface of iron [Ee°Fe2+/Fe) = – 0.44 V] to prevent corrosion and why?
Given: E°(A2+/A) = -2.37 V: E°(B2+/B) = – 0.14 V
Answer:
‘A’ will prevent iron from corrosion. So, we can coat the iron surface with metal A because it has more negative E° value.

OR

(i) The conductivity of 0.001 mol L-1 solution of CH3COOH Is 3.905 × 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α).
Given λ°(H+) = 349.65 cm2 mol-1 and λ°(CH3COO) = 40.9 S cm2 mol-1.
Answer:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 27

(ii) Define electrochemical cell. What happens if external potential applied becomes greater than E°cell of electrochemical cell? (CBSE 2016)
Answer:
Electrochemical cell is a device used for the production of electricity from energy released during spontaneous chemical reaction. Electrochemical cell converts chemical energy into electrical energy.
If E°cell (external) > E°cell, the cell starts acting as an electrolytic cell. In this case, electrical energy is used to carry out non-spontaneous chemical reaction.

Question 15.
(i) Define the following terms:
(a) Limiting molar conductivity
(b) Fuel cell
(ii) Resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L-1 KCl solution. The conductivity of 0.1 mol L-1 KCl solution is 1.29 × 10-2-1 cm-1.

OR

(i) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
(ii) Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) | | Cu2+(0.01)|Cu(s) [Given E°cell = +2.71 V, 1 F = 96500 C mol-1] (CBSE Delhi 2014)
Answer:
(a) The molar conductivity of a solution when concentration approaches zero is called limiting molar conductivity and is expressed as ∧°m.

(b) Fuel cells are voltaic cells in which the reactants are continously supplied to the electrodes and are designed to convert energy from the combustion of fuels such as H2, CO, CH4, etc. directly into electrical energy.

(ii) Let us first calculate cell constant
Cell constant, G* = Conductivity (κ) × Resistance (R)
Resistance of 0.1 M KCl solution = 100 Ω
Conductivity of 0.1 M KCl solution = 1.29 Sm-1
Cell constant = 1.29 (Sm-1) × 100 Ω
= 129 m-1 = 1.29 cm-1
Calculation of conductivity of 0.02 M KCl solution,
Resistance of solution = 520 Ω
Cell constant (G*) = 1.29 cm-1
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 28

OR

(i) Faraday’s first law of electrolysis states that the amount of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolytic solution.
Cu2+ + 2e → Cu(s)
Reduction of 1 mol of Cu2+ ion requires 2F of electricity.

(ii) The electrode reactions and cell reaction are
Mg(s) → Mg2+(aq) + 2e (at anode)
Cu2+(aq) + 2e → Cu(s) at cathode
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 29
= 2.71 – 0.0295
= 2.6805 v.

Question 16.
(i) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO3)2 solution with a current of 2.00 A for 3 hours? [Hg(NO3)2 = 200.6 g mol-1].
Answer:
Hg2+ + 2e → Hg
Quantity of electricity passed = 1 × t(sec)
= 2.0 A × 3.0 × 60 × 60 = 21 600C
2 × 96500 C of electricity produces mercury = 1 mol
21,600 C of electricity will produce mercury
= \(\frac{1}{2 \times 96500}\) × 21600 = 0.112 mol

(ii) A voltaic cell is set up at 25° C with the following half cells:
Al3+ (0.001 M) and Ni2+ (0.50 M).
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given: E°Ni2+/Ni = – 0.25 V, E°Ni2+/Ni = – 1.66V) (CBSE 2011)
Answer:
Cell reaction is:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 30

Question 17.
(i) What type of battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
Answer:
It is secondary cell.
Anode reaction:
Pb (s) + SO42-(aq) → PbSO4(s) + 2e

Cathode reaction:
PbO2 (s) + 4H+(aq) + SO42-(aq) + 2e → PbSO4(s) + 2H2O (l)

Overall reaction:
Pb (s) + PbO2 (s) + 2H2SO4 → PbSO4(s) + 2H2O (l)

(ii) Calculate the potential for half cell containing 0.1 M K2Cr2O7(aq), 0.20 M Cr3+(aq) and 1.0 × 10-4 M H+(aq). The half cell reaction is
Cr2O72-(aq) + 14 H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l),
and the standard electrode potential is given as E° = 1.33 V. (CBSE 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 31
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 32

Question  18.
(i) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.
Answer:
Kohlrausch law states that at infinite dilution, each ion makes a definite contribution towards molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated. This means that the molar conductivity at infinite dilution for a given salt can be expressed as the sum of the individual contributions from the ions of the electrolyte.

For acetic acid, limiting molar conductvity at infinite dilution can be written as:
∧°(CH3COOH) = λ° (CH3COO) + λ° (H+) ……. (i)
This equation can be obtained by the knowledge of molar conductivity at infinite dilution for some electrolytes. For example, consider the strong electrolytes HCl, NaCl and CH3COONa. From Kohlrausch’s law:
∧°(CH3COONa) = λ°(CH3COO) + A°(Na+) …… (i)
∧°(HCl) = λ°(H+) + λ°(Cl) …….. (ii)
∧° (NaCl) = λ° (Na+) + λ° (Cl) …….. (iii)
It is clear that
λ° (CH3COO) + λ° (H+) = [λ°(CH3COO) + A°(Na+)] + [λ°(H+) + λ°(Cl)] – λ° (Na+) + λ° (Cl)
∧°(CH3COOH) = ∧°(CH3COONa) + ∧°(HCl) – ∧° (NaCl)

(ii) Calculate ∧°m for acetic acid
Given that ∧°m(HCl) = 426 S cm2 mol-1
∧°m(NaCl) = 126 S cm2 mol-1
∧°m(CH3COONa) = 91 S cm2 mol-1 (CBSE Delhi 2010)
Answer:
We know
∧°(CH3COOH) = ∧°(CH3COONa) + ∧°(HCl) – ∧° (NaCl)
= 91 + 426 – 126 = 391 S cm2 mol-1

Question 19.
cell for the given redox reaction is 2.71 V.
Mg(s) + Cu2+(0.01 M) → Mg2+(0.001 M) + Cu(s)
Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V
Answer:
Mg(s) + Cu2+(0.01 M) → Mg2+(0.001M) + Cu(s)
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 35
= 2.71 – 0.0295 log 10-1
= 2.71 + 0.0295
= 2.7395 V or approximately 2.74 V
(i) When Eext < 2.71, current flows from Cu to Mg
(ii) When Eext > 2.71, current flows from Mg to Cu.

OR

(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeS04 and ZnS04 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. (Molar mass: Fe = 56 g mol-1, Zn = 65.3 g mol-1, 1F = 96500 C mol-1)
(b) In the plot of molar conductlvtty (∧m) vs square root of concentration (c1/2), the following curves are obtained for two electrolytes A and B:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 34
Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation 0f ∧m to concentration approaching zero for electrolytes A and B? (CBSE Delhi 2019)
Answer:
(a) Cell-X contains FeSO4
FeSO4(aq) → Fe2+ + SO42-
Fe2+ + 2e → Fe (at cathode)
1 mole or 56g of Fe is deposited by 2 × 96500C
2.8g of Fe with be deposited by= 2 × 96500 × 2.8
= 9650 C
Q = I × t
9650 = 2 × t
or t = \(\frac{9650}{2}\) = 4825s
In cell Y
1 mol or 65.3 g of Zn is deposited by 2 × 96500 C
or 2 × 96500 C of electricity will deposit Zn = 65.3 g
9650 C of electricity will deposit Zn = \(\frac{65.3 \times 9650}{2 \times 96500}\) = 3.265 g

(b) (i) Electrolyte A is strong electrolyte.
(ii) Electrolyte B is weak electrolyte.
(iii) For A, the intercept will give ∧°m.
For B, we cannot obtain ∧°m on extrapolation.

Question 20.
(i) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte.
Answer:
Molar conductivity is the conducting power of all the ions produced by dissolving one gram mole of an electrolyte in solution. It is expressed as ∧m and is defined as:
m = \(\frac{\kappa \times 1000}{M}\)
where M is the molarity of the solution. Its units are ohm-1 cm2 mol-1.
The molar conductance of an electrolytic solution decreases with increase in concentration. For weak electrolytes, molar conductivity increases sharply with dilution. On the other hand, for strong electrolytes, molar conductivity increases slowly with dilution.

For weak electrolytes, the increase in molar conductivity is due to increase in degree of ionisation with dilution. For strong electrolytes, the increase in molar conductance with dilution is because of decrease in interactions between ions with dilution. The decrease in molar conductivity for weak electrolyte (CH3COOH) and strong electrolyte (KCl) is shown below:
Class 12 Chemistry Important Questions Chapter 3 Electrochemistry 33

(ii) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1 ? (CBSE Delhi 2012)
Answer:
R = 1500 Ω
Molarity = 0.001 M
K = 0.146 × 10-3S cm-1
\(\frac{1}{A}\) = ?
Conductivity (κ) = Conductance (G) × Cell Constant \(\left(\frac{1}{A}\right)\)
or κ = \(\frac{1}{R} \times \frac{l}{A}\)
or Cell constant \(\frac{1}{A}\) = R × κ
= 1500 Ω × 0.146 × 10-3S cm-1
\(\frac{1}{A}\) = 0.219 cm-1

Question 21.
(i) What type of a battery is the lead storage battery ? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.
Answer:
It is secondary cell.
Anode reaction:
Pb (s) + SO42-(aq) → PbSO4(s) + 2e

Cathode reaction:
PbO2 (s) + 4H+(aq) + SO42-(aq) + 2e → PbSO4(s) + 2H2O (l)

Overall reaction:
Pb (s) + PbO2 (s) + 2H2SO4 → PbSO4(s) + 2H2O (l)

(ii) In the button cell, widely used in watches, the following reaction takes place
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH(aq)
Determine E° and ∆G° for the reaction.
(given: E°Ag2+/Ag = + 0.80V, E°Zn2+/Zn = – 0.76V)
Answer:
Ag2+/Ag = + 0.80V, E°Zn2+/Zn = – 0.76V
cell = E°Ag2+/Ag – E°Zn2+/Zn
cell = + 0.80 – (- 0.76) = 1.56 V
ΔG° = – nFE°
n = 2, F = 96500 C
∴ ΔG° = – 2 × 96500 × 1.56
= – 301080 J = – 301.08 kJ

Hydrocarbons Class 11 Important Extra Questions Chemistry Chapter 13

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 13 Hydrocarbons. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 13 Important Extra Questions Hydrocarbons

Hydrocarbons Important Extra Questions Very Short Answer Type

Question 1.
Give different isomers of C4H10 with their I.U.P.A.C. names.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 1

Question 2.
Give the I.U.P.A.C. name of the lowest molecular weight alkane that contains a quaternary carbon.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 2
It is and its I.U.P.A.C. name is 2, 2-Dimethylpropane.

Question 3.
Which of the following has the highest boiling point?
(i) 2-methylpentane
(ii) 2, 3 – dimethylbutane
(iii) 2, 2-dimethylbutane.
Answer:
(i) 2—methyl pentane has the largest surface area and hence has the highest boiling point.

Alkene-reactions-cheat-sheet-summary-for-organic-chemistry-reactions.

Question 4.
Give the structure of the alkene (C4H8) which adds on HBr in the presence and in the absence of peroxide to give the same product C4H9Br.
Answer:
2-Butene with structure CH3 – CH = CH — CH3 being symmetrical gives the same product, i.e., 2-bromobutane CH3 CH (Br) CH2CH3.

Question 5.
How will you separate propene from propyne?
Answer:
Bypassing the mixture through ammoniacal AgNO3 solution when propyne reacts while propene passes over.

Question 6.
Name two reagents that can be used to distinguish \ between ethene and ethyne.
Answer:
Tollen’s reagent | Ammoniacal AgNO3 | and amm. CuCl solution.

Question 7.
How will you detect the presence of unsaturation in an organic compound?
Answer:
Either by Baeyer’s reagent
Hydrocarbons Class 11 Important Extra Questions Chemistry 3
or by Br, in CC14.

Question 8.
Arrange the following In order of increasing volatility: gasoline, kerosene, and diesel.
Answer:
Diesel, kerosene, gasoline.

Question 9.
Arrange the following: HCl, HBr, HI, HF in order of decreasing reactivity towards alkenes.
Answer:
HF, HCl, HBr, HI.

Question 10.
Out of ethylene and acetylene which is more acidic and why?
Answer:
Acetylene. Ethylene and acetylene have sp2, sp hybridized C atoms respectively. Due to the 50% S character of the C – H bond of acetylene rather than the 33% S-Character of the C – H bond in ethene, acetylene is more acidic.

Question 11.
Write the structure of the alkene which on reductive ozonolysis gives butanone and ethanal.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 4

Question 12.
Write the I.U.P.A.C. names of
Hydrocarbons Class 11 Important Extra Questions Chemistry 5
Answer:
3-methylpent-l— en—4-yne.

(ii) CH2 = CH – CH (CH3) – CH = CH – CH – CH2,
Answer:
3-methylhept-1, 4, 6—triene.

Question 13.
Draw the structures of the following:
(i) Dicyclopropyl methane
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 6

(ii) 2-methyl-3—isopropyl heptane.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 7

Question 14.
What effect the branching of an alkane has on its boiling point?
Answer:
Branching decreases the boiling point.

Question 15. Which of the following polymerizes most readily?
(i) Acetylene
(ii) Ethene
(iii) Buta —1, 3—diene
Answer:
(iii) Buta—1, 3—diene polymerizes most readily, being more reactive.

Question 16.
Arrange the following in increasing order of their release of energy on combustion
Hydrocarbons Class 11 Important Extra Questions Chemistry 8
Answer:
The more the number of C atoms having maximum hydrogen hydrogens, i.e., CH3 groups, the greater is the heat of combustion. Thus the increasing order of heat of combustion is (iii) < (iv) < (i) < (ii).

Question 17.
Arrangement of the following set of compounds in order of their decreasing relative reactivity with an electrophile E+.
(i) Chloro benzene, 2, 4 —dinitrochlorobenzene p—nitrochlorobenzene.
(ii) toluene, p – CH3 – C6H4 – CH3, p – CH3 – C6H4 – NO2, p – O2N – C6H4 – NO2
Answer:
Electron-donating groups increase the reactivity towards an electrophile E+, while electron-withdrawing groups decrease the reactivity. Thus
(i) Chlorobenzene > p — nitrochlorobenzene > 2. 4—di nitrochlorobenzene.
(ii) p – CH3 – C6H4 – CH3 > toluene > p – CH3 – C6H4 – NO2 > p – O2N – C6H4 – NO2.

Question 18.
What is the order of reactivity of halogen and alkyl groups in the dehydrohalogenation of alkyl halides to give alkenes?
Answer:

  • Halogens: Iodine > Bromine > Chlorine
  • Alkyl group: Tert > secondary > primary.

Question 19.
What products are formed when zinc reacts with
(i) vicinal C2H4Br2 and
Answer:
CH2Br – CH2Br + Zn → CH2 = CH2 + ZnBr2.

(ii) CH3CHBr – CH2Br.
Answer:
CH3 – CHBr – CH2Br + Zn → CH3 — CH = CH2 + ZnBr2.

Question 20.
What does L.P.G. stand for?
Answer:
L.P.G. stands for liquefied petroleum gas.

Question 21.
What do the terms (i) CNG and LPG stand for?
Answer:

  • CNG: Compressed natural gas
  • LPG: Liquefied Petroleum gas.

Question 22.
What are sources to obtain:
(i) LPG
Answer:
LPG is obtained by the fractional distillation of petroleum.

(ii) CNG?
Answer:
CNG is obtained by the fractional distillation of coal tar.

Question 23.
Write down the structures and names of all isomers with a molecular formula of C5H12.
Answer:

  1. (i) CH3 — CH2 — CH2 — CH2 — CH3 is n-Pentane or Pentane.
  2. Hydrocarbons Class 11 Important Extra Questions Chemistry 9
    is iso-pentane or 2-Methylbutane.
  3. Hydrocarbons Class 11 Important Extra Questions Chemistry 10
    is neo-pentane or 2, 2-Dimethylpropane.

Question 24.
The sodium salt of which acid will give ethane on heating with soda-lime? Give reaction.
Answer:
Propionic acid.
Hydrocarbons Class 11 Important Extra Questions Chemistry 11

Question 25.
What products are obtained by the acidic dehydration of
(i) Ethanol
Answer:
Ethene

(ii) Propan – 2—ol.
Answer:
Propene.

Question 26.
Out of cis-2-butene and trans-2-butene which is polar and which one is non-polar?
Answer:
Cis – 2 – butene is polar (μ = 0.33 D) and trans – 2 – butene is non-polar (μ = 0). .

Question 27.
Arrange the following in the decreasing order of acidic character.
(i) C2H4, C2H6, C2H2
Answer:
H – C ≡ C H > H2C = CH2 > H3C – CH3

(ii) CH3 – C = CH, C2H2, CH3 – C = C – CH3
Answer:
HC = CH > CH3 — C ≡ CH > > CH3 – C ≡ C – CH3.

Question 28.
Name two industrial sources of hydrocarbons.
Answer:

  1. Petroleum
  2. Coal.

Question 29.
Arrange the following in the increasing order of C – C bond length C2H6, C2H4, C2H22.
Answer:
C2H2 < C2H4 < C2H6.

Question 30.
What type of hydrocarbons is present in high octane gasoline?
Answer:
Branched-chain aliphatic and/or aromatic hydrocarbons.

Question 31.
What are the chief constituents of light oil fraction?
Answer:
Benzene, toluene, and xylenes.

Question 32.
Which of the following shows geometrical isomerism?
(i) CHCl = CHCl
(ii) CH2 = CCl2
(iii) CCl2 = l. Give the structures of cis-and transforms.
Answer:
(i) HC (Cl) = CH (Cl);
Hydrocarbons Class 11 Important Extra Questions Chemistry 12

Question 33.
Name the product formed when methyl bromide is treated with sodium and ether.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 13

Question 34.
Which of the following shows geometrical isomerism?
But-1—ene or but—2—ene.
Answer:
But-2—ene CH3 – CH = CH – CH3.

Question 35.
Write the structure and I.U.P.A.C. name of Acetonitrile.
Answer:
Acetonitrile is CH33 CN. Its I.U.P.A.C. name is Ethane nitrile.

Question 36.
Why does carbon has a larger tendency of catenation than silicon although they have the same number of valance electrons?
Answer:
It is due to the smaller length of the C – C bond which is stronger (335 kJ mol-1) than the Si-Si bond (225.7 kJ mol-1).

Question 37.
Give name atm structure to the first organic compound synthesized in the laboratory.
Answer:
Urea
Hydrocarbons Class 11 Important Extra Questions Chemistry 14

Question 38.
Benzene is highly unsaturated, yet it does not give usual addition reactions readily. Why?
Answer:
Benzene is highly unsaturated, yet, resonance imparts extra-stability to benzene and it does not give additional reactions.
Hydrocarbons Class 11 Important Extra Questions Chemistry 15

Question 39.
What is Lindlar’s catalyst? What is it used for?
Answer:
Pd/BaSO4 poisoned with quinoline. It is used for the partial reduction of alkynes to cis-alkenes.

Question 40.
How will you detect the presence of unsaturation in an organic compound?
Answer:
Either by Baeyer’s reagent or by Br2 in CCl4. The color is discharged.

Question 41.
How can ethylene be converted to ethane?
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 16

Question 42.
Arrange the following in increasing order of C — C bond length C2H6, C2H4, C2H2.
Answer:C2H2 < C2H44 < C2H6.

Question 43.
What type of hybridization is involved in
(i) planar and
Answer:
sp2

(ii) linear molecules?
Answer:
sp.

Question 44.
Name the chain isomer of C55H12 which has a tertiary hydrogen atom.
Answer:
2-Methyl butane (CH3), CHCH2CH3.

Question 45.
What type of isomerism is shown by butane and isobutane.
Answer:
Chain or nuclear isomerism,

Question 46.
What do you mean by cracking?
Answer:
The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking.

Question 47.
Complete the reaction HC ≡ CH
Hydrocarbons Class 11 Important Extra Questions Chemistry 17
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 18

Question 48.
Out of Octane and is heptane, which has a lower octane number?
Answer:
n-Octane.

Question 49.
What are the main components of LPG?
Answer:
Butane and isobutane.

Question 50.
Write a chemical reaction to illustrate the Saytzeffs rule.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 19

Hydrocarbons Important Extra Questions Short Answer Type

Question 1.
The following organic compounds are known by their common names
(i) Neopentane
Answer:
Neopentane is
Hydrocarbons Class 11 Important Extra Questions Chemistry 20
& its h.U.P.A.C. name is 2, 2—dimethyl propane.

(ii) Acetone
Answer:
Acetone is
Hydrocarbons Class 11 Important Extra Questions Chemistry 87
and its I.U.P.A.C. name is Propanone.

(iii) Vinyl chloride
Answer:
Vinyl chloride is CPU = CH — Cl and its I.U.P.A.C. name are chloroethene.

(iv) Tert butyl alcohol. Write their structural formulae and I.U.P.A.C. names.
Answer:
Tert; butyl alcohol
Hydrocarbons Class 11 Important Extra Questions Chemistry 21
and its I.U.P.A.C. name is 2-methyl propan-2-ol.

Question 2.
What are the various products expected when propane reacts with fuming nitric acid?
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 22

Question 3.
How will you convert methane into
(i) Methanol
Answer:
Conversion of methane into methanol:
Hydrocarbons Class 11 Important Extra Questions Chemistry 23

(ii) Methanal.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 24

Question 4.
What is aromatization? How will you convert ^hexane into benzene?
Answer:
Aromatization. It is the process that involves cyclization, isomerization, and dehydrogenation with the application of heat and catalyst to convert alkanes containing six or more carbon atoms into aromatic hydrocarbons.
Hydrocarbons Class 11 Important Extra Questions Chemistry 25

Question 5.
Give the different conformations of ethane with their
(i) Sawhorse representation and
(ii) Newmann Projection formulae.
Answer:
Sawhorse representation Newmann projection models
Hydrocarbons Class 11 Important Extra Questions Chemistry 26
Hydrocarbons Class 11 Important Extra Questions Chemistry 27

Question 6.
What are the relative stabilities of different conforma¬tions of ethane? Is it possible to isolate these at room temperature?
Answer:
The staggering form of ethane is more stable than the eclipsed form because the force of repulsion between hydrogen atoms on adjacent C atoms is minimum. The energy difference between the staggered form and eclipsed form of ethane is just 12.55 kJ mol-1. Therefore, it is not possible to separate these two forms of ethane at room temperature.

Question 7.
What is Saytzeff Rule? What are the expected products when 2-Bromobutane is dehydrohalogenation with ale. KOH?
Answer:
Saytzeff Rule. Whatever two alkenes are theoretically possible during a dehydrohalogenation reaction, it is always the more highly substituted alkene that predominates.
Hydrocarbons Class 11 Important Extra Questions Chemistry 28

Question 8.
What is the order of reactivity of H2C = CH2, (CH3)2, H2C = CH2, CH3 – CH = CH2, CH3 – CH = CH – CH3, (CH3)2 C = C (CH3)2, (CH3)2 C = CH CH3 towards electrophilic addition reactions?
Answer:
The order of reactivity of the above alkenes towards electrophilic addition reactions decreases in the order.
(CH3)2 C = C (CH3)2 > (CH33)2 C = CH CH3 > (CH3)2 C = CH2 > CH3 CH – CH – CH3 > CH3 – CH = CH, > CH2 = CH2.

Question 9.
Define Markownikov rule. Explain it with an example.
Answer:
Markownikov rule states. The negative part of the addendum adding to an unsymmetric alkene goes to that C atom of the double bond which is attached to a lesser number of C atoms.
Hydrocarbons Class 11 Important Extra Questions Chemistry 29

Question 10.
What is the Peroxide effect/Kharasch Effect? Illustrate with an example.
Answer:
In the presence of peroxides such as benzoyl peroxide, the addition of HBr (but not of HCl or HI) to an unsymmetrical alkene takes place contrary to the Markownikov rule. This is known as the peroxide/Kharsch effect.
Hydrocarbons Class 11 Important Extra Questions Chemistry 30

Question 11.
An alkene with the molecular formula C7H14 gives propanone and butanal on ozonolysis. Write down its structural formula and its I.U.P.A.C. name.
Answer:
The structures of the compounds on ozonolysis of C7H14 and
Hydrocarbons Class 11 Important Extra Questions Chemistry 31
Remove the oxygen atoms and connect them by a double bond, the structure of the alkene is
Hydrocarbons Class 11 Important Extra Questions Chemistry 32

Question 12.
How will you prepare propyne and I-Butyne from acetylene;?
Answer:
(i) Preparation of propyne from acetylene
Hydrocarbons Class 11 Important Extra Questions Chemistry 33

(ii) Preparation of 1-butyne from acetylene
Hydrocarbons Class 11 Important Extra Questions Chemistry 34

Question 13.
Describe a method to distinguish between ethane, ethene, ethyne.
Answer:
(i) Ethene (C2H4) and ethyne (C2H2) decolorize bromine in carbon tetrachloride whereas ethane (C2H6) does not.
Hydrocarbons Class 11 Important Extra Questions Chemistry 35

(ii) Ethyne (and not ethane, ethene) reacts with ammoniacal AgNO3 (Tollen’s reagent) to form white precipitates.
Hydrocarbons Class 11 Important Extra Questions Chemistry 36

Question 14.
Give the mechanism of an electrophilic addition of chlorine into propene.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 37
Cl4 (chloronium ion) formed by the heterolytic fission of Cl2 in step (i) being an electrophile attacks propene in (ii) step, propene undergoes electrometric effect combined with + I effect to form carbocation which, is a slow step. In (iii) step Cl ion being a. nucleophile attacks carbocation and forms the product. It is a fast step.

Question 15.
Identify the correct order of reactivity in electrophilic substitution reactions of the following compounds: [I.I.T. 2002]
Hydrocarbons Class 11 Important Extra Questions Chemistry 38
Answer:
— NO2 group in structure IV is an electron attracting group, it deactivates the benzene ring largely towards electrophilic substitution reactions. Cl group in III is also a deactivating group, but its deactivation is lower too — NO2, where—CH3, a group in II is an electron-releasing group and so activates the benzene ring towards electrophilic substitution. Therefore, the order of reactivity is
Hydrocarbons Class 11 Important Extra Questions Chemistry 39

Question 16.
What is meant by (i) delocalization
Answer:
Delocalisation: Delocalisation means that pairs of 7t electrons extend over 3 or more atoms. They belong to the whole molecule. For example, 6n electrons present in benzene are delocalized and are spread on the whole of the ring and this imparts extra stability to the molecule.

(ii) resonance energy.
Answer:
Resonance energy: The difference between the energy of the most stable contributing/canonical structural and the energy of the resonance hybrid is known as resonance energy. In the case of benzene, the resonance hybrid has 147 kJ mol-1 than either A or B below. Thus resonance energy of benzene is 147 kJ mol-1.
Hydrocarbons Class 11 Important Extra Questions Chemistry 40

Question 17.
Describe Friedel’s craft reaction with suitable examples.
Answer:
When an alkyl or acid halide is treated with benzene or its derivative in the presence of anhydrous AlCl3 as a catalyst, we got alkyl or acyl benzene.
Hydrocarbons Class 11 Important Extra Questions Chemistry 41

Question 18.
Classify the following hydrocarbons into alkanes, alkenes, alkynes, and arenes.
(i) (CH3)4C
Answer:
Alkane

(ii) C2H2
Answer:
Alkyne

(iii) C3H6
Answer:
Alkene

Hydrocarbons Class 11 Important Extra Questions Chemistry 42
Answer:
Arene.

Question 19.
Write all possible structures of C5H8 and give their I.U.P.A.C. names.
Answer:
Structural isomers of C5H8 (Pentyne) are
(i) CH3 – CH2 – CH2 – C ≡ CH (Pent-l-yne)
(ii) CH3 – CH2 – C ≡ C – CH3 (Pent-2-yne)
(iii) Hydrocarbons Class 11 Important Extra Questions Chemistry 43
(3-Methylbut-l-yne)

Question 20.
How does ethylene undergo polymerization? What is the use to which the polymer obtained is put?
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 44
Low-density polythene [LDPE] and high-density polyethylene [HDPE] as used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes, radio, and T.V. cabinets, etc.

Question 21.
What is the action of water on
(i) Ethyne
(ii) Propyne? Name the end products obtained.
Answer:
(i) Action of water on ethyne: When ethyne is warmed with dilute H2SO4 at 333K and H8SO4 as catalyst ethanol is obtained.
Hydrocarbons Class 11 Important Extra Questions Chemistry 45
(ii) Action of water on Propyne CH3 — C ≡ CH + H2O
Hydrocarbons Class 11 Important Extra Questions Chemistry 46

Question 22.
Propene reacts with HBr to give Isopropyl bromide but does not give n-propyl bromide. Why?
Answer:
The addition of unsymmetrical addendum (HBr) to unsymmetrical olefines (CH3CH = CH2) takes place according to Markownikov rule, the negative part of reagent (i.e. Br-) adds on the carbon atom having a minimum number of hydrogen atoms. Hence Isopropyl bromide will be formed.
Hydrocarbons Class 11 Important Extra Questions Chemistry 47

Question 23.
An oxidizing agent is needed in the iodination of methane but not in the chlorination or bromination. Give reason.
Answer:
In the iodination of methane, H — I is also formed as the product with iodomethane, since it is a strong reducing agent, it reduces iodomethane back to methane & makes the reaction reversible. In order to destroy HI, an oxidizing agent like HIO3 (or HNO3) is needed. But HCl& HBr formed in the chlorination & bromination reactions of methane & not in a position to react with the monosubstituted products (CH3Cl & CH3Br) since they are comparatively weak reducing agents.

Therefore, no oxidizing agent is needed for these reactions.
CH4 + I2 ⇌ CH3I + HI
5HI + HIO3 → 3H2O + 3I2

Question 24.
The dipole moment of trans 1, 2-chloroethene is less than the cis isomer. Explain.
Answer:
The structure of the trans isomer is more symmetrical to the cis isomer. In the trans isomer, the dipole moments of the polar C—Cl bonds are likely to cancel with each other & the resultant dipole moment of the molecule is nearly zero. But in the cis isomer, these do not cancel. Therefore, the cis isomer has a specific moment but is zero in the case of the trans isomer.
Hydrocarbons Class 11 Important Extra Questions Chemistry 48

Question 25.
Write l.U.P.A.C. names of the products obtained by addition reactions of HBr to hex-l-ene.
(a) In the absence of Peroxide
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 49
(b) In the presence of Peroxide.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 50

Question 26.
How will you distinguish between the following:
(a) Butyne-l & Butyne-2 ,
Answer:
Butyne-1 having an acetylene hydrogen atom will give white ppt. With ammoniacal silver nitrate & red ppt. with ammonical cuprous chloride. On the other hand, butyne-2-having no acetylenic hydrogen atom does not respond to either of the two reagents.

(b) Butene-1 & Butene-2
Answer:
Butene-1 & butene-2 can be distinguished either by ozonolysis or by oxidation with acidic KMnO2 solution with which they give different carbonyl compounds.
Hydrocarbons Class 11 Important Extra Questions Chemistry 51

Question 27.
Alkynes are less reactive than alkenes towards addition reaction even though they contain 2-7t bond. Give reason.
Answer:
This is due to

  1. greater electronegativity of sp-hybridized carbon of alkynes than sp2 hybridized carbon atoms of alkenes which holds the π-electrons of alkynes more tightly and
  2. greater delocalization of π-electrons in alkynes (because of the cylindrical nature of their n electron cloud) than in alkenes. As a result, n electrons of alkynes are less easily available for addition reactions than those of alkenes.

Consequently, alkynes are less reactive than alkenes towards addition reactions.

Question 28.
Why do addition reactions occur more readily with alkenes & alkynes than with aromatic hydrocarbons?
Answer:
The energy gained by forming two sigma bonds (of four sigma bonds) more than compensates for the loss of one or two n bonds when addition occurs to an alkene or alkyne. However, in aromatic hydrocarbons, the aromatic ring is specially stabilized by the delocalization of n electrons about the ring.

It, therefore, requires substantial activation energy to cause the loss of its aromatic character. The most usual reaction in arenes is thus substitution rather than addition, since substitution does not result in loss of aromatic character.

Question 29.
A Hydrocarbon A, adds one mole of hydrogen in presence of platinum catalyst from n-Hexane. When A is oxidized vigorously with KMnO4, a single carboxylic acid, containing three carbon atoms is isolated. Give the structure of A & explain.
Answer:

  1. Since hydrocarbon A adds one molecule of H2 in presence of platinum to form n-hexane. A must be a hexene.
  2. Since A on vigorous oxidation with KMnO4 gives a single carboxylic acid containing three carbon atoms, therefore, A must be asymmetrical hexene i.e. hex-3-ene.
    Hydrocarbons Class 11 Important Extra Questions Chemistry 52
    Thus, the given hydrocarbon A is hex-3-ene.

Question 30.
How would you carry out the following conversion?
Answer:
Propene to Ethyne
Hydrocarbons Class 11 Important Extra Questions Chemistry 53

Hydrocarbons Important Extra Questions Long Answer Type

Question 1.
How would you convert the following compounds to benzene?
(i) Acetylene
Answer:
Acetylene into benzene. Ethyne (Acetylene) in passing through a red hot iron tube at 873 K undergoes cyclic polymerization as shown below.
Hydrocarbons Class 11 Important Extra Questions Chemistry 54

(ii) Benzoic acid
Answer:
Benzoic acid into benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 55

(iii) Hexane
Answer:
Hexane into benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 56

(iv) Benzene diazonium chloride
Answer:
Benzene diazonium chloride into benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 57

Question 2.
How will you convert benzene into
(i) Nitrobenzene
Answer:
Benzene into Nitrobenzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 58

(ii) Benzene sulphonic acid
Answer:
Benzene into benzene sulphonic acid
Hydrocarbons Class 11 Important Extra Questions Chemistry 59

(iii) Toluene
Answer:
Benzene into Toluene
Hydrocarbons Class 11 Important Extra Questions Chemistry 60

(iv) Acetophenone?
Answer:
Benzene into Acetophenone
Hydrocarbons Class 11 Important Extra Questions Chemistry 61

Question 3.
What is the mechanism of nitration of benzene?
Answer:
Nitration of benzene. It is carried out by treating benzene with a mixture of cones. HNO3+ Cone. H2SO4. The various steps involved are:
Step I: Generation of an electrophile, i.e., NOt (nitronium ion)
Hydrocarbons Class 11 Important Extra Questions Chemistry 62

Step II: Formation of complex or carbocation intermediate
Hydrocarbons Class 11 Important Extra Questions Chemistry 63
This step is slow and hence is the rate-determining step of the reaction.

Step III: Loss of a proton from the carbocation intermediate
Hydrocarbons Class 11 Important Extra Questions Chemistry 64
This step is fast and does not affect the rate of the reaction.

Question 4.
(a) What is the general formula of Alkynes?
Answer:
The general formula of alkynes ¡s CnH2n-2.

(b) Give the I.U.P.A.C. names and structure of all alkynes having the molecular formula C2H8.
Answer:
C5H8 has the following isomers.
(i) CH3 — CH2 — CH2 — C ≡ CH Pent-1-yne
(ii) CH3 CH, C ≡ C — CH3 Pent-2-yen
(iii) Hydrocarbons Class 11 Important Extra Questions Chemistry 65
3-Methyl but-1-yen

(c) Give any two methods for preparing acetylene.
Answer:
Acetylene can be prepared by the following two methods
Hydrocarbons Class 11 Important Extra Questions Chemistry 66
(ii) By dehydrohalogenation of dihaloalkanes
Hydrocarbons Class 11 Important Extra Questions Chemistry 67
(d) Discuss any three chemical properties of acetylene.
Answer:
(i) Addition of Hydrogen:
Hydrocarbons Class 11 Important Extra Questions Chemistry 68

(ii) On oxidation with alkaline KMnO4, it gives oxalic acid.
Hydrocarbons Class 11 Important Extra Questions Chemistry 69

(iii) When acetylene is passed through a red hot iron tube, it trickeries to give benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 70

Question 5.
(a) Write structures of different chain isomers of alkanes corresponding to the molecular formula C6H14. Also, write their I.U.P.A.C. names.
(b) Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of —OH groups at different carbons of the chain.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 71
(b) Structures of – C5H11 group Corresponding alcohols
Hydrocarbons Class 11 Important Extra Questions Chemistry 72
Hydrocarbons Class 11 Important Extra Questions Chemistry 73

Question 6.
How will you prepare Alkanes by
(i) Wurtz reaction
Answer:
Methods of preparation of Alkanes
Wurtz reaction. Alkyl halides on treatment with sodium in dry ether give higher alkanes, preferably containing an even number of carbon atoms.
Hydrocarbons Class 11 Important Extra Questions Chemistry 74
(ii) Decarboxylation of sodium salts of fatty acids
Answer:
Sodium salts of fatty acids on heating with-soda lime (a mixture of NaOH and CaO) give alkanes containing one carbon atom less than the carboxylic acid. The process of elimination of carbon dioxide from a carboxylic acid is known as Decarboxylation
Hydrocarbons Class 11 Important Extra Questions Chemistry 75

(iii) Kolbe’s electrolytic method.
Answer:
Kolbe’s electrolytic method. An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives alkanes containing an even number of carbon atoms at the anode.

The probable mechanism for the reaction is
Hydrocarbons Class 11 Important Extra Questions Chemistry 77

(b) How do alkanes udergo:
(i) substitution reactions with halogens
Answer:
Substitution reactions with Halogens. The order of reactivity of halogens is F2 > Cl2 > Br2 > I2
Hydrocarbons Class 11 Important Extra Questions Chemistry 78
Bromination is similar. With fluorine, the reaction is too violent to be controlled. Iodination is very slow and a reversible reaction.
It can proceed in the presence of oxidizing agents like HNO3, HIO3.
CH4 + I2 ⇌ CH3I + HI
HIO3 + 5HI → I3 + 3H2O

Halogenation of alkanes proceeds via a free-radical mechanism which consists of three steps.

  1. Chain initiation step
  2. Chain propagation step
  3. Chain termination step

1. Chain initiating step. Cl2 undergoes hemolysis in the presence of heat and light.
Hydrocarbons Class 11 Important Extra Questions Chemistry 79
2. Chain propagation step
Hydrocarbons Class 11 Important Extra Questions Chemistry 80
3. Chain termination step
Hydrocarbons Class 11 Important Extra Questions Chemistry 81

(ii) oxidation by combustion
Answer:
Oxidation of alkanes by combustion: Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of a large amounts of heat.
Hydrocarbons Class 11 Important Extra Questions Chemistry 82
In the presence of an insufficient amount of air or dioxygen, carbon black is formed.
Hydrocarbons Class 11 Important Extra Questions Chemistry 83

(iii) Isomerisation
Answer:
Isomerization reactions of alkanes, n-Alkanes on heating in the presence of anhydrous aluminum chloride and hydrogen chloride gas isomerize to branched-chain alkanes.
Hydrocarbons Class 11 Important Extra Questions Chemistry 84

(iv) Aromatisation.
Answer:
Aromatisation. n-Alkanes having six or more C atoms on heating to 773 K at 10-20 atmospheric pressure in. the presence of oxides of vanadium, molybdenum get dehydrogenated and cyclized. This reaction is called aromatization or reforming.
Hydrocarbons Class 11 Important Extra Questions Chemistry 85
Hydrocarbons Class 11 Important Extra Questions Chemistry 86

Some Basic Concepts of Chemistry Class 11 Notes Chemistry Chapter 1

By going through these CBSE Class 11 Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry, students can recall all the concepts quickly.

Some Basic Concepts of Chemistry Notes Class 11 Chemistry Chapter 1

Importance of Chemistry: Chemistry plays a central role in science and is often intertwined with branches of science like Physics, Biology, Geology, etc. Chemistry also plays an important role in meeting human needs for food, health care products, etc.

→ Chemistry deals with the composition, structure, and properties of matter. To understand Chemistry, we have to study the basic constituents of matter Atoms & Molecules and their relationship with mass. Chemistry is basically, about chemical transformations.

→ Different manifestations of Chemistry include fertilizers, alkalies, acids, salts, dyes, polymers, drugs, soaps, detergents, metals, alloys, and other inorganic and organic chemicals. Daily new drugs, dyes, polymers, etc. are finding their way from the laboratory to industry. Many life-saving drugs like Cisplatin and Taxol are proving effective in cancer therapy and AZT [Azidothymidine] in helping Aids victims.

Nature of Matter: Anything which has mass and occupies space is called matter. Matter can exist in three physical states-solids, liquids, and gases.

→ Solids: Solids have definite volume and definite shape.

→ Liquids: Liquids have a definite volume, but no definite shape. They take the shape of the container in which they are put.

→ Gases: Gases have neither definite volume nor definite shape. These three forms are interconvertible.
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 1
Matter can be classified as mixtures and pure substances. Mixtures can be both homogeneous as well as heterogeneous. Milk, air are examples of homogeneous mixtures. They are uniform throughout. If they are not having a uniform composition, they are a heterogeneous mixture. Iron and sand is an example of a heterogeneous mixture.

Elements like Cu, Ag, and compounds like Nad, AgNO, constitute pure substances. Elements consist of only one type of particle. When two or more elements combine, they form compounds. The smallest part of an element is an Atom whereas the smallest particle of a compound is a Molecule. Thus, copper is an element composed of Cu atoms whereas sodium chloride is a compound composed of Na+Cl molecules.
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 2
Properties of Matter and their Measurement: Properties can be classified into two categories:

  1. Physical properties and
  2. Chemical properties.

→ Physical properties: Physical properties are those which can be observed or measured without changing the identity or composition of the substance. Some examples of physical properties are color, odor, melting point, boiling point density, etc.

The measurement or observation of chemical properties requires a chemical change to occur. Examples of chemical properties are aridity, basicity, combustibility.

The molar mass calculator with steps uses the chemical formula to determine the number of atoms of each element in the compound.

Base Physical Quantities and their Units:
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 3
Definitions of SI Base Units:
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 4
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 5
In SI, large and small quantities are expressed by using an appropriate prefix with the base units.

S.I. Prefixes:
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 6
Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object. The mass of a substance is constant whereas its weight may vary from one place to another due to change in gravity.
The SI unit of mass is a kilogram However, its fraction gram (1 kg = 1000 g), is used in laboratories due to the smaller amounts of chemicals used in chemical reactions.

Volume has the units of (length)3. So in the SI system, the volume has units of m3. But again, in Chemistry laboratories, the smallest volumes are used. Hence volume is often denoted by cm3 or dm3 units. A common unit, liter (L) which is not a SI unit, is used for the measurement of the volume of liquids.
1 L = 1000 mL, 1000 cm3 = 1 dm3

→ Density: Density of a substance is its amount of mass per unit volume.
∴ SI unit of density = \(\frac{\text { SI unit of mass }}{\text { SI unit of volume }}\)
= \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\) = kg m-3
This unit is very large.

∴ Density is usually expressed is g cm-3.

→ Temperature: SI unit of temperature is kelvin (K)
K = °C + 273.15
where °C is degree Celsius.

It is interesting to note that temperatures below 0°C (i.e., negative values) are possible on the Celsius scale but, in the Kelvin scale, the negative temperature is not possible.

In addition to Kelvin (K – SI unit), there are two more scales to measure temperature.

  1. Celsius scale (°C)
  2. Fahrenheit scale (°F)
    °F = \(\frac{9}{5}\)(°C) + 32

Uncertainty in Measurement: Many a time in the study of Chemistry, one has to deal with experimental data as well as theoretical calculations. There are meaningful ways to handle the numbers conveniently and present the data in a realistic way with certainly to the extent possible.

→ Scientific Notation: We can write 232.508 as 2.32508 × 102 in scientific notation. Similarly, 0.00016 can be written as 1.6 × 10-4.

→ Multiplication and Division
Multiplication: (5.6 × 105) × (6.9 × 108) = (5.6 × 6.9)(105 + 8)
= 5.6 × 6.9 × 1013
= 38.64 × 1013

Division:
\(\frac{2.7 \times 10^{-3}}{5.5 \times 10^{4}}\) = (2.7 ÷ 5.5)(10-3-4)
= 0.4909 × 10-7

→ Addition and Subtraction
Addition: 6.65 × 104 + 8.95 × 103
= 6.65 × 104 + 0.895 × 104
= (6.65 + 0.895) × 104
= 7.545 × 104

Subtraction:2.5 × 10-2 – 4.8 × 10-3
= 2.5 × 10-2 – 0.48 × 10-2
= (2.5-0.48)10-2
= 2.02 × 10-2

→ Significant Figures: All experimental measurements have some degree of uncertainty associated with them. Precision and accuracy are often referred to while we talk about measurement.

→ Precision: Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result.

→ Significant figures: Significant figures are meaningful digits that are known with certainty.

→ Dimensional Analysis: Often in calculations, We have to convert units from one system to another. The method used to accomplish this is called the factor label method or unit factor method or dimensional analysis.

→ Laws of Chemical Combination: The combination of elements to form compounds is governed by the following five basic laws.

  1. Law of Conservation of Mass
  2. Law of Definite Proportions
  3. Law of Multiple Proportions
  4. Gay Lussac’s Law of Gaseous Volumes
  5. Avogadro’s Law

1. Law of conservation of mass: In all physical and chemical changes, the total mass of the reactants is equal to that of the products.
Or
Matter can neither be created nor destroyed.

2. Law of definite proportions or Law of constant composition: The law states “A chemical compound always consists of the same elements combined together in the same fixed proportion by weight”.

3. Law of multiple proportions: When two elements combine to form two or more than two compounds, then the weights of the elements which combine with the fixed weight of the other, bear a simple ratio to one another.

4. Gay-Lussac’s Law of Gaseous Volumes: It states “When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the products if gaseous, provided all measurements of volumes are done under similar conditions of temperature and pressure”.

5. Avogadro’s Hypothesis Or Law: Instates, “Equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules”.

Dalton’s Atomic Theory: The main postulates of this theory are:

  1. Matter consists of indivisible atoms.
  2. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass.
  3. Compounds are formed when atoms of different elements combine in a fixed ratio.
  4. Chemical reactions involve the reorganization of atoms. These are neither created nor destroyed in a chemical reaction.

The main advantage of Dalton’s theory was that it could explain the laws of chemical combination.

Atomic and Molecular Masses:
1. Atomic Mass: The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12.

→ Atomic Mass Unit,(AMU) or u: One atomic mass unit (AMU) or u is equal to \(\frac{1}{12}\)th the mass of an atom of carbon – 12 isotopes.

1 amu = 1.66056 × 10-24 g
Mass of an atom of hydrogen = 1.6736 × 10-24 g

Thus, in terms of amu,
the mass of hydrogen atom = \(\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g}}\)
= 1.0078 amu = 1.0080 amu

Similarly, the mass of oxygen – 16(16O) atom would be 15.995 amu

Today, ‘amu’ has been replaced by u which is known as unified mass.

→ Average Atomic Mass: From different isotopes of an element based on their relative abundance, the average atomic mass of an element can be calculated.

For example, there are three isotopes of C – (12C, 13C, 14C) with their relative abundance (%) – 98.892, 1.108, and 2 × 10-10 with their atomic mass (AMU) as 12, 13.00335, and 14.00317 respectively.
The average atomic mass of carbon will be
=.(0.98892)(12 u) + (0.01108)(13.00335 u) + (2 × 10-12)(14.00317 u)
= 12.011 u.

→ Molecular Mass: Molecular mass is the sum of atomic masses of the elements present in a molecule.

Molecular mass of water (H2O)
= 2 × atomic mass of hydrogen + 1 × atomic mass of oxygen = 2(1.008 u) + 16.00 u = 18.02 u,

Molecular mass of ethane (C2H6) will be = 2 × (12,011 u) + 6 × (1 008 u)
= 24.022 u + 6.048 u = 30 070 u

→ Formula Mass: The formula mass of NaCl which is composed of Na+ and Cl ions in a three-dimensional structure can be calculated
= Atomic mass of sodium + Atomic mass of chlorine
= 23.00 u + 35.5 u = 56.5 u.

Mole Concept and Molar Masses: In the SI system, mole (symbol, mol) was introduced as the base quantity for the amount of a substance.

→ Substance One Mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0. 012 kg) of the 12C Isotope.

1 Mole = 6.0221367 × 1023 atoms
The number of entities in 1 mol is also called Avogadro Constant and denoted by NA.

Therefore, 6:022 × 1023 entities (atoms, molecules, or any other particles) constitute one mole of that particular substance.
∴ 1 Mol of oxygen atoms = 6.022 × 1023 atoms of oxygen.

Similarly, 1 Mol of ammonia molecules
= 6.022 × 1023 molecules of ammonia
where n maybe 1, 2, 3,…..

In the SI system, Mole (symbol, mol) was introduced as the seventh base quantity for the amount of a substance.

One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12g (or 0.012 kg) of the UC isotope.

Stoichiometry and Stoichiometric Calculations: Stoichiometry deals with the calculation of masses and volumes of the reactants and products involved in a chemical reaction.

Let us consider the information available for the combustion of methane from its balanced chemical equation
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Here methane and dioxygen are called reactants and carbon dioxide and water are called products. The letter (g) stands for gases. The coefficient 2 for O2 and H2O are called stoichiometric coefficients. Similarly, the coefficient for CH4 and CO2 is one in each case.

They represent the number of molecules (and moles as well taking part in the reaction or formed in the reaction).
(a) One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two moles of H2O(g).
(b) One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(g) and 2 molecules of H2O(g).
(c) 22.4 L of CH4(g) reacts with 44.8 L of O2(g) at STP to give 22.4 L of CO2(g) and 44,8 L of H2O (g).
(d) 16 g of CH4(g) reacts with 64 g of O2(g) to give 44 g of CO2(g) and 36 g of H2O(g)

From above
mass ⇌ moles ⇌ no. of molecules
\(\frac{\text { Mass }}{\text { Volume }}\) = Density

→ Limiting Reagent: The reacting substance which gets used up first in the reaction is called the limiting reagent. This is because the amount of limiting reagent limits the amount of the product formed. A part of the other reactants which are present in amounts greater than the stoichiometric amounts is left behind as unconsumed reagents.

→ Reactions in Solutions: The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways:

  1. Mass percent or weight percent (w/w%)
  2. Mole fraction
  3. Molarity
  4. Molality

Stoichiometry of Reactions in Solutions:
1. Mass percentage or percent by mass: It is defined as the mass of solute in gram per 100 g of the solution. For example, a 10% solution of sodium chloride means that 10 g of NaCl is present in 100 g of the solution.

Mass % of the solute = \(\frac{\text { Mass of solute }}{\text { Mass of solution }}\) × 100
Both, the mass of the solute and that of the solution must be expressed in the same mass units. viz, both in grams or both in
kilograms, etc.

2. Mole Fraction (X): The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component to the total number of moles of all the components of the solution. Thus if a solution contains A moles of A and n5 moles of B, then.
Some Basis Concept Of Chemistry Class 11 Notes Chemistry 7
If the mole fraction of one component in a binary solution is known, that of the other can be determined.
i.e. XB = 1 – XA.

3. Molarity (M): The molarity of a solution is defined as the number of moles of solute dissolved per dm3 (or liter, L) of the solution. Molarity of any solution depends upon temperature. So, the molarity of any solution is specified for a given temperature.

Mathematically, molarity is defined as
Molarity of solution = \(\frac{\text { No. of moles of the solute }}{\text { Volume of the solution in litres (or in } \mathrm{dm}^{3} \text { ) }}=\frac{n \mathrm{~mol}}{\mathrm{VL}}\)

where n is the number of moles of the solute.
V is the volume of the solution in liters (or dm3).

Since, the number of moles of any substance is related to its mass and the molar mass,
No. of moles of solute = \(\frac{\text { Mass of the solute }}{\text { Molar mass of the solute }}\)

The molarity of a solution then can also be expressed as:
Molarity of the solution = \(\frac{\text { Mass of the solute }}{\text { Molar mass of the solute } \times \text { Volume of the solution in litres }}\)

So, W grams of a substance having molar mass M, dissolved in sufficient solvent, so as to make the total volume of V liter of the substance, the molarity (M) of the solution is given by
Molarity = \(\frac{\mathrm{Wg}}{\mathrm{Mg} \mathrm{mol}^{-1} \times \mathrm{V} \text { litre }}=\frac{\mathrm{W}}{\mathrm{M} \times \mathrm{V}}\) mol/L

4. MoLdity (m): The molality of a solution is defined as the number of moles of solute per kg of the solvent. If a solution is prepared by dissolving n moles of a solute in W kg of the solvent, then.
Molality, (m) = \(\frac{\text { No. of moles of solute }}{\text { Mass of the solvent in } \mathrm{kg} \mathrm{mol}^{-1}}\)

= \(\frac{n_{\text {solute }}}{W_{\text {solvent }} \text { kg }}=\frac{n_{\text {solute }}}{W_{\text {solvent }}}\)mol kg-1
where nsolute = \(\frac{\text { Mass of the solute }}{\text { Molar mass of the solute }}\)

It is to be noted that molality of a solution does not change with temperature since mass remains unaffected with temperature.

→ Element: It consists of only one type of particles. These particles may be atoms or molecules. Sodium, copper, silver, hydrogen, oxygen etc. are some examples of elements.

→ Compound: When two or more atoms of different elements combine in a fixed proportion, the molecules of a compound is obtained.

→ Atomic mass unit (amu): It is defined as a mass exactly equal to one twelfth the mass of one carbon-12.
1 amu = 1.66056 × 10-24 g
Nowadays’ amu’ has been replaced by u which is known as unified mass.

→ Mole: One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g {or 0.012 kg) of the 12C isotope.

→ Avogadro constant: The number of entities in 1 mol is called the Avogadro constant. It is denoted by NA and is = 6.022 × 1023.

→ Molar mass: The mass of one mole of a substance in grams is called its molar mass.

→ Empirical Formula represents the simplest whole-number ratio of various atoms present in a compound.

→ Molecular Formula: It shows the exact number of different types of atoms present in a molecule of a compound.

→ Mass percent:
Mass per cent = \(\frac{\text { Mass of solute }}{\text { Mass of solution }}\) × 100

→ Mole Fraction:
Mole fraction of a component = \(\frac{\text { No. of moles of the component }}{\text { Total no. of moles of solution }}\)

→ Molarity:
Molarity(M) = \(\frac{\text { No. of moles of the solute }}{\text { Volume of solution in litres }}\)

→ Molality
Molality (m) = \(\frac{\text { No. of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}\)

→ There are different laws which govern the combination of elements to form compounds.
They are mentioned below:

  1. Law of conservation of mass.
  2. Law of Definite Proportions.
  3. Law of multiple proportions.
  4. Gay Lussac’s Law of gaseous Volumes.
  5. Avogadro Law.

Note: These laws have been defined earlier in the book.
These Laws led to Dalton’s Atomic Theory which states that atoms are building blocks of matter.