2nd PUC Hindi Previous Year Question Paper June 2017

   

Students can Download 2nd PUC Hindi Previous Year Question Paper June 2017, Karnataka 2nd PUC Hindi Model Question Papers with Answers help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Hindi Previous Year Question Paper June 2017

समय : 3 घंटे 15 मिनट
कुल अंक : 100

I. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (6 × 1 = 6)

प्रश्न 1.
जो मनुष्य सत्य बोलता है, वह किससे दूर भागता है?

प्रश्न 2.
किसका व्यापारीकरण हो रहा है?

प्रश्न 3.
मन्नू भंडारी को प्रभावित करनेवाली हिन्दी प्राध्यापिका का नाम लिखिए।

प्रश्न 4.
चीफ़ की दावत किसके घर पर थी?

KSEEB Solutions

प्रश्न 5.
स्वर्ग या नरक में निवास स्थान अलॉट करनेवाले कौन हैं?

प्रश्न 6.
‘नारा’. का प्रसिद्ध मंदिर कौनसा है?

आ) निम्नलिखित प्रश्नों में से किन्हीं तीन प्रश्नों के उत्तर लिखिएः (3 × 3 = 9)

प्रश्न 7.
घर में सुजान भगत का अनादर कैसे हुआ?

प्रश्न 8.
झूठ की उत्पत्ति और उसके कई रूपों के बारे में लिखिए।

प्रश्न 9.
गंगा मैया का कुर्सी से क्या अभिप्राय है?

प्रश्न 10.
विश्वेश्वरय्या की प्रसिद्धि तथा पदोन्नति देख कुछ इंजीनियर क्यों जलते थे?

प्रश्न 11.
भोलाराम का परिचय दीजिए।

II. अ) निम्नलिखित वाक्य किसने किससे कहे? (4 × 1 = 4)

प्रश्न 12.
“दिन भर एक न एक खुचड़ निकालते रहते हैं।”

प्रश्न 13.
“यू हैव मिस्ड समथिंग।”

KSEEB Solutions

प्रश्न 14.
“वह जरूर बना देंगी। आप उसे देखकर खुश होंगे।”

प्रश्न 15.
“महाराज, रिकार्ड सब ठीक है।”

आ) निम्नलिखित में से किन्हीं दो का ससंदर्भ स्पष्टीकरण कीजिए: (2 × 3 = 6)

प्रश्न 16.
“भगवान की इच्छा होगी, तो फिर रुपये हो जायेंगे। उनके यहाँ किस बात की कमी है?”

प्रश्न 17.
“वे बोलते जा रहे थे और पिताजी के चेहरे का संतोष धीरे-धीरे गर्व में बदला जा रहा था।”

प्रश्न 18.
“मेरी माँ गाँव की रहनेवाली हैं। उमर-भर गाँव में रही हैं।”

प्रश्न 19.
“यहाँ तो डेंटिस्ट मक्खी मारते होंगे।”

III. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (6 × 1 = 6)

प्रश्न 20.
श्रीकृष्ण के अनुसार किसने सब माखन खा लिया?

प्रश्न 21.
बाँसुरी किस रंग की है?

प्रश्न 22.
बेटी सजने-धजने से क्या महसूस करती है?

KSEEB Solutions

प्रश्न 23.
देश को किससे बचाना है?

प्रश्न 24.
भारतीयता कहाँ बहु रूप में सँवरती है?

प्रश्न 25.
दीवार किसकी तरह हिलने लगी?

आ) निम्नलिखित प्रश्नों में से किन्ही दो प्रश्नों के उत्तर लिखिए: (2 × 3 = 6)

प्रश्न 26.
श्रीकृष्ण के रूप सौन्दर्य का वर्णन कीजिए।

प्रश्न 27.
‘गहने’ कविता के द्वारा कवि ने क्या आशय व्यक्त किया है?

प्रश्न 28.
पर्यावरण के संरक्षण के संबंध में कवि कुँवर नारायण के विचार लिखिए।

प्रश्न 29.
दक्षिण प्रदेश की महत्ता को अपने शब्दों में लिखिए।

इ) ससंदर्भ भाव स्पष्ट कीजिए: (2 × 4 = 8)

प्रश्न 30.
अब कैसे छूटै राम रट लागी।
प्रभु जी तुम चंदन हम पानी,
जाकी अंग-अंग बास समानी।
अथवा
धनि रहीम जल पंक को, लघु जिय पिअत अघाय।
उदधि बड़ाई कौन है, जगत पिआसो जाय॥

प्रश्न 31.
ऐसा तेरा लोक, वेदना
नहीं, नहीं जिसमें अवसाद,
जलना जाना नहीं, नहीं
जिसने जाना मिटने का स्वाद!
अथवा
युद्धं देहि कहे जब पामर
दे न दुहाई पीठ फेर कर;
या तो जीत प्रीति के बल पर
या तेरा पद चूमे तस्कर।

IV. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (5 × 1 = 5)

प्रश्न 32.
मिश्रानी कितने वर्षों से मूलराज के परिवार में काम कर रही थी?

प्रश्न 33.
छोटी बहु के मन में किसकी मात्रा जरूरत से ज्यादा है?

KSEEB Solutions

प्रश्न 34.
कवि किस पर शासन करता है?

प्रश्न 35.
शास्त्रार्थों में पंडितों को किसने पराजित किया?

प्रश्न 36.
पितृ-हत्या का दण्ड क्या नहीं है?

आ) निम्नलिखित प्रश्नों में से किन्हीं दो प्रश्नों के उत्तर लिखिए: (2 × 5 = 10)

प्रश्न 37.
इन्दु को अपनी भाभी बेला पर क्यों क्रोध आया?
अथवा
बेला की चारित्रिक विशेषताओं पर संक्षेप में प्रकाश डालिए।

प्रश्न 38.
भारवि अपने पिता से क्यों बदला लेना चाहता था?
अथवा
प्रायश्चित को लेकर पिता और पुत्र के बीच हुए संवाद को लिखिए।

V. अ) वाक्य शुद्ध कीजिएः (4 × 1 = 4)

प्रश्न 39.
i) मेरा तो प्राण निकल गया।
ii) संदीप को पूछो।
iii) यह एक इतिहासिक घटना है।
iv) कोयल डाली में बैठी है।
उत्तरः
i) मेरे तो प्राण निकल गये।
ii) संदीप से पूछो
iii) यह एक ऐतिहासिक घटना है।
iv) कोयल डाली पर बैठी है।

आ) कोष्टक में दिये गए उचित शब्दों से रिक्त स्थान भरिए: (4 × 1 = 4)
(पावन, भला, समाज, समय)

प्रश्न 40.
i) आप …………… तो जग भला।
ii) ……… परिवर्तनशील है।
iii) वह सरस्वती देवी का …………… मंदिर है।
iv) साहित्य …………… का दर्पण है।
उत्तरः
i) भला
ii) समय
iii) पावन
iv) समाज।

इ) निम्नलिखित वाक्यों को सूचनानुसार बदलिए: (3 × 1 = 3)

प्रश्न 41.
i) वसुंधरा गाना गाती थी। (वर्तमानकाल में बदलिए)
ii) वह आग चिता पर रखेगा। (भूतकाल में बदलिए)
iii) पृथ्वीराज ने देश की सेवा की। (भविष्यत्काल में बदलिए)
उत्तरः
i) वसुंधरा गाना गाती है।
ii) उसने आग चिता पर रख दी।
iii) पृथ्वीराज देश की सेवा करेगा।

ई) निम्नलिखित मुहावरों को अर्थ के साथ जोड़कर लिखिए: (4 × 1 = 4)

KSEEB Solutions

प्रश्न 42.
i) खिल्ली उड़ाना a) रूठ जाना
ii) मुँह फुलाना b) सफल न होना
ii) टोपी उछालना c) हँसी उड़ाना
iv) दाल न गलना d) अपमानित करना
उत्तरः
i – c, ii – a, iii – d, iv – b.

उ) अन्य लिंग रूप लिखिए: (3 × 1 = 3)

प्रश्न 43.
i) श्रीमती
ii) गाय
iii) पुत्रवान।
उत्तरः
i) श्रीमान
ii) बैल
iii) पुत्रवती।

ऊ) अनेक शब्दों के लिए एक शब्द बनाइए : (3 × 1 = 3)

प्रश्न 44.
i) नीचे लिखा हुआ।
ii) जो पुत्र गोद लिया हो।
iii) जो छिपाने योग्य हो।
उत्तरः
i) निम्नलिखित
ii) दत्तक
iii) गुप्त/रहस्य।

ए) निम्नलिखित शब्दों के साथ उपसर्ग जोड़कर नए शब्दों का निर्माण कीजिए : (2 × 1 = 2)

प्रश्न 45.
i) परिवार
ii) शासन।
उत्तर:
i) परिवार = स + परिवार = सपरिवार।
ii) शासन = अनु + शासन = अनुशासनं।

KSEEB Solutions

ऐ) निम्नलिखित शब्दों में से प्रत्यय अलग कर लिखिए: (2 × 1 = 2)

प्रश्न 46.
i) बलवान
ii) घुमाव।
उत्तरः
i) बलवान = बल + वान।
ii) घुमाव = घुम + आव।

VI. अ) किसी एक विषय पर निबंध लिखिए : (1 × 5 = 5)

प्रश्न 47.
i) a) इंटरनेट की दुनिया।
b) खेल जगत में क्रिकेट का स्थान।
c) राष्ट्रीय एकता।
अथवा
ii) चरित्र प्रमाण-पत्र प्राप्त करने हेतु अपने कॉलेज के प्राचार्य को आवेदन-पत्र लिखिए।

आ) निम्नलिखित अनुच्छेद पढ़कर उस पर आधारित प्रश्नों के उत्तर लिखिए: (5 × 1 = 5)

प्रश्न 48.
यह कहावत सत्य है कि समय बलवान है, इस पर किसी का वश नहीं चलता। आनेवाले समय में अच्छा-बुरा क्या घट जाए, कोई नहीं जानता। ऐसे अनिश्चित समय के लिए यदि उसके पास कुछ संचित धन है तो उसके काम आ सकता है। कोई भी व्यक्ति दूसरों के सहारे न तो कल रह सका है, न आज रह पा रहा है, न ही भविष्य में रह सकता है। सत्य है। मनष्य का आज का जीवन कई प्रकार की आकस्मिकताओं वाला बन चुका है। अतः उन आकस्मिकताओं का ठीक प्रकार से सामना करने के लिए, प्रत्येक व्यक्ति को चाहिए कि वह प्रतिदिन जितनी भी हो अधिक-से-अधिक बचत करता रहे। इसी में उसकी भलाई है। आज की गई एक-एक पैसे की बचत कल का अनंत सुख सिद्ध हो सकती हैं। धन की कमी से सुखपूर्वक तो क्या सामान्य जीवन कतई संभव नहीं है। मनुष्य के जीवन में हमेशा से धन की आवश्यकता बनी रही है। उसे पूरा करने के लिए बचत करना नितांत आवश्यक है। अतः व्यक्ति अपने सभी तरह के स्रोतों से आज और कल में संतुलन बनाए रखकर ही सुख-चैन से जीवन जी सकता है।
प्रश्नः
i) कौन सी कहावत सत्य है?
ii) कटु सत्य क्या है?
iii) मनुष्य की भलाई किस में है?
iv) धन की कमी से क्या संभव नहीं है?
v) व्यक्ति सुख-चैन से कब जी सकता है?

KSEEB Solutions

इ) हिन्दी में अनुवाद कीजिए: (5 × 1 = 5)

प्रश्न 49.
2nd PUC Hindi Previous Year Question Paper June 2017

KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा

   

Students can Download Hindi Lesson 10 मेरा, हमारा, तेरा, तुम्हारा Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 6 Hindi helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा

मेरा, हमारा, तेरा, तुम्हारा Questions and Answers, Summary, Notes

KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा 1
KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा 2
KSEEB Solutions for Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा 3

KSEEB Solutions

अभ्यास

1. उचित शब्द चुनकर खाली :

  1. मोहन …………… भाई हैं (मेरी / मेरा)
  2. ……………घर बड़ा हैं (हमारा / हमारी)
  3. …………. नाम क्या है ? ( तेरी / तेरा )
  4. ………. गाँव कहाँ है ? (तुम्हारी / तुम्हारा)

उत्तर:

  1. मेरा
  2. हमारा
  3. तेरा
  4. तुम्हारा

2. मिलान करो :

  1. सतीश – a. गाँव के पास हैं
  2. कुत्ता  – b. सविता का बड़ा भाई हैं
  3. खरगोश – c. वफादार होता हैं
  4. विद्यालय – d. प्यारा हैं

उत्तर:

  1. b सविता का बड़ा भाई हैं
  2. c वफादार होता हैं
  3. d प्यारा हैं
  4. a गाँव के पास हैं

KSEEB Solutions

2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company

   

You can Download Chapter 3 Financial Statements of a Company Questions and Answers, Notes, 2nd PUC Accountancy Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company

2nd PUC Accountancy Financial Statements of a Company NCERT Textbook Questions and Answers

2nd PUC Accountancy Financial Statements of a Company Short Answer Type Questions and Answers

Question 1.
What is public company?
Answer:
A public company is defined as a company that offers a part of its ownership in the form of shares, debentures, bonds, securities to the general public through stock market:

Question 2.
What is private limited company?
Answer:
As defined by the Section 3. (1) (iii) of Companies Act 1956, private limited company is defined by the following characteristics:

  • It restricts the right to transfer its shares.
  • There must be at least two and a maximum of 50 members (excluding current and former employees) to form a private company.
  • It ca not invite application from the general public to subscribe its shares, or debentures.
  • It cannot invite or accept deposits from persons other than its members, Directors and their relatives.

Question 3.
Define Government Company?
Answer:
As per the Section 617 of Company Act of 1956, a Government Company means any company in which not less than 51% of the paid up share capital is held by the Central Government, or by any State Government or Governments, or partly the Central Government and partly by one or more State Governments and includes a company which is a subsidiary of a Government Company as thus defined.

Question 4.
What do you mean by a listed company?
Answer:
Those public companies whose shares are listed and can be traded in a recognised stock exchange for public trading like, Tata Moto. s, Reliance, etc are called Listed Company. These companies are also called Quota Cornpanies.

KSEEB Solutions

Question 5.
What are the uses of securities premium?
Answer:
As per the Section 78 of the Companies Act of 1956, the amount of securities premium can be used by the company for the following activities:

  • For paying up unissued shares of the company to be issued to members of the company as fully paid bonus share.
  • For writing off the preliminary expenses of the company.
  • For writing off the expenses of, or the commission paid or discount allowed on, any issue of shares or debentures of the company.
  • For paying up the premium that is to be payable on redemption of preference shares or debentures of the company.
  • Further, as per the Section 77A, the securities premium amount can also be utilised by the company to Buy-back its own shares.

Question 6.
What is buy-back of shares?
Answer:
Buy-back of shares means repurchasing of its own shares by a company from the market for reducing the number of shares in the open market.

Question 7.
Write a brief note on ‘Minimum Subscription’.
Answer:
When shares are issued to the general public, the minimum amount that must be subscribed by the public so that the company can allot shares to the applicants is termed as Minimum Subscription. As per the Company Act of 1956, the Minimum Subscription of share cannot be less than 90% of the issued amount. If the Minimum Subscription is not received, the company cannot allot shares to its applicants and it shall immediately refund the entire application amount received to the public.

2nd PUC Accountancy Financial Statements of a Company Long Answer Type Questions and Answers

Question 1.
Explain the nature of the financial statements.
Answer:
The nature of the financial statements depends upon the following aspects;
1. Recorded facts: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Conventions: The preparation of financial statements is based on some accounting conventions like, Prudence Convention, Materiality Convention, Matching Concept, etc. The adherence to such accounting conventions makes financial statements easy to understand, comparable and reflects the true and fair financial position of the company.

3. Accounting Assumptions: These basic accounting assumptions like Going Concern Concept, Money Measurement Concept, Realisation Concept, etc are called as postulates. While preparing financial statements, certain postulates are adhered to. The nature of these postulates is reflected in the nature of the financial statements.

4. Personal Judgments: Personal value judgments play an important role in deciding the nature of the financial statements. Different judgments are attached to different practices of recording.transactions in the financial statements. Thus, personal judgments determine the nature of the financial statements to a great extent.

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Question 2.
Explain in detail about the significance of the financial statements.
Answer:
The importance of financial statements is mentioned below:
1. Provides Information: Financial statements provide information to various accounting users both internal as well as external users. It acts as a basic platform for different accounting users to derive information according to varying needs. For example, the financial statements on one hand help the shareholders and investors in assessing the viability and return on their investments, while on the other hand, the financial statements help the tax authorities in calculating the amount of tax liability of the company.

2. Cash Flow: Financial statements provide information about the cash flows of the company. The financial statements help the creditors and other investors.in determining solvency of company.

3. Effectiveness of Management: The comparability feature of the financial statements enables management to undertake comparisons like inter-firm and intra-firm comparisons. This not only helps in assessing the viability and performance of the business but also helps in’ designing policies and drafting policies. The financial statements enhance the effectiveness and efficacy of the management.

4. Disclosure of Accounting Policies: Financial statements provide information about the various policies, important changes in the methods, practices and process of accounting by the company. The disclosure of the accounting policies makes financial statements simple, true and enables different accounting users to understand without any ambiguity.

5. Policy Formation by Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policy measures and to address various economic problems like employment, poverty etc.

6. Attracts Investors and Potential Investors: They invest or plan to invest in the business. Hence, in order to assess the’viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

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Question 3.
Explain the limitations of financial statements.
Answer:
The following are the limitations of financial statements.
1. Historical Data: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Ignorance of Qualitative Aspect: A financial statement does not reveal the qualitative aspects of a transaction. The qualitative aspects like colour, size and brand position in the market, employee’s qualities and capabilities are not disclosed by the financial statements.

3. Biased: Financial statements are based on the personal judgments regarding the use of methods of recording. For example, the choice of practice in the valuation of inventory, method of depreciation, amount of provisions, etc. are based on the personal value judgments and may differ from person to person. Thus, the financial statements reflect the personal value judgments of the concerned accountants and clerks.

4. Inter-firm Comparisons: Usually, it is difficult to compare the financial statements of two companies because of the difference in the methods and practices followed by their respective accountants.

5. Window dressing: The possibility of window dressing is probable. This might be because of the motive of the company to overstate or understate the assets and liabilities to attract more investors or to reduce taxable profit. For example, Satyam showed high fixed deposits- in the Assets side of its Balance Sheet for better liquidity that gave false and misleading signals to the investors.

6. Difficulty in Forecasting: Since the financial statements are based on historical data, so they fail to reflect the effect of inflation. This drawback makes forecasting difficult. Prepare the format of income statement and explain its elements
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 1
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 2

Prepare the format of balance sheet and explain the various elements of balance sheet.
Vertical form of balance sheet
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 3
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 4
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 5

1. Share Capital: It is the first item on the Liabilities side. It consists of the following items:

  • Authorised Capital
  • Issued Capital: Equity share and preference share.
  • Subscribed Capital less Call in Arrears add Forfeited Shares

2. Reserve and Surplus: As per the Schedule VI,  it consists of the following items:

  • Capital Reserve
  • Capital Redemption Reserve
  • Security Premium
  • Other Reserve less Debit balance of P & L A/c
  • Credit balance of P & L A/c
  • Proposed Additions.
  • Sinking Fund

3 . Secured Loans

  1. Debentures
  2. Loan and advances from bank etc.

4. Unsecured Loans

  • Fixed Deposits
  • Loan & Advances from subsidiaries

5. Fixed Assets: These are those assets that are used for more than one year, like:

  • Goodwill
  • Land
  • Building
  • Plant ec Machinery
  • Patents, Trade Marks
  • Livestock
  • Vehicles, etc.

6. Current Assets: Assets that can be easily converted into cash or cash equivalents are termed as current assets. These are required to run day to day business activities; for example, cash, debtors, stock, etc.

7. Current Liabilities: Those liabilities that are incurred with an intention to be paid or are payable within a year; for example, bank overdraft creditors, bills payable, outstanding wages, short-term loans, etc are called current liabilities.

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Question 4.
Explain how financial statements are useful to the various parties who are interested in the affairs of an undertaking?
Answer:
The various parties that are directly or indirectly interested in the financial statements of a company can be categorized into the following two categories:
Internal Parties: The following are the various internal accounting users who are directly related to the company.

a) Owner: The owner/s is/’are interested in the profit earned or loss incurred during an accounting period. They are interested in assessing the profitability and viability of the capital invested by them in the business.

b) Management: The financial statements help the management in drafting various policies measures, facilitating planning and decision making process. The financial statements also enable management to exercise various cost controlling measures and to remove inefficiencies.

c) Employees and workers: They are interested in the timely payment of wages and salaries, bonus and appropriate increment in their wages and salaries. With the help of the financial statements they can know the amount of profit earned by the company and can demand reasonable hike in their wages and salaries.

External Parties: There are various external users of accounting who need accounting information for decision making, investment planning and to assess the financial position of the business. The various external users are given below.

a) Banks and other financial institutions: Banks provide finance in the form of loans and advances to various businesses. Thus, they need information regarding liquidity, creditworthiness, solvency and profitability to advance loans.

b) Creditors: These are those individuals and organisations to whom a business owes money on account of credit purchases of goods and receiving services; hence, the creditors require information about credit worthiness of the business.

c) Investors and potential investors: They invest or plan to invest in the business. Hence, in order to assess the viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

d) Tax “authorities: They need information about sales, revenues, profit and taxable income in order to determine the levy various types of tax on the business.

e) Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policies measures and to address various economic problems like employment, poverty etc.

f) Researchers: Various research institutes like NGOs and other independent research institutions like CRISIL, stock exchanges, etc. undertake various research projects and the accounting information facilitates their research work.

g) Consumers: Every business tries to build up reputation in the eyes of consumers, which can be created by the supply of better quality products and post-sale services at reasonable and alfordable prices. Business that has transparent financial records, assists the customers to know the correct cost of production and accordingly assess the degree of reasonability of the price charged by the business for its products and, thus, helps in repo building of the business.

h) Public: Public is keenly interested to know the proportion of the profit that the business spends on various public welfare schemes; for example, charitable hospitals, funding schools, etc. This information is also revealed by the profit and loss account and balance sheet of the business.

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Question 5.
Financial statements reflect a combination of recorded facts, accounting conventions and personal judgments’ discuss.
Answer:
The financial statements are the end-products of the accounting process. The financial statements not only reveal the true financial position of the company but also help various accounting users in decision making and policy designing process. The nature of the financial statements depends upon the following aspects like recorded facts, conventions, concepts, and personal judgment.

1. Recorded facts: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Conventions: The preparation of financial statements is based on some accounting conventions like,: Prudence Convention, Materiality Convention, Matching Concept, etc. The adherence to such accounting conventions makes financial statements easy to understand, comparable and reflects the true and fair financial position of the company.

3. Accounting Assumptions: These basic accounting assumptions like Going Concern Concept, Money Measurement Concept, Realisation Concept, etc are. called as postulates. While preparing financial statements, certain postulates are adhered to, The nature of these postulates is reflected in the nature of the financial statements.

4. Personal Judgments: Personal value judgments play an important role in deciding the nature of the financial statements. Different judgments are attached to different practices of recording transactions in the financial statements. Thus, personal judgments determine the nature of the financial statements to a great extent.

Question 6.
Explain the process of preparing income statement and balance sheet.
Answer:
The process of preparing Horizontal Form of Income Statement is explained below in a chronological order:

  1. Prepare a Trial Balance on the basis of the balances of various accounts in the ledger.
  2. Record Opening Stock, Purchases, Manufacturing Expenses and other direct expenses on the debit side of Trading Account
  3. Record Sales and Closing Stock on the credit side of the Trading Account.
  4. Ascertain the balancing figure by totalling both the sides of the Trading Account If the credit side exceeds the debit side, then the balancing figure is termed as Gross Profit, but if the debit side exceeds the credit side, then the balancing figure is termed as Gross Loss.
  5. Carry forward the Gross Profit (Gross Loss) to the credit (debit) side of the Profit and Loss Account.
  6. Record all current year’s operating and non-operating revenue expenditures with their relevant adjustments on the debit side of the Profit and Loss Account.
  7. Record ail current year’s operating and non-operating revenue incomes with their relevant adjustments on the credit side of the Profit and Loss Account.
  8. Ascertain the balancing figure by totalling both the sides of the Profit and Loss Account. If the credit exceeds the debit side, then the balancing figure is termed as Net Profit, but if the debit side exceeds the credit side, then the balancing figure is termed as Net Loss.

The process of preparing Horizontal Form of Balance Sheet is explained below in a chronological order:

  1. Prepare a Trial Balance on the basis of the balances of various accounts in the ledger.
  2. Record all the debit balances of Real and Personal Accounts on the left hand side (i.e. Assets side) of the Balance Sheet after making all adjustments for provision and other related items.
  3. Record all the credit balances of Real and Personal Accounts on the right hand side (i.e. Liabilities side) of the Balance Sheet after making all adjustments for interest and outstanding items.
  4. Add Net Profit to the Opening Capital and deduct Net Loss, if any from the Opening Capital
  5. Ascertain the total of two sides, jvhich must be equal.

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2nd PUC Accountancy Financial Statements of a Company Numerical Questions and Answers 

Question 1.
Show the following items in the balance sheet as per the provisions of the companies Act, 1956 in (Revised) Schedule VI:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 6
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 7

Question 2.
On 1stApril, 2013, Jumbo Ltd., issued 10,000; 12% debentures of₹ 100 each a discount of 20%; redeemable after 5 years. The company decided to write-off discount on issue of such debentures over the life time of the Debentures. Show the items of the company immediately after the issue of these debentures.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 8

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Question 3.
From the following information prepare the balance sheet of Gitanjali Ltd. Invntories ₹ 14,00,000; Equity Share Capital ₹ 20,00,000; Plant and Machinery ₹ 10,00, 000; Preference Share Capital ₹ 12,00,000; Debenture Redemption Reserve ₹ 6,00,000; Outstanding Expenses ₹ .3,00,000; Proposed-Dividend ₹ 5,00,000; Land and Building ₹ 20,00,000; Current Investment ₹ 8,00,000; Cash. Equivalent ₹ 10,00,000; Short term loan from Zaveri Ltd. (A Subsidiary Company of Twilight Ltd.) ₹ 4,00,000; Public Deposits ₹ 12,00,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 9
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 10

Question 4.
From the following information prepare the balance sheet of Jam Ltd. Inventories ₹ 7,00,000; Equity Share Capital ₹ 16,00,000; Plant and Machinery ₹ 8,00,000; Preference Share Capital ₹ 6,00,000; General Reserves ₹ 6,00,000; Bills payable Rs 1,50,000; Provision for taxation ₹ 2,50,000; Land and Building k 16,00,000; Non-current Investments ₹ 10,00,000; Cash at Bank ₹ 5,00,000; Creditors ₹ 2,00,000; 12% Debentures ₹ 12,00,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 11
Note to account
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 12
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 13

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Question 5.
Prepare the balance sheet of Jyoti Ltd., as at March 31, 2018 from the following information. Building ₹ 10,00,000; Investments in the shares of Metro Tyers Ltd. ₹ 3,00,000; Stores & Spares ₹ 1,00,000; Discount on issue of 10% debentures of ₹ 10,000; Statement of Profit and Loss (Pr.). ₹ 90,000; 5,00,000 Equity Shares of ₹ 20 each fully paid-up; Capital Redemption Reserve ₹ 1,00,000; 10% Debentures ₹ 3,00,000; Unpaid dividends ₹ 90,000; Share options outstanding account ₹ 10,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 14
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 15

Question 6.
Brinda Ltd., has furnished the following information:
(a) 25,000,10% debentures of ₹ 100 each;
(b) Bank Loan of ₹ 10,00,000 repayable after 5 years;
(c) Interest on debentures is yet to be paid.
Show the above items in the balance sheet of the company as at March 31, 2018.
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 16

Question 7.
Prepare a Balance sheet of Black Swan Ltd., as at March 31, 2013 from the following information:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 17
Answer:
2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company - 18

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2nd PUC Chemistry Question Bank Chapter 2 Solutions

   

You can Download Chapter 2 Solutions Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions

2nd PUC Chemistry Solutions NCERT Textbook Questions and Answers

Question 1
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:
A solution may contain two or more substances also called components or constituents. A solution which has two components is known as a binary solution (e.g. a solution of NaCl in water) while a solution with three components is called a ternary solution (e.g. a solution of NaCl and KCl in water). Similarly even more than three components may also be present.

Since the solvent and solute may be either gaseous, liquid, and solid, the number of possible types of binary solutions that can be prepared are given below.
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 0

Question 2.
Suppose a solid solution is formed are very small. What kind of solid solution is this likely to be?
Answer:
The solution of hydrogen in palladium and dissolved gases in minerals.

Question 3.
Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer:
(i) Mole fraction (X): The mole fraction of any component in a solution is the ratio of the number of moles of that component to the sum of the number of moles of all the components present in the solution.
For a binary solution containing A and B, Mole fraction of A,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 1
where nA and nB are the numbers of moles of components A and B respectively.

(ii) Molality (m): Molality is the number of moles of the solute dissolved in 1000 gms (1 kg) of the solvent. It B denoted by ‘m’ mathematically.
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 2

(iii) Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm3) of solution. It is denoted by ‘M’ mathematically.
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 3

(iv) Mass fraction multiplied by 100 gives the mass percentage. E.g.: mass percentage of A
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 4

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Question 4.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mLr-1?
Answer:
68% of nitric acid by mass means that
Mass of nitric acid = 68g
Mass of solution = 100g
Molar mass of HN03 = 63g mol-1?
∴ 68 g HNO3 = \(\frac{68}{63}\) mole = 1.079 mole
Density of solution = 1.504g mL-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 5

Question 5.
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of the solution is 1.2 g mL-1, then what shall be the molarity of the solution?
Answer:
10 g glucose is present in 100 g solution, i.e., 90 g of water = 0.090 kg of H2O
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 6

Question 6.
How many mL of 0.1 M HCl are required to react completely with a 1 g mixture Of Na2CO3 and NaHC03 containing equimolar amounts of both?
Answer:
Step 1: To calculate the number of moles of the components in a mixture
suppose Na2CO3 present in the mixture = X g
NaHCO3 present in the mixture = (1 – x) g
Molar mass of Na2CO3 = 2 x 23+12+3 x 16=106g mol-1
Molar mass of NaHCO3.
= 23 + 1 + 12 + 3 × 16 = 84g mol-1
∴  Moles of Na2CO3 in xg = \(\frac{ x }{ 106}\)
Moles of NaHCO3 in (1-x) g = \(\frac{1-x}{84}\)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 7

Step 2: To calculate the moles of HC1 required.
Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
1 mole of Na2CO3 required HCl = 2 moles
0.00526 mole of Na2CO3 requires HCl
= 0.00526 × 2 moles = 0.01052
1 mole of NaHCO3 required HCl = 1 mole
0.00526 mole of NaHCO3 required HCl
= 0.00526 mole
∴Total HCl required = 0.01052 + 0.00526
=0.01578 moles

Step3: To calculate volume of 0.1M HCl
0.1 mole of 0.1 M HCl are present in 1000 mL of HCl
0.01578 mole of 0.1 M HCl will be present in HCl = \(\frac{1000}{0.1}\) x 0.01578
= 157.8 ml.

Question 7.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
300 g of 25% solutions contains solute = 75g
400 g of 40% solution contains solute = 160 g.
Total solute = 160 + 75 = 235 g
Total solution=300 + 400 = 700 g
% of solute in final solution = \(\frac{235}{700}\) × 100 = 33.5%
% of water in the final solution = 100 – 33.5 = 66.5%

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Question 8.
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?
Answer:
Mass of the solute, C2H4(OH)2 = 222.6g
Molar mass of C2H4 (OH)2 = 62 g mol-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 8
Mass of the solvent = 200 g = 0.200 kg

2nd PUC Chemistry Question Bank Chapter 2 Solutions - 9
Total mass of the solution = 422.6g
Volume of the solutions =

2nd PUC Chemistry Question Bank Chapter 2 Solutions - 10

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Answer:
15 ppm means 15 parts in million (106) parts by mass in solution
∴ % by mass = \(\frac{15}{10^{6}} \times\) × 100=15 × 10-4
Taking 15 g chloroform in 106g
Molar mass of CHCl3 = 12 + 1 + 3 × 35.5 = 119.5 g mol-1
Molality = \(\frac{15 / 119.5}{10^{6}}\) × 1000= 1.25 × 10-4 m

Question 10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
Alcohol and water both have a strong tendency to form intermolecular hydrogen bonding. On mixing the two, a solution is formed as a result of the formation of H-bonds between alcohol and H2 O molecules but these interactions are weaker and less extensive than those in pure H2O. Thus they show a positive deviation from ideal behaviour. As a result of this, the solution of alcohol and water will have higher vapour pressure and lower boiling point than that of water and alcohol.

Question 11.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Dissolution of gas in liquid in an exothermic process (Gas + solvent ⇌ solution + Heat). As the temperature is increased, the equilibrium shifts backward.

Question 12.
State Henry’s law and mention some important applications?
Answer:
The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law. It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas Mathematically, P = KHX where P is the partial pressure of the gas; and X is the mole fraction of the gas in the solution and KH is Henry’s Law constant.

Applications of Henry’s law:

  • In the production of carbonated beverages (as the solubility of CO2 increases at high pressure).
  • In deep-sea diving.
  • For climbers or people living at high altitudes, where low blood 02 causes climbers to become weak and make them unable to think clearly

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Question 13.
The partial pressure of ethane over a solution containing 6.56 x 10-3 g of ethane is 1 bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial pressure of the gas?
Answer:
Applying the relationship m = KH x p
In the first case, 6.56 x 102 g bar-1
In the second case, 5.00 x 10-2 g = (6.56 × 10-2 gbar-1) × p
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 11

Question 14.
What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔmixH related to positive and negative deviations from Raoult’s law?
Answer:
In +ve deviation, A-B interactions are weaker than those between A-A or B-B. In such molecules, A or B will find it easier to escape than in pure state. This increases vapour pressure. In case of -ve deviation, A-B interaction = A-A or B-B. This leads to decrease in vapour pressure.

  • In +ve deviation, ΔmixH is + ve
  • In -ve deviation, ΔmixH is – ve

Question 15.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer:
Vapour pressure of pure water at boiling
point (P°) = 1 atm = 1.013 bar
vapour pressure of solution (ps) = 1.004 bar
Mass of solute = (w2) = 2g
Mass of solution = 100 g
Mass of solvent = 98g
Applying Roault’s law for dilute solution (being 2%)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 12

Question 16.
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer:
Molar mass of heptane (C7 H16) = 100 g mol-1
Molar mass of octane (C8H18) = 114 g mol-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 13
x (octane) = 1 – 0.456 = 0.544
p (heptane) = 0.456 x 105.2 kPa = 47.97 kPa
p (octane) = 0.544 x 46.8 kPa = 25.46 kPa
p total = 47.97 + 25.46 = 73.43 kPa

Question 17.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
1 molal solution means 1 mol of the solute in 1 kg of solvent (water)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 14
or ps = 12.08 kPa

Question 18.
Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer:
Ps = 80% of p° = 0.80 p° solute = \(\frac{w}{40} r\) mol solvent (octane) =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 15
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 16

Question 19.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
(i) the molar mass of the solute
(ii) vapour pressure of water at 298 K.
Answer:
Suppose the molar mass of the solute = Mg mol-1
n2 (solute) = \(\frac{30}{M}\) moles
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 17
After adding 18 g of water,
n(H2O)i.e, n1 = 6 moles
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 18
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 19
Dividing equation (i) by equation (ii), we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 20

(ii) putting M = 23 in equation (i), we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 21

Question 20.
A 5% solution (by mass) of cane sugar in water has a freezing point of 271 A. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.
Answer:
The molality of sugar solution =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 22
∆Tf for sugar solution = 273.15 – 271 = 2.150
∆Tf = Kf × m ∴ Kf = 2.15/0.146
The molality of glucose solution =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 23
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 k

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Question 21.
Two elements A and B form compounds having formulas AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. Tin molar depression – constant for benzene is 5.1 K kg mol-1. Calculate atomic masses of A and B.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 24
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 25
Suppose atomic masses of A and B are ‘a’ and ‘b’ respectively. Then
Molar mass of AB2 = a + 2b = 110.87 g mol-1
Molar mass of AB4 = a + 4b =196.15 g mol-1
Equation (ii) – Equation (i) gives
2b = 85.28 orb = 42.64
substituting in equation (i) we get
a + 2 × 42.64 = 110.87 or a = 25.59
Thus atomic mass A = 25.59 u
Atomic mass of B = 4.64 u.

Question 22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
π = CRT
∴ In the first case,
4.98 = \(\frac{36}{180}\) × R × 300 = 60R
In the second case, 1.52 = C × R × 300
Dividing (ii) by (i), we get C = 0.061 M.

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs.

  1. n-hexane and n-octane
  2. I2 and CCl4
  3. NaClO4 and water
  4. methanol and acetone
  5. acetonitrile (CH3CN) and acetone (C3H6O).

Answer:

  1. London’s forces
  2. London’s forces
  3. Ion-dipole interactions
  4. Intermolecular hydrogen bonding
  5. Dipole-dipole interactions.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer:
(a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions.
(b) KCl is an ionic compound, KCl will not dissolve in n-octane.
(c) CH3OH is polar. CH3OH will dissolve in n-octane.
(d) CH3CN is polar but lesser than CH3OH. Therefore, it will dissolve in n-octane but to a greater extent as compared to CH3OH. Hence, the order is KCl < CH3OH < CH3CN < Cyclohexane.

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) Partially soluble (because phenol has a polar -OH group but aromatic phenyl, C6H5 – group)
(ii) Insoluble because toluene is nonpolar while water is polar
(iii) Highly soluble because formic acid can form hydrogen bonds with water.
(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water
(v) Insoluble chloroform is an organic liquid
(vi) Partially soluble because-OH the group is polar but the large hydrocarbon part (C5H11) is nonpolar.

KSEEB Solutions

Question 26.
If the density of some lake water is 1.25g mL-1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.
Answer:
Number of moles in 92 g of Na+ ions
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 26
As these are present in 1 kg of water, by definition, molality = 4m.

Question 27
If the solubility product of CuS is 6 × 10-16, calculate the maximum molarity of CuS in an aqueous solution.
Answer:
Maximum molarity of CuS in aqueous solution = solubility of CuS in mol L-1
If S in the solubility of CuS in mol L-1 , then
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 27

Question 28.
Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 28

Question 29.
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphine generally given is 1.5 mg. Calculate the mass of 1.5 x 10-3 m aqueous solution required for the above dose.
Answer:
1.5 × 10-3 m solution means that 1.5 × 10-3 mole of nalorphine is dissolved in I kg of water.
Molar mass of C19H21NO3 = 19 × 12 + 21 + 14 + 48 = 3119 mol-1
∴ 1.5 × 10-3 mole of C19H21NO3;
= 1.5 × 10-3 × 3119 = 0.467 g = 467 mg
∴ Mass of solution = 1000 g + 0.467 g = 1000.467g
Thus, for 467 mg of nalorphene, solution required =1000.467 g for 1.5 mg of nalorphene, solution required =\(\frac{1000.467}{467} \times 1.5 = 3.219\)

Question 30.
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer:
0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L
i.e. 1000 mL of the solution.
Molar mass of benzoic acid (C6H5COOH) = 72 + 5 + 12 + 32 + 1 =122 g mol-1
∴ 0.15 mole of benzoic acid = 0.15 × 122g= 18.39
Thus, 1000 mL of the solution contain benzoic acid = 18.39
∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250= 4.575g

KSEEB Solutions

Question 31.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:
The depression in freezing points are in the order:
acetic acid < trichloroacetic acid
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 29
Fluorine, being most electronegative, has the highest electron-withdrawing inductive effect. Consequently, trifluoroacetic acid is the Strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to minimum extent to give ions in their solutions in water. Greater the ions produced, greater is the depression in freezing point. Hence the depression in freezing point is maximum for trifluoroacetic acid and minimum for acetic acid.

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH3CH2 CHCICOOH is added to 250 g of water. Ka = 1.4 × 10-3, Kf= 1.86 K kg mol-1.
Answer:
Molar mass of CH3CH2CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol-1
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 30
If α is the degree of dissociation of CH3CH2CHCICOOH, then CH3CH2CHCLCOOH □ CH3CHXHCICOO + H +
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 31
To calculate vant’s Hoff factor:
CH3CH2CHCICOOH □ CH3CH2HCICOO + H+
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 32

Question 33.
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoro acetic acid. Kf of water is 1.86 K kg mol-1.
Answer:
Here, w2 = 19.5 g, w1, = 500g, Kf = 1.86 K kg mol-1, (ATf) obs = 1.0°
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 33
M2 (Calculated) for CH2 FCOOH = 14+19+45 = 78 g mol-1
Vant Hoff factor (i) =
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 34
calculation of dissociation constant. Suppose the degree of dissociation at the given concentration is a.
Then
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 35
Taking volume of the solutions as 500ml,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 35(i)

KSEEB Solutions

Question 34.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:
Here, P° = 17.535 mm, w2 = 25g,
w1 = 450g
For solute (glucose, C6H12O6, M2 = 180 g mol-1
For solvent (H2O), M1 = 18g mol-1
Applying Raoult’s law,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 36
substituting the given values, we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 37
substituting the given values, we get
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 38

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10s mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer:
Here, KH = 4.27 × 105 mm
p = 760 mm
Applying Henry’s law
P = KH x
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 39
i.e, mole fraction of methane in benzene = 1.78 x 10-3

Question 36.
100 g of liquid A (molar mass 140 g mol-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol-1). The vapour pressure of pure liquid B was found to be 500 torrs. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer:
Number of moles of a liquid
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 40
Number of moles of a liquid B (Solvent)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 41
Mole fraction of B in the solution (xA)
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 42
Mole fraction of B in the solution (xB)
= 1-0.114 = 0.886
Also, given p0B = 500 Torr
Applying Raoult’s law,
PA = xAA = 0.114 × P°A
PB= xBB = 0.886 × 500 = 443 Torr
p Total = pA + pB
475 = 0.114 P°A +443 or
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 43
Substituting this value in equation (i), we get pA = 0.114x 280.7 Torr = 32 Torr

Question 37.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50. 71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer:
Molar mass of benzene (C6H6 = 78 g mol-1
Molar mass of toluene (C6H5CH3) = 92 g mol-1
∴ Number of moles in 80 g of benzene
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 44
Number of moles in 100 g of toluene
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 45
In the solution, mole fraction of benzene
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 46
=0.486
mole fraction of toluene = 1-0,486 = 0.514
p° Benzene = 50.71 mm, p° Toluene = 32.06mm
Applying Raoult’s law,
p Benzene= xTolucno × p°Tolucne =0.514 × 32.06 mm = 16.48mm
Mole fraction of benzene in vapour phase
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 47

Question 38.
The air is a mixture of a number of gases. The major components are oxygen and, nitrogen with the approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.
Answer:
Total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volumes
partial pressure of oxygen (P ) = \(\frac{20}{100}\) x 10 atm = 2atm
= 2 x 760 mm
= 1520 mm
partial pressure of nitrogen \(P_{N_{2}}\)
= \(\frac{79}{100}\) × 10 atm = 7.9 atm = 7.9 x 760 mm
= 6004 mm

KH(O2) = 3.30 x 107mm,KH (N2) = 6.51 × 107 mm
Applying Henry’s law
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 48

Question 39.
Determine the amount of CaCl (i= 2.47) dissolved in 2.51itre of water such that its osmotic pressure is 0.75 atm at 27° C.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 49
Molar mass of CaCl2= 40 + 2x 35.5 = 111 gmol-1
Amount dissolved = 0.0308 xlllg = 3.42g

Question 40.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4in 2 litre of water at 25° C, assuming that it is completely dissociated.
Answer:
K2SO4 dissolved = 0.025g
volume of solution = 2L
T = 250°C = 298 K
Molar mass of K2SO4 = 2 x 39 + 32 + 4 × 16= 174gmol-1
As, K2SO4 dissociates completely as K2SO4 → 2K+ SO42-
i.e., ions produced = 3 ∴ i =3
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 50

KSEEB Solutions

2nd PUC Chemistry Solutions Additional Questions and Answers

Question 1.
How does molarity of a solution change with temperature?
Answer:
Molarity decreases with increase in temperature because volume of solution increases with increase in temperature.

Question 2.
State Raoult’s law.
Ans:
It states that for a solution of volatile liquids its vapour pressure of each component is directly proportional to their mole fraction in the solution.
The relative lowering of vapour pressure is equal to mole fraction of solute in case of non-volatile solute.

Question 3.
Why is liquid ammonia bottle first cooled in ice before opening it?
Answer:
At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out.

Question 4.
Sodium chloride and calcium chloride are used clear show from the roads. Why?
Answer:
Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature.

Question 5.
Two liquids A and B boil at 145°C and 190°C
respectively. Which of them has a higher vapour pressure of 80°C? (CBSE 2006)
Answer:
A being more volatile will have higher vapour pressure at 80°C.

KSEEB Solutions

Question 6.
Calculate the molality of sulphuric acid solution in which the mole fraction is 0.85.
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 51

Question 7.
The molar freezing point depression constant for benzene is 4.90 kg mol’1. Selenium exists as a polymer. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower for pure benzene. Decide the molecular formula of selenium (At Wt. of selenium is 78.8 g mol-1). (CBSE 2002)
Answer:
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 52

Question 8.
CCI4 and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. (CBSE 2003,2001)
Answer:
CCI4 is a non-polar covalent compound.
Water is a polar compound. CCI4 can neither form H bonds with water molecules nor can it break H bonds between water molecules. Therefore, it is insoluble in water.

Ethanol is a polar compound and can form H’ bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions.

Question 9.
The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm3 )
Answer:
Let the solution be 1 litre or 1000 cm3
∴ Number of moles of H2SO4= 1.35
Wt. of solution = 1000 x 1.02 = 1020 g
Wt. of sulphuric acid = 1.35 x 98= 132.3g :
Wt. of water = 1020 – 132.3 = 887.79
Molality of H2SO4= ppp x 1000 = 1.52 m

From (i) M, = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15- 110.82 = B2 85.33 = B2 B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82=Atomic mass of A+85.33 Atomic mass of A = 110.82 – 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669

Question 10.
Two elements A and B form compounds having molecular formulas AB2 and AB4. When dissolved 20 g of C6H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB„4 lower it by 1.3 K. The molar depression constant for benzene is 5.1 kg mol*1. Calculate the mass of A and B. (CBSE2004)
Answer:
Let the molar mass of AB2 and AB4 be M1and M2
Then, for AB2,
2nd PUC Chemistry Question Bank Chapter 2 Solutions - 53
From (I) M1 = 110.82, from (ii) M1 = 196.15
AB4– AB2= B2
196.15 – 110.82 =B2
85.33 = B2
B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B
110.82 = Atomic mass of A+ 85.33
∴ Atomic mass of A = 110.82-85.33 = 25.49
Atomic mass of A = 25 .499
Atomic mass of B = 42.669

KSEEB Solutions

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