125876 + Interview Questions in General QA Page 17 InterviewSolution

801.	Normal profit is calculated to value goodwillA. by deducting abnormal gains (profit)B. by adding abnormal lossesC. by deducting abnormal gains and adding abnormal lossesD. None of the above
Answer» Correct Answer - C

Discussion

802.	Young India Ltd. has a Operating Porfit Ratio of 20%. To maintain this ratio at 25%, management mayA. Increase selling price of Stock-in-tradeB. Reduce Cost of Revenue from OperationC. Increase selling price of Stock-in-Trade and to reduce Cost of Revenue from OperationD. All of the above
Answer» Correct Answer - D

Discussion

803.	A transaction involving a decrease in both Current Ratio and Quick Ratio isA. Sale of Non-current Asset for cashB. Sale of Stock-in-Trade at lossC. Cash payment of a Current LiabilityD. Purchase of Stock-in-Trade on credit
Answer» Correct Answer - D

Discussion

804.	Total capital employed in the firm is Rs 8,00,000, Normal Rate of Return is 15% and Profit for the year is Rs 12,00,000. The value of goodwill of the firm as per capitalisation method would beA. Rs 82,00,000B. Rs 12,00,000C. Rs 72,00,000D. Rs 42,00,000
Answer» Correct Answer - C

Discussion

805.	Revenue from Operations Rs 9,00,000, Gross Profit 25% on Cost, Operating Expenses Rs 90,000, Operating Ratio will beA. 1B. 0.5C. 0.9D. 0.1
Answer» Correct Answer - C

Discussion

806.	Calculate Operating Profit Ratio if Revenue from Operations is Rs 5,00,000, Operating Profit is Rs 75,000.A. 0.25B. 0.12C. `13.33%`D. 0.15
Answer» Correct Answer - D

Discussion

807.	If Current Ratio of a firm is `2.5 : 1` and its Current Liabilities are Rs 2,00,000. Its Working Capital will beA. Rs 3,00,000B. Rs 3,75,000C. Rs 4,00,000D. Rs 7,00,000
Answer» Correct Answer - A

Discussion

808.	From the given items which is a part of Current LiabilitiesA. InventoriesB. Trade payablesC. Cash and Cash EquivalentD. Trade Receivables
Answer» Correct Answer - B

Discussion

809.	State whether the following statements are True or False (i) Receipts and Payments Account shows the receipts and payments of revenue nature only. (ii) Receipt from the sale of an old asset is an income (iii) Receipts and Payments Account is a summary of cash transactions (iv) Surplus of Non-Trading Organisation is destributed among the members (v) Income and Expenditure Account is equivalent to the Profit and Loss Account of a business enterprise. (vi) Proceeds from the sale of old newspaper are taken as income. (vii) Receipts and Payments Account shows total income and expenditure (viii) Only revenue nature items are shown in the Income and Expenditure Account. (ix) Scholarships granted to students provided by government is debited to Income and Expenditure Account. (x) Payment of honorarium is a revenue expenditure (xi) Any expenditure relating to special fund is deducted from the special fund. (xii) Loss on sale of fixed asset is debited to income and Expenditure Account. Whereas gain (profit) on sale of fixed asset is credited to Income and Expenditure Account
Answer» Correct Answer - (i) False; (ii) False; (iii) True; (iv) False; (v) True (vi) True (vii) False (viii) True (ix) False (x) True (xi) True (xii) True

Discussion

810.	From the given items which is not shown under Current LiabilitiesA. Trade PayablesB. Short-term ProvisionsC. Short-term BorrowingsD. Inventories
Answer» Correct Answer - D

Discussion

811.	From the following information, Calculate Proprietary Ratio: Share Capital Rs 5,00,000, Non-Current Assets Rs 22,00,000 Reserves and Surplus Rs 3,00,000 Current Assets Rs 10,00,000A. 1B. 0.7C. 0.4D. 0.25
Answer» Correct Answer - D

Discussion

812.	Premium on issue of shares can be used forA. issue of fully paid bonus sharesB. writing off preliminary expensesC. writing off discount/loss on issue of debenturesD. All the above
Answer» Correct Answer - D

Discussion

813.	Name the account which calculates Surplus/Deficit ofNPO?A. Cash BookB. Income and Expenditure AccountC. Receipts and Payments AccountD. Balance Sheet
Answer» Correct Answer - B

Discussion

814.	Bills Payable is shown asA. Long-term BorrowingsB. Short-term BorrowingsC. Other Current LiabilitiesD. Trade Payables
Answer» Correct Answer - D

Discussion

815.	Surplus, i.e., Balance in Statement of Profit and Loss is shown asA. Share CapitalB. Reserves and SurplusC. Other Long-term LiabilitiesD. None of these
Answer» Correct Answer - B

Discussion

816.	If the loss on reissue of shares is less than the amount forfeited, the surplus is transferred toA. Capital ReserveB. An assetC. Revenue ReserveD. None of these
Answer» Correct Answer - A

Discussion

817.	Share Application Money is shown asA. Other Long -term LiabilitiesB. Other Current LiabilitiesC. Short-term ProvisionsD. Trade Payables
Answer» Correct Answer - B

Discussion

818.	If a share of Rs 10 issued at a premium of Rs 2 on which the full amount has been called and Rs 8 (including premium) paid is forfeited, the Share Capital Account should be debited withA. Rs 12B. Rs 10C. Rs 8D. Rs 6
Answer» Correct Answer - B

Discussion

819.	When shares are forfeited, share Capital Account is debited withA. nominal value of sharesB. called-up value of sharesC. paid-up value of sharesD. market value of shares
Answer» Correct Answer - B

Discussion

820.	Debentures redeemable after 10 years from the date of issue are shown asA. Long term BorrowingsB. Other Long-tern LiabilitiesC. Short-term BorrowingsD. Other Short-term Liabilities
Answer» Correct Answer - A

Discussion

821.	If total Assets are Rs 1,25,000, Total Debts, i.e., external debts are Rs 1,00,000 and Current Liabilities are Rs 50,000, Debt-Equity Ratio will beA. `1 : 1`B. `1 : 2`C. `2 : 1`D. None of these
Answer» Correct Answer - C

Discussion

822.	Mohar Ltd. Forfeited 160 shares of Rs 10 each on which the holder had paid only the application money of Rs2 per share. Out of these, 40 shares were reissued to Gaurav as fully paid for Rs 9 per share. The gain on rejssue isA. Rs 320B. Rs 160C. Rs 40D. None of these
Answer» Correct Answer - C

Discussion

823.	A company has issued 1,00,000 Equity Shares of Rs 10 each. It has called the total nominal value of the share. It has received the cells made except the final call of Rs 3 on 1,000 shares. Subscribed capital will be shown as follows:A. Subscribed and not fully paid-upB. `{:("subscribed and fully paid-up",Rs,),("Rs 1,00,000 Equity Shares of Rs 10 each","10,00,000",),("Less: Calls-in-Arrears",ul("3,000"),),(,ul("9,97,000"),):}`C. `{:("Subscribed and fully paid-up",,Rs),("99,000 Equity Shares of Rs 10 each",,"9,90,000"),("Subscribed but not fully paid-up",,),("1,000 Equity Shares of Rs 10 each","10,000",),("Less: Calls-in-Arrears",ul("3,000"),ul("7,000")),(,,ul("9,97,000")):}`D. Can be shown as (b) or as (c)
Answer» Correct Answer - C

Discussion

824.	Gopal Ltd. Purchased machine of Rs 1,15,000 from Indian Traders, payment of Rs 10,000 was made by issuing cheque and the remaining amount by issue of equity shares of the face value of Rs 10 each fully paid at an issue price of Rs 10.50 each. Amount of securities premium will beA. Rs 6,000B. Rs 7,000C. Rs 5,000D. Rs 4,000
Answer» Correct Answer - C

Discussion

825.	Anothony Ltd. Issued 40,000 equity shares of Rs 20 each payable as Rs 5 on application, Rs 7 on allotment and Rs 8 on final call. Company received the due amount but one shareholder holding 250 shares did not pay the allotment money and another shareholder holding 150 shares failed to pay the amount due on final call. Total amount of Calls-in-Arrears isA. Rs 1,750B. Rs 3,200C. Rs 6,000D. Rs 4,950
Answer» Correct Answer - D

Discussion

826.	Green Ltd. Had allotted 10,000 shares to the applicants of 14,000 shares on pro rata basis. The amount payable on application is Rs 2 per share. Mohan applied for 420 shares. The number of shares allotted and the amount carried forward for adjustment against allotment money due from Mohan areA. 60 shares Rs 120B. 320 shares, Rs 200C. 340 shares, Rs 100D. 300 shares, Rs 240
Answer» Correct Answer - D

Discussion

827.	Why definition of an atom given by Dalton is no longer valid?
Answer» Atoms contains protons, neutrons, and electrons. Dalton was not able to explain this concept. According to him, atoms are indivisible particles. Hence, the definition of an atom given by Dalton is no longer valid.

Discussion

828.	How can you appreciate John Dalton for proposing his atomic theory?
Answer» We know, development of science and technology is a combined effort of Roman scientists. The zeal to know or discover something new, leads the scientists for their discoveries. One of such is the atomic theory proposed by Dalton, based on Lavoisier’s law of conservation of mass and Proust’s law of constant proportions. Dalton said atom is indivisible. This proposal lead to discover many new things by various scientists and to unveil the complete structure of atom. Hence Dalton’s contribution is highly appreciable.

Discussion

829.	State the law of conservation of mass. Give one example to illustrate this law.
Answer» Antonie Lavoisier proposed the Law of conservation of mass which states that ‘Mass can be neither created nor destroyed’ so in a chemical reaction, the total mass of reactants must be equal to the total mass of products. For example: 100gs of calcium carbonate decomposes to produce 56gms of calcium oxide and 44gms of carbon dioxide Mass of reactant = 100gm Mass of Product = 56+44gm = 100gm Since mass of the reactant is equal to the mass of the product, there is no change in the chemical reaction.

829.

State the law of conservation of mass. Give one example to illustrate this law.

Answer»

Antonie Lavoisier proposed the Law of conservation of mass which states that ‘Mass can be neither created nor destroyed’ so in a chemical reaction, the total mass of reactants must be equal to the total mass of products.

For example: 100gs of calcium carbonate decomposes to produce 56gms of calcium oxide and 44gms of carbon dioxide

Mass of reactant = 100gm

Mass of Product = 56+44gm = 100gm

Since mass of the reactant is equal to the mass of the product, there is no change in the chemical reaction.

Discussion

830.	What is the law of constant proportions ?
Answer» A given chemical substance always contains the same elements combined in a fixed proportion by weight. This is the law of constant proportions.

Discussion

831.	State the following. a) Law of conservation of mass b) Law of constant proportions
Answer» a) Law of conservation of mass : Matter is neither created nor destroyed during a chemical reaction. (or) The mass of the reactants is equal to the mass of the products of chemical reaction. b) Law of constant proportions : A given chemical substance always contains the same elements combined in a fixed proportions by weight.

831.

State the following. a) Law of conservation of mass b) Law of constant proportions

Answer»

a) Law of conservation of mass : Matter is neither created nor destroyed during a chemical reaction.

(or)

The mass of the reactants is equal to the mass of the products of chemical reaction.

b) Law of constant proportions : A given chemical substance always contains the same elements combined in a fixed proportions by weight.

Discussion

832.	What is the proposal of Indian sage Kanada, about atom?
Answer» 1. About 2600 years ago, an Indian sage called Kanada, postulated atoms in his Vaishesika sutra”. 2. He proposed that all forms of matter are composed of very small particles known as “Anu”. 3. Each “Anu” may be made up of still smaller particles called “Paramanu”.

832.

What is the proposal of Indian sage Kanada, about atom?

Answer»

1. About 2600 years ago, an Indian sage called Kanada, postulated atoms in his Vaishesika sutra”.

2. He proposed that all forms of matter are composed of very small particles known as “Anu”.

3. Each “Anu” may be made up of still smaller particles called “Paramanu”.

Discussion

833.	Dalton’s atomic theory says that atoms are indivisible. Is this statement still valid? Give reasons for your answer.
Answer» Dalton’s atomic theory states that atoms are indivisible. This statement is not valid as, atoms can be further divided into electrons, protons and neutrons. This is one of the major drawbacks of Dalton’s atomic theory of matter.

Discussion

834.	What are the Dalton’s proposals about the nature of matter?
Answer» John Dalton proposed the basic theory about the nature of matter. His proposals are : If mass was to be conserved, then all elements must be made up of extremely small particles called atoms. If law of constant proportions is to be followed, the particles of same substance couldn’t be dissimilar.

834.

What are the Dalton’s proposals about the nature of matter?

Answer»

John Dalton proposed the basic theory about the nature of matter. His proposals are :

If mass was to be conserved, then all elements must be made up of extremely small particles called atoms.
If law of constant proportions is to be followed, the particles of same substance couldn’t be dissimilar.

Discussion

835.	The formula of water molecule is H2O. What information do you get from this formula?
Answer» Water is a combination of hydrogen and oxygen. Two hydrogen atoms and one oxygen atom combine to form one water molecule. Molecular weight of water molecule is 18. [Hydrogen 1, Oxygen 16. H₂O ⇒ 2 × 1 + 16=18] 18 g of water molecule contains 6.022 × 10²³ particles in it. Valency of hydrogen is 1 and oxygen is 2.

Discussion

836.	Molecules in water is H2O, then molecules in the Hydrogen is A) H B) H2C) A or B D) No molecules
Answer» Correct option is B) H₂

Discussion

837.	Which of the following correctly represents 360 g of water?i. 2 moles of H2Oii. 20 moles of wateriii. 6.022 x 1023 molecules of wateriv. 1.2044 x 1025 molecules of waterA. iB. I and ivC. ii and iiiD. ii and iv
Answer» 20 moles of water: Given: Number of moles = 20 Atomic mass of H₂O = 2 × 1 + 16 = 18g/mol To find out the mass of water, apply the formula given: Number of moles = \(\frac{Mass}{Molar\,mass}\) ⇒20 = \(\frac{mass}{18\,g/mol}\) ⇒ Mass = 20 mol × 18g/mol ⇒ Mass = 360 g 1.2044 x 10²⁵ molecules of water: 1 mole = 6.022 × 10²³ 1.2044 x 10²⁵ molecules = \(\frac{1.2044\times10^{23}}{6.022\times10^{23}}\) ⇒20 moles Mass of 20 moles of water is 360g. Thus, option D is correct.

837.

Which of the following correctly represents 360 g of water?i. 2 moles of H2Oii. 20 moles of wateriii. 6.022 x 1023 molecules of wateriv. 1.2044 x 1025 molecules of waterA. iB. I and ivC. ii and iiiD. ii and iv

Answer»

20 moles of water:

Given: Number of moles = 20

Atomic mass of H₂O = 2 × 1 + 16 = 18g/mol

To find out the mass of water, apply the formula given:

Number of moles = \(\frac{Mass}{Molar\,mass}\)

⇒20 = \(\frac{mass}{18\,g/mol}\)

⇒ Mass = 20 mol × 18g/mol

⇒ Mass = 360 g

1.2044 x 10²⁵ molecules of water:

1 mole = 6.022 × 10²³

1.2044 x 10²⁵ molecules = \(\frac{1.2044\times10^{23}}{6.022\times10^{23}}\)

⇒20 moles

Mass of 20 moles of water is 360g.

Thus, option D is correct.

Discussion

838.	Which of the following will have maximum mass?A. 1022 molecules of H2OB. 1022 molecules of NH3C. 2 moles of NH3D. 2 moles of H2O
Answer» 2 moles of H₂O has less number of moles as well as less molecular mass among all the given options. Hence, it has maximum mass. Thus, option D is correct.

838.

Which of the following will have maximum mass?A. 1022 molecules of H2OB. 1022 molecules of NH3C. 2 moles of NH3D. 2 moles of H2O

Answer»

2 moles of H₂O has less number of moles as well as less molecular mass among all the given options. Hence, it has maximum mass.

Thus, option D is correct.

Discussion

839.	Give one major drawback of Dalton’s atomic theory of matter.
Answer» The major drawback of Dalton’s atomic theory is that atoms were thought to be indivisible. But, it is not true since atoms are divisible.

Discussion

840.	Which of the following has maximum numbers of atoms?A. 18 g of H2OB. 18 g of O2C. 18 g of CO2D. 18 g of CH4
Answer» When the mass is same, the number of atoms is inversely proportional to the molecular mass. This means less will be the molecular mass, more will be the number of atoms. H₂O = 2× 1 + 16 = 18u O₂ = 2× 16 = 32u CO₂ = 12 × 2 × 16 = 44u CH₄ = 12 + 4× 1 = 16u Hence, 18 g of CH₄ has maximum numbers of atoms.

840.

Which of the following has maximum numbers of atoms?A. 18 g of H2OB. 18 g of O2C. 18 g of CO2D. 18 g of CH4

Answer»

When the mass is same, the number of atoms is inversely proportional to the molecular mass. This means less will be the molecular mass, more will be the number of atoms.

H₂O = 2× 1 + 16 = 18u

O₂ = 2× 16 = 32u

CO₂ = 12 × 2 × 16 = 44u

CH₄ = 12 + 4× 1 = 16u

Hence, 18 g of CH₄ has maximum numbers of atoms.

Discussion

841.	Name the scientist who gave atomic theory of matter.
Answer» John Dalton is the scientist who gave atomic theory of matter.

Discussion

842.	What are the materials used in “Conservation of mass” experiment?
Answer» Material required for Conservation of mass: Sodium sulphate, Barium chloride, distilled water, conical flask, spring balance, small test tube, rubber cork, thread, retort stand.

842.

What are the materials used in “Conservation of mass” experiment?

Answer»

Material required for Conservation of mass:

Sodium sulphate, Barium chloride, distilled water, conical flask, spring balance, small test tube, rubber cork, thread, retort stand.

Discussion

843.	Name the law of chemical combination: (a) Which was given by Lavoisier (b) Which was given by Proust
Answer» (a) Law of conservation of mass. (b) Law of constant proportions.

843.

Name the law of chemical combination: (a) Which was given by Lavoisier (b) Which was given by Proust

Answer»

(a) Law of conservation of mass.

(b) Law of constant proportions.

Discussion

844.	3.42 g of glucose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution is A. 6.68 × 1023 B. 6.09 × 1022 C. 6.022 × 1023 D. 6.022 × 1021
Answer» Molar mass of sucrose = C₁₂H₂₂O₁₁ = 12× 12 + 1× 22 + 16 × 11 = 342g/mol Number of moles = \(\frac{Mass\,of\,glucos}{Molar\,mass\,of\,sucrose}\) ⇒ Number of moles = \(\frac{3.42g}{342g/mol}\) ⇒ Number of moles = 0.01 Sucrose (C₁₂H₂₂O₁₁) contains 11 oxygen atoms ⇒ 11 × 6.022 × 10²³ For 0.01 moles of sucrose ⇒ 0.01 × 11 × 6.022 × 10²³= 6.6 × 10 22 Now, Molar mass of water = H₂O = 2× 1 + 16 = 18g/mol Number of moles =\(\frac{Mass\,of\,water}{Molar\,mass\,of\,water}\) ⇒ Number of moles = \(\frac{18g}{18g/mol}\) ⇒ Number of moles = 1 Sucrose (H₂O) contains 1 oxygen atom = 6.022 × 10²³ For 1 mole of water = 6.022 × 10²³ Now, add the both values: 6.6 × 10²²+ 6.022 × 10²³ We get 6.68 × 10²³ atoms. Hence, the option (a) is correct.

844.

3.42 g of glucose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution is A. 6.68 × 1023 B. 6.09 × 1022 C. 6.022 × 1023 D. 6.022 × 1021

Answer»

Molar mass of sucrose = C₁₂H₂₂O₁₁

= 12× 12 + 1× 22 + 16 × 11 = 342g/mol

Number of moles = \(\frac{Mass\,of\,glucos}{Molar\,mass\,of\,sucrose}\)

⇒ Number of moles = \(\frac{3.42g}{342g/mol}\)

⇒ Number of moles = 0.01

Sucrose (C₁₂H₂₂O₁₁) contains 11 oxygen atoms

⇒ 11 × 6.022 × 10²³

For 0.01 moles of sucrose

⇒ 0.01 × 11 × 6.022 × 10²³= 6.6 × 10 22

Now, Molar mass of water = H₂O = 2× 1 + 16 = 18g/mol

Number of moles =\(\frac{Mass\,of\,water}{Molar\,mass\,of\,water}\)

⇒ Number of moles = \(\frac{18g}{18g/mol}\)

⇒ Number of moles = 1

Sucrose (H₂O) contains 1 oxygen atom = 6.022 × 10²³

For 1 mole of water = 6.022 × 10²³

Now, add the both values: 6.6 × 10²²+ 6.022 × 10²³ We get 6.68 × 10²³ atoms.

Hence, the option (a) is correct.

Discussion

845.	Which of the following would weigh the highest?A. 0.2 mole of sucrose (C12H22O11)B. 2 moles of CO2C. 2 moles of CaCO3D. 10 moles of H2O
Answer» A. 0.2 mole of sucrose (C₁₂H₂₂O₁₁): Given: Number of moles = 0.2 Molar mass of C₁₂H₂₂O₁₁ = 342 g/mol To find out the mass of sucrose, apply the formula given: Number of moles = \(\frac{Mass}{Molar\,mass}\) ⇒ 0.2 = \(\frac{mass}{342\,g/mol}\) ⇒ Mass = 0.2 mol × 342g/mol ⇒ Mass = 68.4g Thus, the mass of sucrose is 68.4 g. B. 2 moles of CO₂: Given: Number of moles = 2 Molar mass of CO₂ = 44 g/mol To find out the mass of sucrose, apply the formula given: ⇒ Mass = 2 mol × 44g/mol ⇒ Mass = 88g Thus, the mass of CO₂ is 88 g. C. 2 moles of CaCO₃: Given: Number of moles = 2 Molar mass of CaCO₃ = 40 + 12 + 3× 16 = 100 g/mol To find out the mass of sucrose, apply the formula given: Number of moles = \(\frac{Mass}{Molar\,mass}\) ⇒ 2 = \(\frac{mass}{100\,g/mol}\) ⇒ Mass = 2 mol × 100g/mol ⇒ Mass = 200g Thus, the mass of CaCO₃ is 200 g. D. 10 moles of H₂O: Given: Number of moles = 10 Molar mass of H₂O = 18 g/mol To find out the mass of sucrose, apply the formula given: Number of moles = \(\frac{Mass}{Molar\,mass}\) ⇒ 10 = \(\frac{mass}{18\,g/mol}\) ⇒ Mass = 10 mol × 18g/mol ⇒ Mass = 180g Thus, the mass of H₂O is 180 g. Hence, the option C is correct which has higher mass.

845.

Which of the following would weigh the highest?A. 0.2 mole of sucrose (C12H22O11)B. 2 moles of CO2C. 2 moles of CaCO3D. 10 moles of H2O

Answer»

A. 0.2 mole of sucrose (C₁₂H₂₂O₁₁):

Given: Number of moles = 0.2

Molar mass of C₁₂H₂₂O₁₁ = 342 g/mol

To find out the mass of sucrose, apply the formula given:

Number of moles = \(\frac{Mass}{Molar\,mass}\)

⇒ 0.2 = \(\frac{mass}{342\,g/mol}\)

⇒ Mass = 0.2 mol × 342g/mol

⇒ Mass = 68.4g

Thus, the mass of sucrose is 68.4 g.

B. 2 moles of CO₂:

Given: Number of moles = 2

Molar mass of CO₂ = 44 g/mol

To find out the mass of sucrose, apply the formula given:

⇒ Mass = 2 mol × 44g/mol

⇒ Mass = 88g

Thus, the mass of CO₂ is 88 g.

C. 2 moles of CaCO₃:

Given: Number of moles = 2

Molar mass of CaCO₃ = 40 + 12 + 3× 16 = 100 g/mol

To find out the mass of sucrose, apply the formula given:

Number of moles = \(\frac{Mass}{Molar\,mass}\)

⇒ 2 = \(\frac{mass}{100\,g/mol}\)

⇒ Mass = 2 mol × 100g/mol

⇒ Mass = 200g

Thus, the mass of CaCO₃ is 200 g.

D. 10 moles of H₂O:

Given: Number of moles = 10

Molar mass of H₂O = 18 g/mol

To find out the mass of sucrose, apply the formula given:

Number of moles = \(\frac{Mass}{Molar\,mass}\)

⇒ 10 = \(\frac{mass}{18\,g/mol}\)

⇒ Mass = 10 mol × 18g/mol

⇒ Mass = 180g

Thus, the mass of H₂O is 180 g.

Hence, the option C is correct which has higher mass.

Discussion

846.	One of the precautions to be taken in the conduction of experiment on law of conservation of mass is A) Test tube should not be tilted B) Test tube should be tilted C) Test tube should be immersed in the conical flask D) Test tube should be kept outside the conical flask
Answer» B) Test tube should be tilted

Discussion

847.

Describe the experiment conducted by Joseph L. Proust, which lead him to propose law of constant proportions.

Answer»

Proust took two samples of copper carbonate, one from nature and another prepared in the lab.
These two samples are chemically decomposed to find percentage of copper, carbon and oxygen.
The results obtained are given in the table.

Weight percentage of	Natural sample	Synthetic sample
Copper	51.35	51.35
Carbon	38.91	38.91
Oxygen	9.74	9.74

From the above table we observe that the percentage of copper, carbon and oxygen atoms in two samples are same.
Based on this observation Proust proposed the law of constant proportions as “A given chemical substance always contains the same elements combined in a fixed proportions by weight”.

Discussion

848.	Nitrogen and hydrogen combine together to form ammonia. N2 + 3H2→ 2NH3 [Relative atomic masses of N = 14 u, H = 1 u] The mass of nitrogen and hydrogen which combine together to form 6.8 g ammonia is A. N2 = 2.8g, H2 = 4.0g B. N2 = 5.6g, H2 = 1.2g C. N2 = 4.0g, H2 = 2.8g D. N2 = 12g, H2 = 5.6g
Answer» If N₂ = 5.6g, H₂ = 1.2g According to the law of conservation of mas, 5.6g + 1.2g = 6.8g of ammonia Hence, the option (b) is correct

848.

Nitrogen and hydrogen combine together to form ammonia. N2 + 3H2→ 2NH3 [Relative atomic masses of N = 14 u, H = 1 u] The mass of nitrogen and hydrogen which combine together to form 6.8 g ammonia is A. N2 = 2.8g, H2 = 4.0g B. N2 = 5.6g, H2 = 1.2g C. N2 = 4.0g, H2 = 2.8g D. N2 = 12g, H2 = 5.6g

Answer»

If N₂ = 5.6g, H₂ = 1.2g

According to the law of conservation of mas, 5.6g + 1.2g = 6.8g of ammonia

Hence, the option (b) is correct

Discussion

849.	In the above given experiment, this precaution must be takenA) put the cork tightly B) measure components accurately C) while measuring leave the apparatus freely D) above all
Answer» Correct option is D) above all

Discussion

850.

During an experiment hydrogen (H2) and oxygen (O2) gases reacted in an electric are to produce water asfollows: 2H +O2 \(\overset{Electicity}{\longrightarrow}\) 2H2OThe experiment is repeated three times and data tabulated as shown below:Experiment numberMass of H2 reactedMass of O2 reactedMass f H2O produced12g16g18g24g32g36g3--9gDuring 3rd experiment the researcher forgot to list masses of H2 and O2 used. So, if the law of constant proportion is correct then ind mass of O2 used during 3rd experiment. A. 4g B. 8g C. 16g D. 32g

Answer»

In experiment (i), the ratio is 2:16, i.e., 1:8

In experiment (ii), the ratio is 4:32, i.e., 1:8

According to the law of constant proportions, the ratio of third experiment should be the same (1:8)

Therefore, the mass of O₂ used during 3^rd experiment is 8g.

Discussion

Explore topic-wise InterviewSolutions in .

Normal profit is calculated to value goodwillA. by deducting abnormal gains (profit)B. by adding abnormal lossesC. by deducting abnormal gains and adding abnormal lossesD. None of the above

Young India Ltd. has a Operating Porfit Ratio of 20%. To maintain this ratio at 25%, management mayA. Increase selling price of Stock-in-tradeB. Reduce Cost of Revenue from OperationC. Increase selling price of Stock-in-Trade and to reduce Cost of Revenue from OperationD. All of the above

A transaction involving a decrease in both Current Ratio and Quick Ratio isA. Sale of Non-current Asset for cashB. Sale of Stock-in-Trade at lossC. Cash payment of a Current LiabilityD. Purchase of Stock-in-Trade on credit

Total capital employed in the firm is Rs 8,00,000, Normal Rate of Return is 15% and Profit for the year is Rs 12,00,000. The value of goodwill of the firm as per capitalisation method would beA. Rs 82,00,000B. Rs 12,00,000C. Rs 72,00,000D. Rs 42,00,000

Revenue from Operations Rs 9,00,000, Gross Profit 25% on Cost, Operating Expenses Rs 90,000, Operating Ratio will beA. 1B. 0.5C. 0.9D. 0.1

Calculate Operating Profit Ratio if Revenue from Operations is Rs 5,00,000, Operating Profit is Rs 75,000.A. 0.25B. 0.12C. `13.33%`D. 0.15

If Current Ratio of a firm is `2.5 : 1` and its Current Liabilities are Rs 2,00,000. Its Working Capital will beA. Rs 3,00,000B. Rs 3,75,000C. Rs 4,00,000D. Rs 7,00,000

From the given items which is a part of Current LiabilitiesA. InventoriesB. Trade payablesC. Cash and Cash EquivalentD. Trade Receivables

From the given items which is not shown under Current LiabilitiesA. Trade PayablesB. Short-term ProvisionsC. Short-term BorrowingsD. Inventories

From the following information, Calculate Proprietary Ratio: Share Capital Rs 5,00,000, Non-Current Assets Rs 22,00,000 Reserves and Surplus Rs 3,00,000 Current Assets Rs 10,00,000A. 1B. 0.7C. 0.4D. 0.25

Premium on issue of shares can be used forA. issue of fully paid bonus sharesB. writing off preliminary expensesC. writing off discount/loss on issue of debenturesD. All the above

Name the account which calculates Surplus/Deficit ofNPO?A. Cash BookB. Income and Expenditure AccountC. Receipts and Payments AccountD. Balance Sheet

Bills Payable is shown asA. Long-term BorrowingsB. Short-term BorrowingsC. Other Current LiabilitiesD. Trade Payables

Surplus, i.e., Balance in Statement of Profit and Loss is shown asA. Share CapitalB. Reserves and SurplusC. Other Long-term LiabilitiesD. None of these

If the loss on reissue of shares is less than the amount forfeited, the surplus is transferred toA. Capital ReserveB. An assetC. Revenue ReserveD. None of these

Share Application Money is shown asA. Other Long -term LiabilitiesB. Other Current LiabilitiesC. Short-term ProvisionsD. Trade Payables

If a share of Rs 10 issued at a premium of Rs 2 on which the full amount has been called and Rs 8 (including premium) paid is forfeited, the Share Capital Account should be debited withA. Rs 12B. Rs 10C. Rs 8D. Rs 6

When shares are forfeited, share Capital Account is debited withA. nominal value of sharesB. called-up value of sharesC. paid-up value of sharesD. market value of shares

Debentures redeemable after 10 years from the date of issue are shown asA. Long term BorrowingsB. Other Long-tern LiabilitiesC. Short-term BorrowingsD. Other Short-term Liabilities

If total Assets are Rs 1,25,000, Total Debts, i.e., external debts are Rs 1,00,000 and Current Liabilities are Rs 50,000, Debt-Equity Ratio will beA. `1 : 1`B. `1 : 2`C. `2 : 1`D. None of these

Mohar Ltd. Forfeited 160 shares of Rs 10 each on which the holder had paid only the application money of Rs2 per share. Out of these, 40 shares were reissued to Gaurav as fully paid for Rs 9 per share. The gain on rejssue isA. Rs 320B. Rs 160C. Rs 40D. None of these

Why definition of an atom given by Dalton is no longer valid?

How can you appreciate John Dalton for proposing his atomic theory?

State the law of conservation of mass. Give one example to illustrate this law.

What is the law of constant proportions ?

State the following. a) Law of conservation of mass b) Law of constant proportions

What is the proposal of Indian sage Kanada, about atom?

Dalton’s atomic theory says that atoms are indivisible. Is this statement still valid? Give reasons for your answer.

What are the Dalton’s proposals about the nature of matter?

The formula of water molecule is H2O. What information do you get from this formula?

Molecules in water is H2O, then molecules in the Hydrogen is A) H B) H2C) A or B D) No molecules

Which of the following correctly represents 360 g of water?i. 2 moles of H2Oii. 20 moles of wateriii. 6.022 x 1023 molecules of wateriv. 1.2044 x 1025 molecules of waterA. iB. I and ivC. ii and iiiD. ii and iv

Which of the following will have maximum mass?A. 1022 molecules of H2OB. 1022 molecules of NH3C. 2 moles of NH3D. 2 moles of H2O

Give one major drawback of Dalton’s atomic theory of matter.

Which of the following has maximum numbers of atoms?A. 18 g of H2OB. 18 g of O2C. 18 g of CO2D. 18 g of CH4

Name the scientist who gave atomic theory of matter.

What are the materials used in “Conservation of mass” experiment?

Name the law of chemical combination: (a) Which was given by Lavoisier (b) Which was given by Proust

3.42 g of glucose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution is A. 6.68 × 1023 B. 6.09 × 1022 C. 6.022 × 1023 D. 6.022 × 1021

Which of the following would weigh the highest?A. 0.2 mole of sucrose (C12H22O11)B. 2 moles of CO2C. 2 moles of CaCO3D. 10 moles of H2O

One of the precautions to be taken in the conduction of experiment on law of conservation of mass is A) Test tube should not be tilted B) Test tube should be tilted C) Test tube should be immersed in the conical flask D) Test tube should be kept outside the conical flask

Describe the experiment conducted by Joseph L. Proust, which lead him to propose law of constant proportions.

Nitrogen and hydrogen combine together to form ammonia. N2 + 3H2→ 2NH3 [Relative atomic masses of N = 14 u, H = 1 u] The mass of nitrogen and hydrogen which combine together to form 6.8 g ammonia is A. N2 = 2.8g, H2 = 4.0g B. N2 = 5.6g, H2 = 1.2g C. N2 = 4.0g, H2 = 2.8g D. N2 = 12g, H2 = 5.6g

In the above given experiment, this precaution must be takenA) put the cork tightly B) measure components accurately C) while measuring leave the apparatus freely D) above all