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Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 2 Introduction to Analytical Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 2 Introduction to Analytical Chemistry

Question 1.
Explain the statement: Analytical chemistry provides physical or chemical information about a sample.
Answer:

• Analytical chemistry facilitates the investigation of the chemical composition of substances.
• It uses the instruments and methods to separate, identify and quantify a sample under study.

Thus, analytical chemistry provides chemical or physical information about a sample.

Question 2.
What is the difference between qualitative analysis and quantitative analysis?
Answer:

1. Qualitative analysis deals with the detection of the presence or absence of elements in compounds and of chemical compounds in mixtures.
2. Quantitative analysis deals with the determination of the relative proportions of elements in compounds and of chemical compounds in mixtures.

Question 3.
Explain the importance of chemical analysis.
Answer:

• Chemical analysis is one of the most important methods of monitoring the composition of raw materials, intermediates and finished products, and also the composition of air in streets and premises of industrial plants.
• In agriculture, chemical analysis is used to determine the composition of soils and fertilizers.
• In medicine, it is used to determine the composition of medicinal preparations.

Question 4.
Write a note on the applications of analytical chemistry.
Answer:

• Analytical chemistry has applications in forensic science, engineering and industry.
• Analytical chemistry is also useful in the field of agriculture and pharmaceutical industry.
• Industrial process as a whole and the production of new kinds of materials are closely associated with analytical chemistry.

Question 5.
What is semi-microanalysis?
Answer:

• When the amount of a solid or liquid sample taken for analysis is a few grams, the analysis is called semi-microanalysis.
• It is of two types: qualitative and quantitative analysis.

Question 6.
What does classical qualitative analysis method include?
Answer:
Classical qualitative analysis method includes separation and identification of compounds.

• Separations may be done by methods such as precipitation, extraction and distillation.
• Identification may be based on differences in colour, odour, melting point, boiling point, and reactivity.

Question 7.
Name two methods of classical quantitative analysis.
Answer:

1. Volumetric analysis (Titrimetric analysis)
2. Gravimetric analysis (i.e., decomposition, precipitation)

Question 8.
What are the two stages involved in the chemical analysis of a sample?
Answer:
The chemical analysis of a sample is carried out mainly in two stages: by the dry method and by the wet method. In dry method, the sample under test is not dissolved and in wet method, the sample under test is first dissolved and then analysed to determine its composition.

Question 9.
Explain: Qualitative analysis of organic compounds
Answer:

• The majority of organic compounds are composed of a relatively small number of elements.
• The most important ones are: carbon, hydrogen, oxygen, nitrogen, sulphur, halogen, phosphorus.
• Elementary qualitative analysis is concerned with the detection of the presence of these elements.
• The identification of an organic compound involves tests such as detection of functional group, determination of melting/boiling points, etc.

Question 10.
What does qualitative analysis of inorganic compounds involve?
Answer:
The qualitative analysis of simple inorganic compounds involves detection and confirmation of cationic and anionic species (basic and acidic radical) in them.

Question 11.
Explain: Chemical methods of quantitative analysis
Answer:

• Quantitative analysis of organic compounds involves methods such as determination of percentage of constituent elements, concentrations of a known compound in the given sample, etc.
• Quantitative analysis of simple inorganic compounds involves methods such as gravimetric analysis (i.e., decomposition, precipitation, etc.) and the titrimetric or volumetric analysis (i.e., progress of reaction between two solutions till its completion).
• The quantitative analytical methods involve measurement of quantities such as mass and volume using some equipment/apparatus such as weighing machine, burette, etc.

Question 12.
Why is accurate measurement crucial in science?
Answer:

• The accuracy of measurement is of a great concern in analytical chemistry. This is because faulty equipment, poor data processing or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in the analytical measurement.
• When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.
• Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
• Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation and properly express the quantitative error in the result.

Question 13.
Why are scientific notations (exponential notations) used?
Answer:
A chemist has to deal with numbers as large as 602,200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g. that is, mass of a H atom. To avoid the writing of so many zeros in mathematical operations, scientific notations i.e. exponential notations are used.

Question 14.
How are numbers expressed in scientific notations (exponential notations)?
Answer:
In scientific notations, numbers are expressed in the form of N × 10ⁿ, where ‘n’ is an exponent with positive or negative values and N can have a value between 1 to 10.
e.g. i. The number, 602,200,000,000,000,000,000,000 is expressed as 6.022 × 10²³.
ii. The mass of a H atom, 0.00000000000000000000000166 g is expressed as 1.66 × 10^-24 g.
iii. The number 123.546 is written as 1.23546 × 10².
iv. The number 0.00015 is written as 1.5 × 10^-4.

Question 15.
Express the following quantities in scientific notations (exponential notations).
i. 0.0345
ii . 0.08
iii. 653.00
iv. 34.768
Answer:
i. 0.0345 = 3.45 × 10^-2
ii. 0.08 = 8 × 10^-2
iii. 653.00 = 6.5300 × 10²
iv. 34.768 = 3.4768 × 10¹

Question 16.
Define: Accuracy of measurement
Answer:
Nearness of the measured value to the true value is called the accuracy of measurement.

Question 17.
Explain with the help of a diagram how accuracy depends upon the sensitivity or least count of the measuring equipment.
Answer:
A burette reading of 10.2 mL is as shown in the diagram below:

• For all the three situations in the above figure, the reading would be noted is 10.2 mL.
• It means that there is an uncertainty about the digit appearing after the decimal point in the reading 10.2 mL because the least count of the burette is 0.1 mL.
• The meaning of the reading 10.2 mL is that the true value of the reading lies between 10.1 mL and 10.3 mL.
• This is indicated by writing 10.2 ± 0.1 mL.
• Here, the burette reading has an error of ± 0.1 mL.
• Smaller the error, higher is the accuracy.

Question 18.
How is absolute error calculated?
Answer:
Absolute error is calculated by subtracting true value from observed value.
Absolute error = Observed value – Tme value

Question 19.
Explain the term: Relative error
Answer:

1. Relative error is the ratio of an absolute error to the true value.
2. Relative error is generally a more useful quantity than absolute error.
3. Relative error is expressed as a percentage and can be calculated as follows:
   Relative error = \(\frac{\text { Absolute error }}{\text { True value }} \times 100 \%\)

Question 20.
Explain the term: Precision in measurement
Answer:

• Multiple readings of the same quantity are noted to minimize the error.
• If the multiple readings of the same quantity match closely, they are said to have high precision.
• High precision implies reproducibility of the readings.
• High precision is a prerequisite for high accuracy.
• Precision is expressed in terms of deviation (i.e. absolute deviation and relative deviation).

Question 21.
Explain the following terms with respect to precise measurement:
i. Absolute deviation
ii. Mean absolute deviation
iii. Relative deviation
Answer:
i. Absolute deviation: An absolute deviation is the modulus of the difference between an observed value and the arithmetic mean for the set of several measurements made in the same way. It is a measure of absolute error in the repeated observation. It is expressed as follows:
Absolute deviation = |Observed value – Mean|

ii. Mean absolute deviation: Arithmetic mean of all the absolute deviations is called the mean absolute deviation in the measurements.

iii. Relative deviation: The ratio of mean absolute deviation to its arithmetic mean is called relative deviation. It is expressed as follows:
Relative deviation = \(\frac{\text { Mean absolute deviation }}{\text { Mean }}\) × 100%

Question 22.
Explain the need of significant figures in measurement.
Answer:

• Uncertainty in measured value leads to uncertainty in calculated result.
• Uncertainty in a value is indicated by mentioning the number of significant figures in that value. e.g. Consider, the column reading 10.2 ± 0.1 mL recorded on a burette having the least count of 0.1 mL. Here, it is said that the last digit ‘2’ in the reading is uncertain, its uncertainty is ±0.1 mL. On the other hand, the figure ‘10’ is certain.
• The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
• In a scientific experiment, a result is obtained by doing calculation in which values of a number of quantities measured with equipment of different least counts are used.

Question 23.
How many significant figures are present in the following measurements?
i. 4.065 m
ii. 0.32 g
iii. 57.98 cm³
iv. 0.02 s
v. 4.0 × 10^-4 km
vi. 604.0820 kg
vii. 307.100 × 10^-5 cm
Answer:
i. 4
ii. 2
iii. 4
iv. 1
v. 2
vi. 7
vii. 6

Question 24.
How many significant figures are present in each of the following?
i. 45.0
ii. 0.001
iii. 2.10 × 10^-8
iv. 340000
v. 0.0100
vi. 7890320
vii. 100.00
viii. 100
Answer:
i. 3
ii. 1
iii. 3
iv. 2
v. 3
vi. 6
vii. 5
viii. 1

Question 25.
State the rules used to round off a number to the required number of significant figures.
Answer:
The following rules are used to round off a number to the required number of significant figures:

• If the digit following the last digit to be kept is less than five, the last digit is left unchanged, e.g. 46.32 rounded off to two significant figures is 46.
• If the digit following the last digit to be kept is five or more, the last digit to be kept is increased by one. e.g. 52.87 rounded to three significant figures is 52.9.

Question 26.
Round off each of the following to the number of significant digits indicated:
i. 1.223 to two digits
ii. 12.56 to three digits
iii. 122.17 to four digits
iv. 231.5 to three digits
Answer:
i. 1.223 to two digits = 1.2
This is because the third digit is less than 5, so we drop it and all the other digits to its right.
ii. 12.56 to three digits = 12.6
This is because the fourth digit is greater than 5, so we drop it and add 1 to the third digit.
iii. 122.17 to four digits = 122.2
This is because the fifth digit is greater than 5, so we drop it and add 1 to the fourth digit.
iv. 231.5 to three digits = 232
This is because the fourth digit is 5, so we drop it and add 1 to the third digit.

Question 27.
Add 5.55 × 10⁴ and 6.95 × 10³ and express the result in scientific notation.
Solution:
To perform addition operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added.
(5.55 × 10⁴) + (6.95 × 10³)= (5.55 × 10⁴) + (0.695 × 10⁴)
= (5.55 +0.695) × 10⁴
= 6.245 × 10⁴

Question 28.
Add 1.77 × 10² and 2.23 × 10³ and express the result in scientific notation.
Solution:
To perform addition operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added.
(1.77 × 10²) + (2.23 × 10³) = (0.177 × 10³) + (2.23 × 10³)
= (0.177 + 2.23) × 10³
= 2.407 × 10³

Question 29.
Subtract 5.8 × 10^-3 from 3.5 × 10^-2 and express result in scientific notation.
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
(3.5 × 10^-2)- (5.8 × 10^-3) = (3.5 × 10^-2) – (0.58 × 10^-2)
= (3.5 – 0.58) × 10^-2
= 2.92 × 10^-2

Question 30.
Subtract 6.90 × 10^-5 from 5.11 × 10^-4 and express the result in scientific notation.
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
(5.11 × 10^-4) – (6.90 × 10^-5) = (5.11 × 10^-4) – (0.690 × 10^-4)
= (5.11 – 0.690) × 10^-4
= 4.42 × 10^-4

Question 31.
Perform following calculations and express results in scientific notations (exponential notations),
i. (1.5 × 10^-6) – (5.8 × 10^-7)
ii. (9.8 × 10^-3) – (8.8 × 10^-3)
iii. (6.5 × 10^-8) – (5.5 × 10^-9)
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
i. (1.5 × 10^-6) – (5.8 × 10^-7) = (1.5 × 10^-6) – (0.58 × 10^-6)
= (1.5 – 0.58) × 10^-6
= 0.92 × 10^-6
= 9.2 × 10^-7

ii. (9.8 × 10^-3) – (8.8 × 10^-3) = (9.8 – 8.8) × 10^-3
= 1.0 × 10^-3

iii. (6.5 × 10^-8) – (5.0 × 10^-9) = (6.5 × 10^-8) – (0.50 × 10^-8)
= (6.5 – 0.50) × 10^-8
= 6.0 × 10^-8

Question 32.
Multiply 5.6 × 10⁵ and 6.9 × 10⁸ and express result in scientific notation.
Solution:
(5.6 × 10⁵) × (6.9 × 10⁸) = (5.6 × 6.9) (10⁵⁺⁸)
= 38.64 × 10¹³
= 3.864 × 10¹⁴

Question 33.
Multiply 9.8 × 10^-2 and 2.5 × 10^-6 and express result in scientific notation.
Solution:
(9.8 × 10^-2) × (2.5 × 10^-6) = (9.8 × 2.5) (10^-2+(-6))
= (9.8 × 2.5) × (10^-2-6)
= 24.5 × 10^-8
= 2.45 × 10^-7

Question 34.
Perform following calculations and express results in scientific notations (exponential notations).
i. (2.5 × 10^-6) × (1.8 × 10^-7)
ii. (4.5 × 10^-3) × (1.8 × 10³)
iii. (8.5 × 10⁷) × (3.5 × 10⁹)
Solution:
i. (2.5 × 10^-6) × (1.8 × 10^-7) = (2.5 × 1.8) (10^-6+(-7))
= (2.5 × 1.8) × (10^-6-7)
= 4.5 × 10^-13

ii. (4.5 × 10-3) × (1.8 × 103) = (4.5 × 1.8) (10^-3+3)
= (4.5 × 1.8) × (10⁰)
= 8.1

iii. (8.5 × 10⁷) × (3.5 × 10⁹) = (8.5 × 3.5) (10⁷⁺⁹)
= (8.5 × 3.5) × 10¹⁶
= 29.75 × 10¹⁶
= 2.975 × 10¹⁷
[Note: To express number in scientific notation, the number has to be greater than or equal to 10 or less than 1. The number, 8.1 is greater than 1, but less than 10 and hence, it cannot be expressed in scientific notation.]

Question 35.
In laboratory experiment, 10 g potassium chlorate sample on decomposition gives following data: The sample contains 3.8 g of oxygen and the actual mass of oxygen in the quantity of potassium chlorate is 3.92 g. Calculate absolute error and relative error.
Solution:
The observed value is 3.8 g and accepted (true) value is 3.92 g.
i. Absolute error = Observed value – True value
= 3.8 – 3.92
= -0.12 g

[Note: The negative sign indicates that experimental result is lower than the true value.]
Ans: i. Absolute error = -0.12 g
ii. Relative error = -3.06%

Question 36.
12.15 g of magnesium gives 20.20 g of magnesium oxide on burning. The actual mass of magnesium oxide that should be produced is 20.15 g. Calculate absolute error and relative error.
Solution:
The observed value is 20.20 g and accepted (true) value is 20.15 g.
i. Absolute error = Observed value – True value
= 20.20 – 20.15
= 0.05 g

Ans: i. Absolute error = 0.05 g
ii. Relative error = 0.25 %

Question 37.
The three identical samples of potassium chlorate are decomposed. The mass of oxygen is determined to be 3.87 g, 3.95 g and 3.89 g for the set. Calculate absolute deviation and relative deviation.
Solution:
Mean = \(\frac{3.87+3.95+3.89}{3}\) = 3.90

Ans: i. Absolute deviation in each observation = 0.03 g, 0.05 g, 0.01 g
Relative deviation = 0.8%

Question 38.
In repeated measurements in volumetric analysis, the end-points were observed as 11.15 mL, 11.17 mL, 11.11 mL and 11.17 mL. Calculate mean absolute deviation and relative deviation.
Solution:
Mean = \(\frac{11.15+11.17+11.11+11.17}{4}\) = 11.15

Ans: i. Mean absolute deviation = ±0.02 mL
ii. Relative deviation = 0.2%

Question 39.
Calculate the mass percentages of H, P and O in phosphoric acid if atomic masses are H = 1, P = 31 and O = 16.
Solution:
Atomic mass of H = 1, P = 31 and 0=16
The mass percentage of hydrogen, phosphorus, oxygen in H₃PO₄
Formula:

Calculation: Molecular formula of phosphoric acid: H₃PO₄
∴ Molar mass of H₃PO₄ = 3 × (1) + 1 × (31) + 4 × (16)
= 3 + 31 + 64
= 98 g mol^-1
Percentage of H = \(\frac {3}{98}\) × 100 = 3.06%
Percentage of P = \(\frac {31}{98}\) × 100 = 31.63%
Percentage of O = \(\frac {64}{98}\) × 100 = 65.31%
Ans: Mass percentage of H, P and O in phosphoric acid are 3.06%, 31.63% and 65.31% respectively.

Question 40.
Calculate the percentage composition of the elements in HNO₃ (H = 1, N = 14, O = 16).
Solution:
Atomic mass of H = 1, N = 14 and O = 16
The mass percentage of H, N and O in HNO₃
Formula:

Calculation: Molecular formula of HNO₃
∴ Molar mass = 1 × (1) + 1 × (14) + 3 × (16) = 1 + 14 + 48 = 63 g mol^-1
∴ Percentage of H = \(\frac {1}{63}\) × 100 = 1.59%
Percentage of N = \(\frac {14}{63}\) × 100 = 22.22%
Percentage of O = \(\frac {48}{63}\) × 100 = 76.19%
Ans: Mass percentage of H, N and O in HNO₃ are 1.59%, 22.22% and 76.19% respectively.

Question 41.
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass. Its molar mass is 98.96 g mol^-1. What is its empirical formula and molecular formula? Atomic masses of hydrogen, carbon and chlorine are 1.008,12.000 and 35.453 u, respectively
Solution:
Given: Percentage of H, C and Cl = 4.07% , 24.27% and 71.65% by mass respectively.
To find: Empirical formula and molecular formula
Calculation: Step I:
Check whether the sum of all the percentages is 100.
4.07 + 24.27 + 71.65 = 99.99 ≈ 100
Therefore, no need to consider presence of oxygen atom in the molecule.

Step II:
Conversion of mass percent to grams. Since we are having mass percent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07 g hydrogen 24.27 g carbon and 71.65 g chlorine are present.

Step III:
Convert into number of moles of each element. Divide the masses obtained above by respective atomic masses of various elements.
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}\) = 4.04 mol
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.000 \mathrm{~g}}\) = 2.0225 mol
Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}\) = 2.021 mol

Steps IV:
Divide the mole values obtained above by the smallest value among them.

Hence, the ratio of number of moles of 2 : 1 : 1 for H : C : Cl.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step V:
Write empirical formula by mentioning the numbers after writing the symbols of respective elements. CH₂Cl is thus, the empirical formula of the above compound.

Step VI:
Writing molecular formula
a. Determine empirical formula mass: Add the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass = 12.000 + 2 × 1.008 + 35.453 = 49.469 g mol^-1
b. Divide molar mass by empirical formula mass

c. Multiply empirical formula by r obtained above to get the molecular formula:
Molecular formula = r × empirical formula
∴ Molecular formula is 2 × CH₂Cl i.e. C₂H₄Cl₂.
Ans: The empirical formula of the compound is CH₂Cl and the molecular formula of the compound is C₂H₄Cl₂.
[Note: The question is modified to include the determination of molecular formula of the compound.]

Question 42.
A compound with molar mass 159 was found to contain 39.62% copper and 20.13% sulphur. Suggest molecular formula for the compound (Atomic masses: Cu = 63, S = 32 and O = 16).
Solution:
Given: Atomic mass of Cu = 63, S = 32, and O = 16
Percentage of copper and sulphur = 39.62% and 20.13% respectively.
To find: The molecular formula of the compound
Calculation: % copper + % sulphur = 39.62 + 20.13 = 59.75
This is less than 100 %. Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, % of oxygen = 100 – 59.75 = 40.25%

Hence, empirical formula is CuSO₄.
Empirical formula mass = 63 + 32 +16 × 4 = 159 g mol^-1
Hence,
Molar mass = Empirical formula mass
∴ Molecular formula = Empirical formula = CuSO₄
Ans: Molecular formula of the compound = CuSO₄

Question 43.
An inorganic compound contained 24.75% potassium and 34.75% manganese and some other common elements. Give the empirical formula of the compound. (K = 39 u, Mn = 54.9 u, O = 16 u)
Solution:
Given: Atomic mass of K = 39 u, Mn = 59 u, and O = 16 u.
Percentage of potassium and manganese = 24.75 % and 34.75% respectively
To find: The empirical formula of the given inorganic compound
Calculation: Percentage of potassium = 24.75%
Percentage of manganese = 34.75%
Total percentage = 59.50%
∴ Remaining must be that of oxygen
∴ Percentage of oxygen = 100 – 59.50 = 40.50%
Moles of K = \(\frac{\% \text { of } \mathrm{K}}{\text { Atomic mass of } \mathrm{K}}=\frac{24.75}{39}\) = 0.635 mol

Empirical formula = KMnO₄
Ans: The empirical formula of given inorganic compound is KMnO₄.

Question 44.
An organic compound contains 40.92% carbon by mass, 4.58% hydrogen and 54.50% oxygen. Determine the empirical formula of the compound.
Solution:
Given: Percentage mass of carbon = 40.92%; Percentage mass of hydrogen = 4.58%
Percentage mass of oxygen = 54.50%
To find: The empirical formula of compound
Calculation:

∴ Ratio = 1 : 1.34 : 1
Multiply by 3 to get whole number
∴ Ratio = 3 : 4.02 : 3 ≈ 3 : 4 : 3
∴ The empirical formula of the compound is C₃H₄O₃.
Ans: Empirical formula of the compound is C₃H₄O₃.

Question 45.
Define stoichiometric calculations.
Answer:
Calculations based on a balanced chemical equation are known as stoichiometric calculations.

Question 46.
Give reason: Balanced chemical equation is useful in solving problems based on chemical equations.
Answer:
Balanced chemical equation is symbolic representation of a chemical reaction. It provides the following information, which is useful in solving problems based on chemical equations:

• It indicates the number of moles of the reactants involved in a chemical reaction and the number of moles of the products formed.
• It indicates the relative masses of the reactants and products linked with a chemical change, and
• It indicates the relationship between the volume/s of the gaseous reactants and products, at STP.

Hence, balanced chemical equation is useful in solving problems based on chemical equations.

Question 47.
What are different types of stoichiometric problems? Write steps involved in solving stoichiometric problems.
Answer:
i. Generally, problems based on stoichiometry are of the following types:

• Problems based on mass-mass relationship
• Problems based on mass-volume relationship
• Problems based on volume-volume relationship.

ii. Steps involved in problems based on stoichiometric calculations:

• Write down the balanced chemical equation representing the chemical reaction.
• Write the number of moles and the relative masses or volumes of the reactants and products below the respective formulae.
• Relative masses or volumes should be calculated from the respective formula mass referring to the condition of STP.
• Apply the unitary method to calculate the unknown factors) as required by the problem.

Question 48.
Calculate the mass of carbon dioxide and water formed on complete combustion of 24 g of methane gas. (Atomic masses, C = 12 u, H = 1 u, O = 16 u)
Solution:
Mass of methane consumed in reaction = 24 g
Atomic mass: C = 12 u, H = 1 u, O = 16 u
To find: Mass of carbon dioxide and water formed
Calculation: The balanced chemical equation is,

Hence, 16 g of CH₄ on complete combustion will produce 44 g of CO₂.
∴ 24 g of CH₄ = \(\frac {24}{16}\) × 44 = 66 g of CO₂
Similarly, 16 g of CH₄ will produce 36 g of water.
∴ 24 g of CH₄ = \(\frac {24}{16}\) × 36 = 54 g of water
Ans: Mass of carbon dioxide and water formed respectively are 66 g and 54 g.

Question 49.
How much CaO will be produced by decomposition of 5 g CaCO₃?
Solution:
Given: Mass of CaCO₃ consumed in reaction = 5 g
To find: Mass of CaO produced
Calculation: Calcium carbonate decomposes according to the balanced equation,

So, 100 g of CaCO₃ produce 56 g of CaO.

Ans: Mass of CaO produced = 2.8 g

Question 50.
How many litres of oxygen at STP are required to burn completely 2.2 g of propane, C₃H₈ ?
Solution:
Given: Mass of propane used up in reaction = 2.2 g
To find: Volume of oxygen required at STP
Calculation:
The balanced chemical equation for the combustion of propane is,

(∵ 1 mol of ideal gas occupies 22.4 L of volume at STP)
Thus, 44 g of propane require 112 litres of oxygen at STP for complete combustion.
∴ 2.2 g of propane will require
\(\frac {112}{44}\) × 2.2 = 5.6 litres of O₂ at STP for complete combustion.
Ans: Volume of O₂ (at STP) required to bum 2.2 g propane = 5.6 litres

Question 51.
A piece of zinc weighing 0.635 g when treated with excess of dilute H₂SO₄ liberated 200 cm³ hydrogen at STP. Calculate the percentage purity of the zinc sample.
Solution:
Given: Mass of zinc = 0.635 g, volume of H₂ liberated = 200 cm³
To find: % purity of zinc sample
Calculation: The relevant balanced chemical equation is,
Zn + H₂SO₄ → ZnSO₄ + H₂
It indicates that 22.4 L of hydrogen at STP = 65 g of Zn.
(where, atomic mass of Zn = 65 u)
∴ 0.200 L of hydrogen at STP
= \(\frac{65 \mathrm{~g}}{22.4 \mathrm{~L}}\) × 200 L
= 0.5803 g of Zn
∴ Percentage purity of Zn = \(\frac{0.5803}{0.635}\) × 100
= 91.37 % (by using log tables)
Ans: Percentage purity of Zn sample = 91.37%

[Calculation using log table:
\(\frac{65 \times 0.200}{22.4}\)
= Antilog₁₀ [log₁₀ (65) + log₁₀ (0.200) – log₁₀ (22.4)]
= Antilog₁₀ [1.8129 + \(\overline{1} .3010\) – 1.3502]
= Antilog₁₀ [latex]\overline{1} .7637[/latex] = 0.5803
\(\frac{0.5803}{0.635} \times 100=\frac{58.03}{0.635}\)
= Antilog₁₀ [log₁₀ (58.03) – log₁₀ (0.635)]
= Antilog₁₀ [1.7636 – \(\overline{1} .8028\)]
= Antilog₁₀ [1.9608] = 91.37]

Question 52.
Explain with the help of a chemical reaction how limiting reagent works.
Answer:

• Consider the formation of nitrogen dioxide (NO₂) from nitric oxide (NO) and oxygen.
  2NO_(g) + O_2(g) → 2NO_2(g)
• Suppose initially, we take 8 moles of NO and 7 moles of O₂.
• To determine the limiting reagent, calculate the number of moles NO₂ produced from the given initial quantities of NO and O₂. The limiting reagent will yield the smaller amount of the product.
• Starting with 8 moles of NO, the number of NO₂ produced is,
  8 mol NO × \(\frac{2 \mathrm{~mol} \mathrm{NO}_{2}}{2 \mathrm{~mol} \mathrm{NO}}\) = 8 mol NO₂
• Starting with 7 moles of O₂, the number of moles NO₂ produced is,
  7 mol O₂ × \(\frac{2 \mathrm{~mol} \mathrm{NO}_{2}}{1 \mathrm{~mol} \mathrm{NO}}\) = 14 mol NO₂
• Since, 8 moles NO result in a smaller amount of NO₂, NO is the limiting reagent, and O₂ is the excess reagent, before reaction has started.

Question 53.
Urea [(NH₂)₂CO] is prepared by reacting ammonia with carbon dioxide.
2NH_3(g) + CO_2(g) → (NH₂)₂CO_(aq) + H₂O_(l)
In one process, 637.2 g of NH₃ are treated with 1142 g of CO₂.
i. Which of the two reactants is the limiting reagent?
ii. Calculate the mass of (NH₂)₂CO formed.
iii. How much excess reagent (in grams) is left at the end of the reaction?
Solution:
i. If 637.2 g of NH₃ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation.
Mass of NH₃ → Moles of NH₃ → Moles of (NH₂)₂CO

If 1142 g of CO₂ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation.
Mass of CO₂ → Moles of CO₂ → Moles of (NH₂)₂CO

Since NH₃ produces smaller amount of (NH₂)₂CO, the limiting reagent is NH₃.

iii. Starting with 18.71 moles of (NH₂)₂CO, we can determine the mass of CO₂ that reacted using the mole ratio from the balanced equation and the molar mass of CO₂ by the following relation.
Moles of (NH₂)₂CO → Moles of CO₂ → Grams of CO₂

The amount of CO₂ remaining = 1142 g – 823.4 g = 318.6 g ≈ 319 g CO₂ remaining
Ans: i. Limiting reagent = NH₃
ii. Mass of (NH₂)₂CO produced = 1124 g
iii. Mass of CO₂ remaining = 319 g

Question 54.
6 g of H₂ reacts with 32 g of O₂ to yield water. Which is the limiting reactant? Find the mass of water produced and the amount of excess reagent left.
Solution:
i. The reaction is: 2H_2(g) + O_2(g) → 2H₂O_(l)
If 6 g of H₂ react completely, calculate the number of moles of H₂O, that could be produced, by the following relation.
Mass of H₂ → Moles of H₂ → Moles of H₂O

If 32 g of O₂ react completely, calculate the number of moles of H₂O, that could be produced, by the following relation.
Mass of O₂ → Moles of O₂ → Moles of H₂O

Since O₂ produces smaller amount of H₂O, the limiting reagent is O₂.

iii. Starting with 2 moles of H₂O, we can determine the mass of H₂ that reacted using the mole ratio from the balanced equation and the molar mass of H₂ by the following relation.
Moles of H₂O → Moles of H₂ → Grams of H₂

The amount of H₂ remaining = 6g – 4g = 2g H₂ remaining
Ans: i. Limiting reagent = O₂
ii. Mass of H₂O produced = 36 g
iii. Mass of H₂ remaining = 2 g

Question 55.
Name different ways to express the concentration of a solution (or the amount of substance in a given volume of solution).
Answer:

• Mass percent or weight percent (w/w %)
• Mole fraction
• Molarity (M)
• Molality (m)

Question 56.
State the formula to obtain mass percent.
Answer:
Mass percent (w/w %) = \(\frac{\text { Mass of Solute }}{\text { Massof solution }} \times 100\)

Question 57.
Why is molality NOT affected by temperature?
Answer:

• Molality is the number of moles of solute present in 1 kg of solvent. Therefore, molality is mass dependent.
• Mass remains unaffected with temperature.

Hence, molality is not affected by temperature.

Question 58.
State TRUE or FALSE. Correct the statement if false.
i. A majority of reactions in the laboratory are carried out in in gaseous forms.
ii. Molarity is the most widely used to express the concentration of solution.
iii. Molality of a solution changes with temperature.
Answer:
i. False
A majority of reactions in the laboratory are carried out in solution forms.
ii. Tme
iii. False
Molality of a solution does not change with temperature.

Question 59.
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass percent of the solute.
Solution:
Given: Mass of substance = 2 g, mass of water = 18 g
To find: Mass % of solute

Ans: Mass percent of A = 10 % w/w

Question 60.
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Solution:
Given: Mass of solute (NaOH) = 4 g, volume of solution = 250 mL = 0.250 L
To find: Molarity of the solution

Ans: Molarity of the NaOH solution = 0.4 M

Question 61.
The density of 3 M solution of NaCl is 1.25 g mL^-1. Calculate molality of the solution.
Solution:
Given: Molarity of the solution = 3 M, density of the solution = 1.25 g mL^-1
To find: Molality of the solution
Formula:

Calculation: Molarity = 3 mol L^-1
∴ Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1 L solution = 1000 × 1.25 = 1250 g (∵ Density = 1.25 g mL^-1)
Mass of water in solution = 1250 – 175.5 = 1074.5 g = 1.0745 kg

Ans: Molality of the NaCl solution = 2.790 m

[Calculation using log table:
\(\frac{3}{1.0745}\)
= Antilog₁₀ [log₁₀ (3) – log₁₀ (1.0745)]
= Antilog₁₀ [0.4771 – 0.0315]
= Antilog₁₀ [0.4456] = 2.790]

Question 62.
Calculate the molarity of 1.8 g HNO₃ dissolved in 250 mL aqueous solution.
Solution:
Mass of HNO₃ = 1.8 g,
volume of solution = 250 mL = 0.250 L
To find: Molarity
Formulae:

Calculation: Molar mass of HNO₃ = 63 g mol^-1
Using formula (i),

Ans: Molarity of the HNO₃ solution = 0.114 M

Question 63.
What volume in mL of a 0.1 M H₂SO₄ solution will contain 0.5 moles of H₂SO₄?
Solution:
Given: Molarity of H₂SO₄ solution = 0.1 M
Moles of H₂SO₄ = 0.5 mol
To find: Volume of H₂SO₄ solution

= 5 L
= 5000 L
Ans: Volume of a 0.1 M H₂SO₄ solution = 5000 mL

Question 64.
A mixture has 18 g water and 414 g ethanol. What are the mole fractions of water and ethanol?
Solution:
Given: Mass of water = 18 g, mass of ethanol = 414 g
To find: Mole fractions of water and ethanol
Formulae:


Ans: Mole fraction of water = 0.1
Mole fraction of ethanol = 0.9

Question 65.
Explain in brief: Use of graphs in analytical chemistry.
Answer:
i. Analytical chemistry often involves deducing some relation between two or more properties of matter under study.
ii. For example, the relation between temperature and volume of a given amount of gas.
iii. A set of experimentally measured values of volume and temperature of a definite mass of a gas upon plotting on a graph paper appears as in the figure below:

iv. When the points are directly connected, a zig zag pattern results as shown below:

From the above pattern, no meaningful result can be deduced.
v. A smooth curve (or average curve) passing through these points can be drawn as shown below. This straight line is consistent with the V ∝ T .

vi. While fitting the points into a smooth curve, all the plotted points should be evenly distributed. This can be verified mathematically, by drawing a perpendicular from each point to the curve. The perpendicular represents deviation of each point from the curve. Take sum of all the perpendiculars on side of the line and sum of all the perpendiculars on another side of the line separately. If the two sums are equal (or nearly equal), the curve drawn shows the experimental points in the best possible representation

Question 66.
1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO₄ was obtained as dry precipitate. Calculate the percentage purity of the sample.
Solution:

The chemical equation representing the reaction is:

To calculate the mass of Na₂SO₄ from which 1.74 g of BaSO₄ is obtained:
233 g of BaSO₄ is produced from 142 g of Na₂SO₄.
∴ Mass of Na₂SO₄ from which 1.74 g of BaSO₄ would be obtained = \(\frac {142}{233}\) × 1.74 = 1.06 g
∴ The mass of pure Na₂SO₄ present in 1.5 g of impure sample = 1.06 g
To calculate the percentage purity of the impure sample:
1.5 g of impure sample contains 1.06 g of pure Na₂SO₄
∴ 100 g of the impure sample will contain = \(\frac{1.06}{1.5}\) × 100 = 70.67 g of pure Na₂SO₄
Ans: Percentage purity of the sample is 70.67 %.

Question 67.
Calculate the amount of lime Ca(OH)₂, required to remove hardness of 50,000 L of well water which has been found to contain 1.62 g of calcium bicarbonate per 10 L .
Solution:
Calculation of total Ca(HCO₃)₂ present:
10 L of water contains 1.62 g of Ca(HCO₃)
∴ 50,000 L of water will contain \(\frac{1.62}{10}\) × 50,000 = 8100 g of Ca(HCO₃)
Calculation of lime required:
The balanced equation for the reaction:

∴ 162 g of Ca(HCO₃) requires 74 g of lime.
Mass of lime required by 8100 g of Ca(HCO₃) = \(\frac {74}{162}\) × 8100 g = 3700 g = 3.7 kg
Ans: The amount of lime required to remove hardness of 50,000 L of well water, with 1.62 g of calcium bicarbonate per 10 L, is 3.7 kg.

Multiple Choice Questions

1. The number of significant figures in 0.0110 is …………….
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

2. For the following measurements in which the true value is 4.0 g, find the CORRECT statement.

(A) Results of both the students are neither accurate nor precise.
(B) Results of student A are both precise and accurate.
(C) Results of student A are neither precise nor accurate.
(D) Results of student B are both precise and accurate.
Answer:
(B) Results of student A are both precise and accurate.

3. 18.238 is rounded off to four significant figures as ………….
(A) 18.20
(B) 18.23
(C) 18.2360
(D) 18.24
Answer:
(D) 18.24

4. The % of H₂O in Fe(CNS)₃.3H₂O is ……………..
(A) 34
(B) 11
(C) 19
(D) 46
Answer:
(C) 19

5. The molecular mass of an organic compound is 78 g mol^-1. Its empirical formula is CH. The molecular formula is ………….
(A) C₂H₄
(B) C₂H₂
(C) C₆H₆
(D) C₄H₄
Answer:
(C) C₆H₆

6. The percentage of oxygen in NaOH is ……………
(A) 40%
(B) 60%
(C) 8%
(D) 10%
Answer:
(A) 40%

7. 1.2 g of Mg (At. mass 24) will produce MgO equal to …………….
(A) 0.05 mol
(B) 0.03 mol
(C) 0.01 mol
(D) 0.02 mol
Answer:
(A) 0.05 mol

8. …………. reagent is the reactant that reacts completely but limits further progress of the reaction.
(A) Oxidizing
(B) Reducing
(C) Limiting
(D) Excess
Answer:
(C) Limiting

9. If a solution is made up of 1 mol ethanol and 9 mol water, then mole fraction of water in the solution is ……………
(A) 0.1
(B) 0.5
(C) 0.9
(D) 1.0
Answer:
(C) 0.9

10. 0.9 glucose (C₆H₁₂O₆) is present in 1 L of solution. Find molarity.
(A) 5 M
(B) 50 M
(C) 0.005 M
(D) 0.5 M
Answer:
(C) 0.005 M

11. Molality of a solution is the ……………
(A) number of moles of solute present in 1 kg of solvent
(B) number of moles of solute present in 1 L of solution
(C) mass of solute present in 1 kg of solvent
(D) number of moles of solute present in 1 kg of solution
Answer:
(A) number of moles of solute present in 1 kg of solvent

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 7 Depreciation Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 7 Depreciation

1. Answer in One Sentence only.

Question 1.
What is the Reducing Balance Method of charging depreciation?
Answer:
A method of charging depreciation in which depreciation is charged on fixed assets at a fixed percentage on its opening balance every year is called the reducing balance method.

Question 2.
Under which method of depreciation the amount of depreciation remains constant every year?
Answer:
Under the Fixed Instalment Method of depreciation, an amount of depreciation remains constant every year.

Question 3.
Under which method of depreciation does the number of depreciation change every year?
Answer:
Under the Reducing Balance Method of depreciation amount of depreciation changes every year.

Question 4.
Which method of depreciation would you suggest for depreciating a five years lease?
Answer:
For depreciating a five years lease fixed installment Method of depreciation is suggested.

Question 5.
What is meant by the cost of assets?
Answer:
The sum total of the purchase price of a fixed asset and its installation charges are called the cost of the asset.

2. Write the word/term/phrase which can substitute each of the following statements:

Question 1.
The method of charging depreciation under which depreciation is calculated on the original cost of an asset.
Answer:
Fixed Instalment Method

Question 2.
The method of charging depreciation under which depreciation is calculated on the balance amount.
Answer:
Reducing Balance Method

Question 3.
The Latin word for reduction or decline in the value of a fixed asset due to its use.
Answer:
Depretium

Question 4.
An amount to which the balance in the depreciation account is transferred.
Answer:
Profit and Loss A/c

Question 5.
Transport charges, coolie charges, charges for electrification, etc. incurred for the erection of machinery.
Answer:
Installation Charges

3. Select the most appropriate answers from the alternatives given below and rewrite the sentences.

Question 1.
Wages paid for installation of machinery is debited to ____________ account.
(a) Profit and Loss
(b) Trading
(c) Wages
(d) Machinery
Answer:
(d) Machinery

Question 2.
Depreciation arises because of ____________
(a) wear and tear
(b) inflation
(c) loss in business
(d) profit in business
Answer:
(a) wear and tear

Question 3.
The profit on sale of an asset is debited to ____________ A/c.
(a) Profit and Loss
(b) Reserve
(c) Asset
(d) Balance sheet
Answer:
(c) Asset

Question 4.
By the amount of depreciation the value of asset ____________
(a) decreases
(b) increases
(c) becomes zero
(d) remains constant
Answer:
(a) decreases

Question 5.
In fixed instalment system the amount of depreciation is ____________ every year.
(a) constant
(b) fluctuating
(c) increased
(d) decreased
Answer:
(a) constant

Question 6.
Under ____________ method depreciation is calculated on written down value.
(a) Fixed Instalment
(b) Reducing Balance
(c) Revaluation
(d) Depletion
Answer:
(b) Reducing Balance

Question 7.
Under the ____________ system of depreciation, the amount of depreciation does not change from year to year.
(a) Fixed Instalment
(b) Reducing Balance
(c) Depletion
(d) Machine Hour Rate
Answer:
(a) Fixed Instalment

Question 8.
Depreciation = \(\frac{Cost of Asset (-) ………….}{Estimated life of Asset}\)
(a) Purchase price
(b) Scrap value
(c) Installation charges
(d) months
Answer:
(b) Scrap value

4. State whether the following statements are True or False with reasons.

Question 1.
There is no need to provide depreciation if the asset is maintained with care.
Answer:
This statement is False.
There is no relation between good maintenance of assets and depreciation working life of fixed assets decreases with the passage of time and the introduction of new technology. So even assets are maintained with care depreciation is provided.

Question 2.
Under the Reducing Balance, method depreciation is charged on the original cost.
Answer:
This statement is False.
Underwritten down value method depreciation is calculated at a certain fixed rate of percentage every on the balance of the asset which is brought forward from the previous year.

5. Complete the following sentence.

Question 1.
____________ is the major cause for Depreciation.
Answer:
Normal and Natural wear of Tear

Question 2.
Depreciation is ____________ to the business.
Answer:
Loss

Question 3.
Depreciation is necessary to make provision for ____________ of old assets.
Answer:
replacement

Question 4.
Depreciation enables the business to compute and pay the correct amount of ____________ to the Government.
Answer:
Tax

Question 5.
____________ cost concept is use for depreciation of Assets.
Answer:
Historical

Question 6.
Fixed Installment Method of depreciation is also known as ____________ cost method.
Answer:
Original

Question 7.
____________ method of depreciation is recognised by Tax/Law.
Answer:
Written Down

Question 8.
____________ account is credited when incidental expenses paid on acquiring Assets.
Answer:
Cash/Bank

Question 9.
Balance of depreciation A/c is transferred to ____________ A/c at the end of financial year.
Answer:
profit and loss

Question 10.
Depreciation is ____________ expenses.
Answer:
non cash

6. Do you agree or disagree with the following statements.

Question 1.
Depreciation is charged on current Assets only.
Answer:
Disagree

Question 2.
Depreciation is not charged on Intangible assets.
Answer:
Disagree

Question 3.
No depreciation is charged on Land.
Answer:
Agree

Question 4.
Written Down value of the asset is calculated by adding depreciation for a period of use of assets.
Answer:
Disagree

Question 5.
No depreciation is charged on assets purchased on credit.
Answer:
Disagree

7. Correct the following statement and rewrite the statement.

Question 1.
Depreciation is cash expenses.
Answer:
Depreciation is a non-cash expense.

Question 2.
The depreciation account is a Real account.
Answer:
The depreciation account is a Nominal account.

Question 3.
Wages paid on the installation of Machinery is Debited to wages A/c.
Answer:
Wages paid on the installation of Machinery is Debited to Machinery A/c.

Question 4.
Depreciation is charged only when the business is making a loss.
Answer:
Depreciation is charged whether a business is making a profit or loss.

Question 5.
Depreciation increases the value of an asset.
Answer:
Depreciation reduces the value of the Asset.

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Question 1.
Define chemistry.
Answer:
Chemistry is the study of matter, its physical and chemical properties and the physical and chemical changes it undergoes under different conditions.

Question 2.
Why is chemistry called a central science?
Answer:

1. Knowledge of chemistry is required in the studies of physics, biological sciences, applied sciences, and earth and space sciences.
2. Chemistry is involved in every aspect of day-to-day life, i.e. the air we breathe, the food we eat, the fluids we drink, our clothing, transportation and fuel supplies, etc.

Hence, chemistry is called a central science.

Question 3.
Give reason: Although chemistry has ancient roots, it has developed as a modern science.
Answer:
Technological development in sophisticated instruments have expanded knowledge of chemistry which, now, has been used in applied sciences such as medicine, dentistry, engineering, agriculture and in daily home use products. Hence, due to development and advancement in science and technology, chemistry has developed as modem science.

Question 4.
How is chemistry traditionally classified?
Answer:
Chemistry is traditionally classified into five branches:

• Organic chemistry
• Inorganic chemistry
• Physical chemistry
• Biochemistry
• Analytical chemistry

Question 5.
Explain the following terms:
i. Organic chemistry
ii. Inorganic chemistry
iii. Physical chemistry
Answer:
i. Organic chemistry: It deals with properties and reactions of compounds of carbon.
ii. Inorganic chemistry: It deals with the study of all the compounds which are not organic.
iii. Physical chemistry: It deals with the study of properties of matter, the energy changes and the theories, laws and principles that explain the transformation of matter from one form to another. It also provides basic framework for all the other branches of chemistry.

Question 6.
Distinguish between
i. Mixtures and pure substances
ii. Mixtures and compounds
Answer:
i.

ii.

Question 7.
What is the difference between element and compound?
Answer:
Elements cannot be broken down into simpler substances while compounds can be broken down into simpler substances by chemical changes.

Question 8.
Explain: States of matter
Answer:
There are three different states of matter as follows:

1. Solid: Particles are held tightly in perfect order. They have definite shape and volume.
2. Liquid: Particles are close to each other but can move around within the liquid.
3. Gas: Particles are far apart as compared to that of solid and liquid.

These three states of matter can be interconverted by changing the conditions of temperature and pressure.

Question 9.
Explain: Physical and chemical properties
Answer:
i. Physical properties: These are properties which can be measured or observed without changing the identity or the composition of the substance. e.g. Colour, odour, melting point, boiling point, density, etc.

ii. Chemical properties: These are properties in which substances undergo change in chemical composition. e.g. Coal bums in air to produce carbon dioxide, magnesium wire bums in air in the presence of oxygen to form magnesium oxide, etc.

Question 10.
How are properties of matter measured?
Answer:

• Measurement involves comparing a property of matter with some fixed standard which is reproducible and unchanging.
• Properties such as mass, length, area, volume, time, etc. are quantitative in nature and can be measured.
• A quantitative measurement is represented by a number followed by units in which it is measured.
• These units are arbitrarily chosen on the basis of universally accepted standards. e.g. Length of class room can be expressed as 10 m. Here, 10 is the number and ‘m’ is the unit ‘metre’ in which the length is measured.

Question 11.
Define: Units
Answer:
The arbitrarily decided and universally accepted standards are called units.
e.g. Metre (m), kilogram (kg).

Question 12.
What are the various systems in which units are expressed?
Answer:
Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.

Question 13.
What are SI units? Name the fundamental SI units.
Answer:
SI Units: In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI system (abbreviated from its French name).
The seven fundamental SI units are as given below:

[Note: Units for other quantities such as speed, volume, density, etc. can be derived from fundamental SI units.]

Question 14.
What is the basic unit of mass in the SI system?
Answer:
The basic unit of mass in the SI system is kilogram (kg).

Question 15.
Name the following:
i. Full form of CGS unit system
ii. Full form of FPS unit system
iii. The SI unit of length
iv. Symbol used for Candela unit
v. SI unit of electric current
vi. SI unit of electric current
Answer:
i. Centimetre Gram Second
ii. Foot Pound Second
iii. Metre (m)
iv. Cd
v. Kelvin (K)
vi. Ampere (A)

Question 16.
Give reason: The mass of a body is more fundamental property than its weight.
Answer:

• Mass is an inherent property of matter and is the measure of the quantity of matter of a body.
• The mass of a body does not vary with respect to its position.
• On the other hand, the weight of a body is a result of the mass and gravitational attraction
• Weight varies because the gravitational attraction of the earth for a body varies with the distance from the centre of the earth.

Hence, the mass of a body is more fundamental property than its weight.

Question 17.
How is gram related to the SI unit kilogram?
Answer:
The SI unit kilogram (kg) is related to gram (g) as 1 kg = 1000 g= 10³ g.
[Note: ‘Gram’ is used for weighing small quantities of chemicals in the laboratories.
Other commonly used quantity is ‘milligram’. 1 kg = 1000 g = 10⁶ mg]

Question 18.
Why are fractional units of the SI units of length often used? Give two examples of the fractional units of length. How are they related to the SI unit of length?
Answer:
i. Some properties such as the atomic radius, bond length, wavelength of electromagnetic radiation, etc. are very small and therefore, fractional units of the SI unit of length are often used to express these properties.
ii. Fractional units of length: Nanometre (nm), picometre (pm), etc.
iii. Nanometre (nm) and picometre (pm) are related to the SI unit of length (m) as follows:
1 nm = 10^-9 m, 1 pm = 10^-12 m

Question 19.
Define: Volume
Answer:
Volume is the amount of space occupied by a three-dimensional object. It does not depend on shape.

Question 20.
State the common unit used for the measurement of volume of liquids and gases.
Answer:
The common unit used for the measurement of volume of liquids and gases is litre (L).

Question 21.
How is the SI unit of volume expressed?
Answer:
The SI unit of volume is expressed as (metre)³ or m³.

Question 22.
Name some glassware that are used to measure the volume of liquids and solutions.
Answer:

• Graduated cylinder
• Burette
• Pipette

Question 23.
What is a volumetric flask used for in laboratory?
Answer:
A volumetric flask is used to prepare a known volume of a solution in laboratory.

Question 24.
What is density of a substance? How is it measured?
Answer:
Density:

• Density of a substance is its mass per unit volume. It is the characteristic property of any substance.
• It is determined in the laboratory by measuring both the mass and the volume of a sample.
• The density is calculated by dividing mass by volume.

Question 25.
How is the SI unit of density derived? State CGS unit of density.
Answer:
i. The SI unit of density is derived as follows:
Density = \(\frac{\text { SI unit mass }}{\text { SI unit volume }}\)
= \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
= kg m^-3

ii. CGS unit of density: g cm^-3
[Note: The CGS unit, g cm^-3 is equivalent to \(\frac{\mathrm{g}}{\mathrm{mL}}\) or g mL^-1.]

Question 26.
State three common scales of temperature measurement.
Answer:

1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)

Question 27.
State the temperatures in Fahrenheit scale that corresponds to 0 °C and 100 °C.
Answer:
The temperature that corresponds to 0 °C is 32 °F and the temperature that corresponds to 100 °C is 212 °F.

Question 28.
Write the expression showing the relationship between:
i. Degree Fahrenheit and Degree Celsius
ii. Kelvin and Degree Celsius
Answer:
i. The relationship between degree Fahrenheit and degree Celsius is expressed as,
°F = \(\frac {9}{5}\) (°C) + 32
ii. The relationship between Kelvin and degree Celsius is expressed as,
K = °C + 273.15

Question 29.
Convert the following degree Fahrenheit temperature to degree Celsius.
i. 50 °F ii. 10 °F
Answer:
Given: Temperature in degree Fahrenheit = 50 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 50 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
50 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(50-32) \times 5}{9}\)
= 10 °C

ii. Given: Temperature in degree Fahrenheit = 10 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 10 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
10 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(10-32) \times 5}{9}\)
= -12.2 °C
Ans: i. The temperature 50 °F corresponds to 10 °C.
ii. The temperature 10 °F corresponds to -12.2 °C.

Question 30.
What is a chemical combination?
Answer:

• The process in which the elements combine with each other to form compounds is called chemical combination.
• The process of chemical combination is governed by five basic laws which were discovered before the knowledge of molecular formulae.

Question 31.
State and explain the law of definite proportions.
Answer:
Law of definite proportions:
i. The law states that “A given compound always contains exactly the same proportion of elements by weight”.
ii. French chemist, Joseph Proust worked with two samples of cupric carbonate; one of which was naturally occurring cupric carbonate and other was synthetic sample. He found the composition of elements present in both the samples was same as shown below:

iii. Thus, irrespective of the source, a given compound always contains same elements in the same proportion.

Question 32.
State and explain the law of multiple proportions.
Answer:
Law of multiple proportions:
i. John Dalton (British scientist) proposed the law of multiple proportions in 1803.
ii. It has been observed that two or more elements may combine to form more than one compound.
iii. The law states that, “ When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.
e.g. Hydrogen and oxygen combine to form two compounds, water and hydrogen peroxide.

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.

Question 33.
Show that NO and NO₂ satisfy the law of multiple proportions.
Answer:
Nitrogen and oxygen combine to form two compounds, nitric oxide (NO) and nitrogen dioxide (NO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of nitrogen (14 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 34.
Show that carbon monoxide and carbon dioxide satisfy the law of multiple proportions.
Answer:
Chemical reaction of carbon with oxygen gives two compounds, carbon monoxide (CO) and carbon dioxide (CO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of carbon (12 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 35.
Show that SO₂ and SO₃ satisfy the law of multiple proportions.
Answer:
Chemical reaction of sulphur with oxygen gives two compounds, sulphur dioxide (SO₂) and sulphur trioxide (SO₃).

Here, the two masses of oxygen (32 g and 48 g) which combine with the fixed mass of sulphur (32 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 32 : 48 or 2 : 3.
This is in accordance with the law of multiple proportions.

Question 36.
State and explain Gay Lussac’s law of gaseous volume with two examples.
Answer:
Gay Lussac’s law:
i. Gay Lussac proposed the law of gaseous volume in 1808.
ii. Gay Lussac’s law states that, “ When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
e.g. a. Under identical conditions of temperature and pressure, 100 mL of hydrogen gas combine with 50 mL of oxygen gas to produce 100 mL of water vapour.

Thus, the simple ratio of volumes is 2 : 1 : 2.

b. Under identical conditions of temperature and pressure, 1 L of nitrogen gas combine with 3 L of hydrogen gas to produce 2 L of ammonia gas.

Thus, the simple ratio of volumes is 1 : 3 : 2.

Question 37.
Give two examples which support the Gay Lussac’s law of gaseous volume.
Answer:
i. Under identical conditions of temperature and pressure, 1 L of hydrogen gas reacts with 1 L of chlorine gas to produce 2 L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2
This is in accordance with Gay Lussac’s law.

ii. Under identical conditions of temperature and pressure, 200 mL sulphur dioxide combine with 100 mL oxygen to form 200 mL sulphur trioxide.

Thus, the ratio of volumes is 2 : 1 : 2.
This is in accordance with Gay Lussac’s law.

Question 38.
Match the following:

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

Question 39.
32 g of oxygen reacts with some carbon to make 56 grams of carbon monoxide. Find out how much mass must have been used.
Answer:
Given: Mass of oxygen (reactant) = 32 g, mass of carbon monoxide (product) = 56 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 16 g oxygen to form 28 g of carbon monoxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 16 = 32 g) of oxygen to give (2 × 28 = 56 g) carbon monoxide.
Ans: Mass of carbon used = 24 g

Question 40.
Calculate the mass of sulphur trioxide produced by burning 64 g of sulphur in excess of oxygen. (Average atomic mass: S = 32 u, O = 16 u).
Solution:
Given: Mass of sulphur (reactant) = 64 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 48 g oxygen to form 80 g of sulphur trioxide as follows:

Hence, (2 × 32 = 64 g) of sulphur will combine with (2 × 48 = 96 g) of oxygen to give (2 × 80 = 160 g) sulphur trioxide.
Ans: Mass of sulphur trioxide produced = 160 g

Question 41.
Explain Dalton’s atomic theory.
Answer:
John Dalton published “A New System of chemical philosophy” in the year of 1808. He proposed the following features, which later became famous as Dalton’s atomic theory.

• Matter consists of tiny, indivisible particles called atoms.
• All the atoms of a given elements have identical properties including mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.

Dalton’s atomic theory could explain all the laws of chemical combination.

Question 42.
Give reason: Dalton’s atomic theory explains the law of conservation of mass.
Answer:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• According to Dalton’s atomic theory, chemical reactions involve only the reorganization of atoms. Therefore, the total number of atoms in the reactants and products should be same and mass is conserved during a reaction.

Hence, Dalton’s atomic theory explains the law of conservation of mass.

Question 43.
Give reason: Dalton’s atomic theory explains the law of multiple proportion.
Answer:

• The law of multiple proportion states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
• According to Dalton’s atomic theory, compounds are formed when atoms of different elements combine in fixed ratio.

Hence, Dalton’s atomic theory explains the law of multiple proportion.

Question 44.
Define: Atomic mass unit (amu).
Answer:
Atomic mass unit or amu is defined as a mass exactly equal to one twelth of the mass of one carbon-12 atom.

Question 45.
How is relative atomic mass of an atom measured?
Answer:

• The mass of a single atom is extremely small, i.e. the mass of a hydrogen atom is 1.6736 × 10^-24 g. Hence, it is not possible to weigh a single atom.
• In the present system, mass of an atom is determined relative to the mass of an atom of carbon-12 as the standard. This was decided in 1961 by international agreement.
• The atomic mass of carbon-12 is assigned as 12.00000 atomic mass unit (amu).
• The masses of all other elements are determined relative to the mass of an atom of carbon-12 (C-12).
• The atomic masses are expressed in amu which is exactly equal to one twelth of the mass of one carbon-12 atom.
• The value of 1 amu is equal to 1.6605 × 10^-24 g.

Question 46.
What is meant by Unified Mass unit?
Answer:

• Presently, instead of amu, Unified Mass has now been accepted as the unit of atomic mass.
• It is called Dalton and its symbol is ‘u’ or ‘Da’.

Question 47.
What is average atomic mass?
Answer:
The atomic mass of an element which exists as mixture of two or more isotopes is the average of atomic masses of its isotopes. This is called average atomic mass.

Question 48.
Define: Molecular mass
Answer:
Molecular mass of a substance is the sum of average atomic masses of the atoms of the elements which constitute the molecule.
OR
Molecular mass of a substance is the mass of one molecule of that substance relative to the mass of one carbon-12 atom.

Question 49.
How is molecular mass of a substance calculated? Give example.
Answer:
Molecular mass is calculated by multiplying average atomic mass of each element by the number of its atoms and adding them together.
e.g. Molecular mass of carbon dioxide (CO₂) is calculated as follows:
Molecular mass of CO₂ = (1 × average atomic mass of C) + (2 × average atomic mass of O)
= (1 × 12.0 u) + (2 × 16.0 u)
= 44.0 u

Question 50.
Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g. What is the atomic mass of hydrogen in u?
Solution:
Given: Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g.
To find: Atomic mass of hydrogen in u
Calculation: 1.66056 × 10^-24 g = 1 u
∴ 1.6736 × 10^-24 g = x
x = \(\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g} / \mathrm{u}}\) = 1.008u
Ans: The atomic mass of hydrogen in u = 1.008 u

Question 51.
The mass of an atom of one carbon atom is 12.011 u. What is the mass of 20 atoms of the same isotope?
Solution:
Mass of l atom of carbon = 12.011 u
∴ Mass of 20 atoms of same carbon isotope = 20 × 12.011 u = 240.220 u
Ans: The mass of 20 atoms of same carbon isotope = 240.220 u

Question 52.
Calculate the average atomic mass of neon using the following data:

Solution:
Average atomic mass of Neon (Ne)

Ans: Average atomic mass of neon = 20.1707 u

Question 53.
Calculate the average atomic mass of argon from the following data:

Solution:
Average atomic mass of argon (Ar)

Ans: Average atomic mass of argon = 39.974 g mol^-1

Question 54.
Calculate the molecular mass of the following in u:
i. H₂O ii. C₆H₅Cl iii. H₂SO₄
Solution:
i. Molecular mass of H₂O = (2 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 1.0u) + (1 × 16.0 u)
= 18 u

ii. Molecular mass of C₆H₅Cl = (6 × Average atomic mass of C) + (5 × Average atomic mass of H) + (1 × Average atomic mass of Cl)
= (6 × 12.0 u) + (5 × 1.0 u) + (1 × 35.5 u)
= 112.5 u

iii. Molecular mass of H₂SO₄ = (2 × Average atomic mass of H) + (1 × Average atomic mass of S) + (4 × Average atomic mass of O)
= (2 × 1.0 u) + (1 × 32.0 u) + (1 × 16.0 u)
= 98 u
Ans: i. The molecular mass of H₂O = 18 u
ii. The molecular mass of C₆H₅Cl = 112.5 u
iii. The molecular mass of H₂SO₄ = 98 u

Question 55.
Find the mass of 1 molecule of oxygen (O₂) in amu (u) and in grams.
Solution:
Molecular mass of O₂ = 2 × 16 u
∴ Mass of 1 molecule = 32 u
∴ Mass of 1 molecule of O₂= 32 × 1.66056 × 10^-24 g = 53.1379 × 10^-24 g
Ans: Mass of 1 molecule in amu = 32 u
Mass of 1 molecule in grams = 53.1379 × 10^-24 g

Question 56.
Find the formula mass of
i. NaCl ii. Cu(NO₃)₂
Solution:
i. Formula mass of NaCl
= Average atomic mass of Na + Average atomic mass of Cl
= 23.0 u + 35.5 u = 58.5 u

ii. Formula mass of Cu(NO₃)₂
= Average atomic mass of Cu + 2 × (Average atomic mass of N + Average atomic mass of three O)
= 63.5 + 2 × [14 + (3 × 16)] = 187.5 u
Ans: i. Formula mass of NaCl = 58.5 u
ii. Formula mass of Cu(NO₃)₂ = 187.5 u

Question 57.
Find the formula mass of
i. KCl
ii. AgCl
Atomic mass of K = 39 u, Ag =108 u and Cl = 35.5 u.
Solution:
i. Formula mass of KCl
= Average atomic mass of K + Average atomic mass of Cl
= 39 u + 35.5 u = 74.5 u

ii. Formula mass of AgCl
= Average atomic mass of Ag + Average atomic mass of Cl
= 108 + 35.5 = 143.5 u
Ans: i. Formula mass of KCl = 74.5 u
ii. Formula mass of AgCl = 143.5 u

Question 58.
Calculate the number of moles and molecules of urea present in 5.6 g of urea.
Solution:
Given: Mass of urea = 5.6 g
To find: The number of moles and molecules of urea
Formulae: i. Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Mass of urea = 5.6 g
Molecular mass of urea, NH₂CONH₂
= (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O)
= (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) = 60 u
∴ Molar mass of urea = 60 g mol^-1
∴ Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{5.6 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.09333 mol

[Calculation using log table:
\(\frac{5.6}{60}\)
= Antilog₁₀ [log₁₀ (5.6) – log₁₀ (60)]
= Antilog₁₀ [0.7482 – 1.7782]
= Antilog₁₀ [latex]\overline{2} .9700[/latex]
= 0.09333]

Now,
Number of molecules of urea
= Number of moles × Avogadro’s constant
= 0.09333 mol × 6.022 × 10²³ molecules/mol
= 0.5616 × 10²³ molecules (by using log table)
= 5.616 × 10²² molecules
Ans: Number of moles of urea = 0.0933 mol
Number of molecules of urea = 5.616 × 10²² molecules

[Calculation using log table:
0.09333 × 6.022
= Antilog₁₀ [log₁₀ (0.09333) + log₁₀ (6.022)]
= Antilog₁₀ [\(\overline{2} .9698\) + 0.7797]
= Antilog₁₀ [latex]\overline{1} .7495[/latex]
= 0.5616]

Question 59.
Calculate the number of atoms in each of the following:
i. 64 u of oxygen (O)
ii. 42 g of nitrogen (N)
Solution:
i. 64 u of oxygen (O) = x atoms
Atomic mass of oxygen (O) = 16 u
∴ Mass of one oxygen atom = 16 u
∴ x = \(\frac{64 \mathrm{u}}{16 \mathrm{u}}\) = 4 atoms

ii. 42 g of nitrogen (N)
Atomic mass of nitrogen = 14 u
∴ Molar mass of nitrogen = 14 g mol^-1
Now,
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{42 \mathrm{~g}}{14 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3 mol
Now,
Number of atoms = Number of moles × Avogadro’s constant
= 3 mol × 6.022 × 10²³ atoms/mol
= 18.07 × 10²³ atoms
= 1.807 × 10²⁴ atoms
Ans: i. Number of oxygen atoms in 64 u = 4 atoms
ii. Number of nitrogen atoms in 42 g = 1.807 × 10²⁴ atoms

Question 60.
Calculate the number of atoms in each of the following.
i. 52 moles of Argon (Ar)
ii. 52 u of Helium (He)
iii. 52 g of Helium (He)
Solution:
i. 52 moles of Argon
1 mole Argon atoms = 6.022 × 10²³ atoms of Ar
∴ Number of atoms = 52 mol × 6.022 × 10²³ atoms/mol
= 313.144 × 10²³ atoms of Argon

ii. 52 g of He
Molar mass of He = mass of 1 atom of He = 4.0 u
4.0 u = 1 He
∴ 52 u = x
∴ x = 52 u × \(\frac{1 \text { atom of He }}{4.0 \mathrm{u}}\) = 13 atoms of He

iii. 52 g of He
Molar mass of He = 4.0 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{52 \mathrm{~g}}{4.0 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 13 mol
Number of atoms of He = Number of moles × Avogadro’s constant
= 13 mol × 6.022 × 10²³ atoms/mol
= 78.286 × 10²³ atoms of He
Ans. i. Number of argon atoms in 52 moles = 313.144 × 10²³ atoms of Argon
ii. Number of helium atoms in 52 u = 13 atoms of He
iii. Number of helium atoms in 52 g = 78.286 × 10²³ atoms of He

Question 61.
Calculate the number of atoms of ‘C’, ‘H’ and ‘O’ in 72.5 g of isopropanol, C₃H₇OH (molar mass 60 g mol^-1).
Solution:
Mass of isopropanol(C₃H₇OH) = 72.5 g
The number of atoms of C, H, O
Calculation: Molecular formula of isopropanol, is C₃H₇OH.
Molar mass of C₃H₇OH = 60 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{72.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 1.208 mol
∴ Moles of isopropanol = 1.21 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of isopropanol has total 12 atoms, out of which 8 atoms are of H, 3 of C and 1 of O.
∴ Number of C atoms in 72.5 g isopropanol = (3 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 2.182 × 10²⁴ atoms of carbon
∴ Number of ‘H’ atoms in 72.5 g isopropanol = (8 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 5.819 × 10²⁴ atoms of hydrogen
∴ Number of ‘O’ atoms in 72.5 g isopropanol = (1 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 7.274 × 10²³ atoms of oxygen
Ans. 72.5g of isopropanol contain 2.182 ×10²⁴ atoms of H and 7.274 × 10²³ atoms of O.

Question 62.
Calculate the number of moles and molecules of ammonia (NH₃) gas in a volume 67.2 dm³ of it measured at STP.
Solution:
Given: Volume of ammonia at STP = 67.2 dm³
To find: Number of moles and molecules of ammonia
Formulae: i. Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
ii. Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Number of moles (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Number of moles of NH₃ = \(\frac{67.2 \mathrm{dm}^{3}}{22.4 \mathrm{dm}^{3} \mathrm{~mol}^{-1}}\)
Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
3.0 mol × 6022 × 10²³ molecules mol^-1
= 18.066 × 10²³ molecules
Ans: Number of moles of ammonia = 3.0 mol
Number of molecules of ammonia = 18.066 × 10²³ molecules

Question 63.
3.40 g of ammonia at STP occupies volume of 4.48 dm³. Calculate molar mass of ammonia.
Solution:
Given: Mass of ammonia = 3.40 g
Volume at STP = 4.48 dm³
To Find: Molar mass of ammonia
Calculation: Let ‘x’ grams be the molar mass of NH₃.
Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Volume occupied by 3.40 g of NH₃ at S.T.P = 4.48 dm³
Volume occupied by ‘x’ g of NH3 at S.T.P = 22.4 dm³
∴ x = \(\frac{22.4 \times 3.40}{4.48}\) = 17.0 g mol^-1
Ans: Molar mass of ammonia is 17.0 g mol^-1.

Question 64.
Veg puffs from a particular bakery have an average mass of 27.0 g, whereas egg puffs from the same bakery have an average mass of 40 g.
i. Suppose a person buys 1 kg of veg puff from the bakery. Calculate the number of veg puffs he receives.
ii. Determine the mass of egg puffs (in kg) that will contain the same number of eggs puffs as in one kilogram of veg puffs.
Solution:
i. Mass of a veg puff = 27.0 g = 0.027 kg
∴ Number of veg puffs in 1 kg = 1 / 0.027 = 37
ii. One kilogram of veg puffs contains 37 veg puffs.
Mass of 37 egg puffs = 37 × 0.040 = 1.48 kg
Ans: i. 37 veg puffs in 1 kg of puff.
ii. Mass of 37 egg puffs is 1.48 kg

Multiple Choice Questions:

1. The branch of chemistry which deals with carbon compounds is called ……………. chemistry.
(A) organic
(B) inorganic
(C) carbon
(D) bio
Answer:
(A) organic

2. A/an is a simple combination of two or more substances in which the constituent substances retain their separate identities.
(A) compound
(B) mixture
(C) element
(D) All of these
Answer:
(B) mixture

3. Which one of the following is NOT a mixture?
(A) Paint
(B) Gasoline
(C) Liquefied Petroleum Gas (LPG)
(D) Distilled water
Answer:
(D) Distilled water

4. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with ………………
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of reciprocal proportion
Answer:
(C) law of conservation of mass

5. A sample of calcium carbonate (CaCO₃) has the following percentage composition: Ca = 40 %; C = 12 %; O = 48 %. If the law of definite proportions is true, then the weight of calcium in 4 g of a sample of calcium
carbonate from another source will be ……………..
(A) 0.016 g
(B) 0.16 g
(C) 1.6 g
(D) 16 g
Answer:
(C) 1.6 g

6. Two elements, A and B, combine to form two compounds in which ‘a’ g of A combines with ‘b₁’ and ‘b₂’g of B respectively. According to law of multiple proportion ………………
(A) b₁ = b₂
(B) b₁ and b₂ bear a simple whole number ratio
(C) a and b₁ bear a whole number ratio
(D) no relation exists between b₁ and b₂
Answer:
(B) b₁ and b₂ bear a simple whole number ratio

7. At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour. This is in accordance with ……………….
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of gaseous volumes
Answer:
(D) law of gaseous volumes

8. One mole of oxygen molecule weighs …………….
(A) 8 g
(B) 32 g
(C) 16 g
(D) 6.022 × 10²³ g
Answer:
(B) 32 g

9. The mass of 0.002 mol of glucose (C₆H₁₂O₆) is ………………
(A) 0.20 g
(B) 0.36 g
(C) 0.50 g
(D) 1.80 g
Answer:
(B) 0.36 g

10. Which of the following is CORRECT?
(A) 1 mole of oxygen atoms contains 6.0221367 × 10²³ atoms of oxygen.
(B) 1 mole of water molecules contains 6.0221367 × 10²³ molecules of water.
(C) 1 mole of sodium chloride contains 6.0221367 × 10²³ formula units of NaCl.
(D) All of these
Answer:
(D) All of these

11. 180 g of glucose (C₆H₁₂O₆) contains ……………. carbon atoms.
(A) 1.8 × 10²³
(B) 1.8 × 10²⁴
(C) 3.6 × 10²³
(D) 3.6 × 10²⁴
Answer:
(C) 3.6 × 10²³

12. The number of molecules present in 8 g of oxygen gas is …………….
(A) 6.022 × 10²³
(B) 3.011 × 10²³
(C) 12.044 × 10²³
(D) 1.505 × 10²³
Answer:
(D) 1.505 × 10²³

13. The number of molecules in 22.4 cm³of ozone gas at STP is ……………….
(A) 6.022 × 10²⁰
(B) 6.022 × 10²³
(C) 22.4 × 10²⁰
(D) 22.4 × 10²³
Answer:
(A) 6.022 × 10²⁰

14. 11.2 cm³ of hydrogen gas at STP, contains …………….. moles.
(A) 0.0005
(B) 0.01
(C) 0.029
(D) 0.5
Answer:
(A) 0.0005

15. The mass of 224 mL of hydrogen gas at STP is
(A) 0.02 g
(B) 0.224 g
(C) 2.24 g
(D) 20.0 g
Answer:
(A) 0.02 g

16. 4.4 g of an unknown gas occupies 2.24 L of volume under STP conditions. The gas may be ………………
(A) CO₂
(B) CO
(C) O₂
(D) SO₂
Answer:
(A) CO₂

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 5 Subsidiary Books Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 5 Subsidiary Books

Objective Type Questions & Answers

1. Answer the following questions in one sentence.

Question 1.
Which transactions are recorded in the Return Outward Book?
Answer:
Return of goods purchased on credit to the suppliers is only recorded in the Return Outward Book.

Question 2.
Which transactions are recorded in the Return Inward Book?
Answer:
Goods sold on credit when returned by the customers due to certain reasons are only recorded in the Return Inward Book.

Question 3.
Who is a payee in case of cheque?
Answer:
A person to whom the amount of cheque is payable or paid is called a payee.

Question 4.
What is a dishonour of cheque?
Answer:
When the bank refuses to pay the amount of cheque on its presentation to the payee for certain reasons, the cheque is said to be dishonoured.

Question 5.
What is Pass Book?
Answer:
A bank passbook is a small booklet in which the details of all ledger entries in respect of banking transactions appearing in the books of the bank are entered for the knowledge of the account holder.

Question 6.
What is a Bank Book?
Answer:
A book maintained by businessmen to record only banking transactions is called Bank Book.

Question 7.
What do you mean by crossed cheque?
Answer:
A cheque on which two parallel transverse lines are drawn on its face at the left-hand top corner with or without any word is called crossed cheque.

Question 8.
What is meant by Bank Overdraft?
Answer:
The amount withdrawn by the account holder from his current account in excess of the balance standing in that account up to the specified limit is known as ‘Bank overdraft’.

Question 9.
What do you mean by analytical petty cash book?
Answer:
A petty cash book that had many sub-columns on the payment side for recording expenses that are repetitive in nature is called an analytical/columnar petty cash book.

Question 10.
What are opening entries?
Answer:
Entries passed at the beginning of the financial year to bring the assets and liabilities in the books of account are known as opening entries.

Question 11.
What are closing entries?
Answer:
Entries passed at the end of the financial year to close the accounts of expenses and incomes are called closing entries.

Question 12.
What are transfer entries?
Answer:
Entries passed to transfer the balance of one account to another account are known as transfer entries.

Question 13.
What are adjustment entries?
Answer:
The accounting entries are passed into the books of accounts to bring certain items which do not appear in the trial balance.
e.g. depreciation, further bad debts, closing stock, incomes receivable, etc. are called adjustment entries.

Question 14.
What is a cheque?
Answer:
A cheque is a written order given by the account holder to the bank to pay the sum of money specified there into himself or to the person in whose favour the cheque is issued.

Question 15.
What is a ‘Pay-in slip’?
Answer:
Pay-in-slip is a document used by the account holder for depositing cash as well as cheque into the bank.

Question 16.
What is a ‘Debit Note’?
Answer:
A document that is prepared to give debit if less debit is given earlier or to cancel the wrong credit or extra credit given is called a debit note.

Question 17.
What is a ‘Credit Note’?
Answer:
A document that is prepared to give credit if less credit is given earlier or to cancel the wrong debit or extra debit given is called a credit note.

2. Give a word/term or phrase for each of the following statements:

Question 1.
Signing on the backside of the cheque to transfer its ownership.
Answer:
Endorsement of Cheque

Question 2.
An extract of customer’s account maintained by the bank.
Answer:
Pass Book

Question 3.
A type of cheque whose payment is done across the counter of the bank.
Answer:
Bearer Cheque

Question 4.
Bank on whom a cheque is drawn.
Answer:
Drawee

Question 5.
Document used for depositing cash or cheque into the bank.
Answer:
Pay-in-slip

Question 6.
Subsidiary book in which only credit sale of goods is recorded.
Answer:
Sales Book

Question 7.
Subsidiary book in which return of goods purchased on credit is recorded.
Answer:
Purchase Return

Question 8.
Source document on the basis of which Purchase Book is prepared.
Answer:
Inward Invoice

Question 9.
Source document on the basis of which Sales Book is prepared.
Answer:
Outward Invoice

Question 10.
Source document on the basis of which Purchase Return Book is prepared.
Answer:
Debit Note

Question 11.
Source document on the basis of which Sales Return Book is prepared.
Answer:
Credit Note

Question 12.
Accounting entries passed in the journal proper in the beginning of the financial year.
Answer:
Opening Entries

Question 13.
Accounting entries are passed in the journal proper at the end of the financial year.
Answer:
Closing Entries

Question 14.
Bank’s refusal to pay the amount of cheque to the payee.
Answer:
Dishonour of Cheque

Question 15.
Book in which small payments are recorded.
Answer:
Petty Cash Book

Question 16.
Type of bank account in which money is deposited for a fixed period of time.
Answer:
Fixed Deposit Account

Question 17.
A written order is issued by the account holder to the bank to pay a certain amount from his account.
Answer:
Cheque

Question 18.
A cheque on which two parallel transverse lines are drawn on its face at the left-hand top corner with or without certain words.
Answer:
Crossed Cheque

Question 19.
A person who endorses the cheque.
Answer:
Endorser

Question 20.
A person in whose favour the cheque is endorsed.
Answer:
Endorsee

3. State whether the following statements are True or False with reasons:

Question 1.
The total of the payment side can be greater than the total of the receipt side in the case of the bank column.
Answer:
This statement is True.
It is an overdraft on the current account.

Question 2.
A cheque received and deposited into the bank on the same day is a contra entry.
Answer:
This statement is False.
A cheque received and deposited on the same day is not a contra entry.

Question 3.
Interest on overdraft charged by the bank is recorded in the cash column.
Answer:
This statement is False.
Interest on overdraft charged by the bank is recorded in the bank column.

Question 4.
It is not necessary to record dishonour of cheques in the cash book.
Answer:
This statement is False.
It is necessary to record dishonour of cheques in the cash book.

Question 5.
Businessman generally operates savings bank account.
Answer:
This statement is False.
Businessman generally operates a current account with the bank.

4. Do you agree with the following statements.

Question 1.
Issue of Bearer cheque record on payments side of cash book in Bank Column.
Answer:
Agree

Question 2.
Subsidiary books are the books of secondary entry.
Answer:
Disagree

Question 3.
Cash Discount is recorded in the cash book with cash and Bank columns.
Answer:
Disagree

Question 4.
Credit sale of goods to Mr. Harsh ₹ 12,000 at 5% cash discount recorded on the receipts side of cash book.
Answer:
Disagree

Question 5.
All Business transactions are recorded in the cash book only.
Answer:
Disagree

Question 6.
Subsidiary books are useful for small trades only.
Answer:
Disagree

Question 7.
The big amount of payment of the organization is recorded in Petty Cash Book.
Answer:
Disagree

Question 8.
Sale of the old computer on credit recorded in the sales journal.
Answer:
Disagree

Question 9.
Cashbook cash column debit side is always greater than credit side.
Answer:
Disagree

Question 10.
When goods are returned by the buyer to the seller, Seller sends a credit note.
Answer:
Agree

5. Complete the following sentences:

Question 1.
Cash book is a ___________ book.
Answer:
Subsidiary

Question 2.
Only ___________ transactions are recorded in Simple Cash Book.
Answer:
Cash

Question 3.
Purchase book always shows ___________ balance.
Answer:
Debit

Question 4.
Sales book shows ___________ balance.
Answer:
Credit

Question 5.
Cash column of cash book can never have a ___________ balance.
Answer:
Credit

Question 6.
The source document for recording sales book is ___________
Answer:
Outward Invoice

Question 7.
___________ columns of a cash book are never balanced.
Answer:
discount

Question 8.
Credit purchase of goods are recorded in ___________
Answer:
purchase book

Question 9.
A person who draws a cheque is called ___________
Answer:
drawer

Question 10.
All expenses are recorded on the ___________ side of Cashbook.
Answer:
payment

Question 11.
When a bank refuses to make the payment of the cheque, it is said to be ___________
Answer:
dishonoured

Question 12.
The person who keeps a record of small payments is known as ___________
Answer:
petty cashier

Question 13.
In ___________ small cash payment of expenses are recorded.
Answer:
petty cash book

Question 14.
In ___________ petty cash book, on payment side there are many sub columns.
Answer:
analytical

Question 15.
___________ advance is given to the petty cashier.
Answer:
Petty

Question 16.
___________ means fixed amount of petty advance.
Answer:
Imprest

Question 17.
___________ petty cash book is not popular in the business world.
Answer:
Simple

Question 18.
Petty cash book is balanced at the end of every ___________
Answer:
month

Question 19.
Only ___________ discount is recorded in the books of accounts.
Answer:
cash

Question 20.
Cash book is book of ___________ entry.
Answer:
prime

Question 21.
___________ book is an extract of customer’s account in the ledger of bank.
Answer:
pass

Question 22.
The debit side of the passbook is called ___________ side.
Answer:
receipt

Question 23.
To draw 2 parallel lines on the face of the cheque is called ___________ of the cheque.
Answer:
crossing

Question 24.
Discount received is recorded on ___________ side of cash book.
Answer:
credit

Question 25.
The documents from which the returns to suppliers are recorded is known as ___________
Answer:
debit note

Question 26.
Sales Return book is also called as ___________
Answer:
return inward book

Question 27.
Purchase of goods on credit basis are recorded in ___________
Answer:
purchase book

Question 28.
Account in which there is restriction on withdrawals are called ___________ account.
Answer:
savings

Question 29.
___________ Entry is recorded on both the sides of cash with and bank columns.
Answer:
Contra

Question 30.
In sale journal goods sold on ___________ are recorded.
Answer:
Credit

Question 31.
Machinery purchase on credit will be recorded in ___________ subsidiary book.
Answer:
Journal Proper

Question 32.
Return of goods sold on credit by the customers will be entered in ___________ journal.
Answer:
Sales Return

Question 33.
Cash discount is recorded in ___________ book.
Answer:
Journal Proper

Question 34.
___________ Note is prepared when goods are returned to supplier.
Answer:
Debit

Question 35.
___________ is used to deposit cash on cheque in the bank.
Answer:
Pay-in-slip

Question 36.
___________ Cheque is more safer for payments.
Answer:
Cross Account Payee

6. Correct the following sentences and rewrite them same.

Question 1.
Credit sale of goods recorded in cash book receipts side.
Answer:
Cash sale of goods recorded in cash book receipts side.

Question 2.
Journal proper book is used to record cash transactions.
Answer:
Cashbook is used to record cash transactions.

Question 3.
Pay-in-slip is us to withdraw money from Bank.
Answer:
Pay-in-slip is us to deposit the money into the Bank.

Question 4.
The dividend received is a loss for the business.
Answer:
The dividend received is profit for the business.

7. Complete the following Table.

Question 1.

Answer:
62,500

Question 2.

Answer:
3,500

Question 3.

Answer:
2,05,000

Question 4.

Answer:
26,500

Question 5.

Answer:
1,25,000

Question 6.

Answer:
3,50,000

Question 7.

Answer:
4,50,000

Question 8.

Answer:
550

Question 9.

Answer:
1,990

Solved Problems

Simple/Single Column Cash Book

Question 1.
Prepare a simple cash book from the following information:
2019 January
1 Mr. Ashvin started the business with cash of ₹ 95,000.
3 Purchased furniture for office use ₹ 17,500.
4 Purchased goods worth ₹ 11,000.
7 Purchased machinery for ₹ 15,000.
10 Sold goods of ₹ 14,300 to Sandeep Traders for cash.
13 Received from Suyash ₹ 9,000.
15 Withdrew ₹ 2,500 from business for personal use.
19 Borrowed loan from Mr. Tilak ₹ 40,000.
22 Purchased goods of ₹ 14,000 at 5% trade discount.
26 Paid salary to staff ₹ 18,000.
29 Paid carriage on goods purchased ₹ 1,400.
30 Paid electricity bill ₹ 4,980.
31 Deposited into bank ₹ 7,000.
Solution:
In the books of Mr. Ashvin

Working Notes:
1. 22nd Jan.:
Net Purchase Price = Purchase Price – 5% Trade Discount
= 14,000 – 5% of 14,000
= 14,000 – 700
= ₹ 13,300

2. Balance of Cash on 31st Jan. = Total of Receipt side – Total of Payment side
= 1,58,300 – 90,680
= ₹ 67,620

Question 2.
Prepare a simple cash book of Mr. Shriram from the following details:
2019 Feb.
1 Balance of cash ₹ 65,500.
2 Brought additional capital ₹ 12,000.
4 Received from Nilesh on account ₹ 3,800.
7 Purchased goods from United Ltd. ₹ 16,000 and paid half the amount immediately.
9 Received commission from Aarti Traders ₹ 2,650.
10 Paid to Prithviraj Traders on account ₹ 17,800.
13 Purchased stationery for office use ₹ 2,460.
17 Received rent ₹ 4,900.
20 Paid ₹ 1,750 for insurance premium.
22 Sold goods for cash ₹ 13,800.
25 Purchased 10% debentures of ₹ 20,000
27 Paid telephone bill ₹ 3,000 and electricity bill ₹ 2,340.
28 Deposited into the bank all cash in excess of ₹ 11,000.
Solution:
In the books of Mr. Shriram

Working Note:
Dates: 28th February 2011:
Excess Cash Deposited = Total of Receipt Side – Total of Payment side including Closing Cash balance
= 1,02,650 – 66,350
= ₹ 36,300

Two Column Cash Book

Question 3.
Enter the following transaction in a cash book with cash and bank columns for the month of April 2019:
2019 April
1 Started business with cash ₹ 30,000.
2 Opened a current account with the bank and deposited ₹ 9,000.
5 Received ₹ 4,850 for Cash Sales and discount allowed ₹ 50.
6 Purchased goods of ₹ 3,000 @ 10% cash discount and half the amount was paid by cash and remaining by cheque.
9 Received a crossed cheque for ₹ 7,775 from Megha Associates and discount allowed ₹ 25.
12 Purchased machinery of ₹ 13,300 and paid for installation ₹ 700.
15 Received a bearer cheque of ₹ 735 from Rahul in full settlement of his account ₹ 800.
17 Deposited into the bank ₹ 2,000.
20 Transferred ₹ 8,000 from Savings Account to Current Account.
22 Received a hearer cheque from Vinod ₹ 2,230.
24 Deposited the cheque received from Vinod into the bank.
29 Bank informed that cheque received from Vinod is dishonoured.
30 Paid for printing bill book ₹ 660.
31 Withdrew from the bank for office use ₹ 1,000.
Solution:
In the books of ___________

Working Notes:
1. Transaction dated 6th April 2019.
Cash Discount = 10% of Goods Purchased
= 10% of 3,000
= ₹ 300
Net Cash Payable = 3,000 – 300 = ₹ 2,700
Amount paid in cash = \(\frac{1}{2}\) of 2,700 = ₹ 1,350
Amount paid by cheque = \(\frac{1}{2}\) of 2,700 = ₹ 1,350

2. Dated: 12th April, 2019.
Cost of Machinery = Purchase Price + Installation Charges
= 13,300 + 700
= ₹ 14,000

3. Transaction dated: 29th April, 2019.
Entry for dishonour of Vinod’s Cheque:
Earlier Entries:

Question 4.
Enter the following transaction of Hariram Bros, in a cash book with cash and bank column for the month of May 2019.
2019 May
1 Cash Balance ₹ 33,000
Bank Balance ₹ 8,000
3 Received from Prerna cash ₹ 2,750 and a bearer cheque for ₹ 645.
6 Paid to Arjun ₹ 5,300 by cheque and discount received ₹ 125.
8 Cash sales ₹ 5,645 and discount allowed ₹ 55.
10 Cheque received on 3 May 2011 deposited into the bank for collection.
12 Deposited into bank ₹ 4,000.
15 Cheque received from Prerna returned dishonoured.
18 Purchased goods from Indrajit ₹ 4,000 at an 8% trade discount and paid half the amount immediately.
20 Bank paid insurance premium under our standing instruction ₹ 1,950 and collected interest on investment ₹ 3,650.
22 Cheque issued to Arjun was dishonoured.
24 Lucky Stores directly deposited into our bank ₹ 9,500.
25 Paid salaries by cheque ₹ 5,700.
26 Withdrew by cheque ₹ 2,000 for office use and ₹ 1,500 for personal use.
31 Deposited into the bank all cash in excess of ₹ 5,000.
Solution:
In the books of Hariram Bros.

Working Notes:
1. Transaction dated 3rd May 2019:
Receipt of bearer cheque of ₹ 645 is debited to Cash A/c
Hence, Cash A/c is debited by 2,750 + 645 = ₹ 3,395

2. Transaction dated 10th May 2019:
Entry for deposit of cheque ₹ 645
Bank A/c ……………. Dr. 645
To Cash A/c 645

3. Transaction dated 15th May 2019:
Entry for dishonour of Prerna’s cheque.

When reverse entries are reconciled Cash A/c’s gets cancelled.
? Entry for dishonour of cheque is
Prerna’s A/c…………..Dr. 645
To Bank A/c 645

4. Transaction dated 18th May, 2019:
Net Price of goods purchased = Purchase Price – Trade discount
= 4,000 – 8% of 4,000
= 4,000 – 320
= ₹ 3,680
Amount Paid = \(\frac{1}{2}\) 3,680 = ₹ 1,840

5. Date: 31st May 2019:
Excess Cash deposited = Total of a debit column of cash – Total of credit column of cash including the closing balance
= 44,040 – 11,485
= ₹ 32,555

Question 5.
Enter the following transaction of Mr. Chavan in a cash book with cash and bank column for the month of April 2019.
2019 April
1 Cash Balance ₹ 56,000
Bank Overdraft ₹ 11,000
3 Purchased goods for ₹ 13,000 for cash at 2% cash discount and amount paid by cheque.
6 Received a bearer cheque for ₹ 13,250 in full settlement of ₹ 13,500 from Govind Traders.
9 Purchased 100 shares of Amar Ltd. of ₹ 100 each at ₹ 110 each and paid by cheque.
11 Sold goods of ₹ 7,000 at a 5% cash discount to Pramod and he paid half the amount immediately.
14 Deposited into bank ₹ 11,000.
17 Received a crossed cheque for ₹ 18,000 from Gajanan Traders.
20 Bank paid our telephone bill ₹ 3,230.
21 Bank charged ₹ 540 as interest on overdraft.
22 Paid by cheque to Urmila ₹ 8,000.
25 Deposited into the bank the cheque received from Govind Traders.
27 Received a bearer cheque for ₹ 3,460 for rent which was deposited into the bank.
29 Bank informed that the cheque received for rent was dishonoured.
30 Paid life insurance premium of Mr. Chavan ₹ 4,250 by cash and electricity bill ₹ 7,400 by cheque.
Solution:
In the books of Mr. Chavan

Working Notes:
1. Transaction dated 3rd April 2019:
Net payment made on cash purchases = Purchase Price – Cash discount
= 13,000 – 2% on 13,000
= 13,000 – 260
= ₹ 12,740

2. Transaction dated 9th April 2019:
Amount paid by cheque for purchase of shares = No. of shares purchased × Market value
= 100 × 110
= ₹ 11,000

3. Closing balance of Cash book
Cash balance = 72,575 – 28,500 = ₹ 44,075
Bank overdraft balance = 57,370 (Cr. column) – 45,710 (Dr. column)
Overdraft = ₹ 11,660

Petty Cash Book

Question 6.
Enter the Following transaction in a petty cash book having analysis columns for the month of October 2019.
2019 Oct.
1 Received cash from head cashier ₹ 1,350.
3 Paid packing charges ₹ 102.
6 Paid for postage ₹ 43.
9 Purchased 3 office files of ₹ 25 each.
13 Gave a tip to watchman ₹ 20.
18 Gave advance to clerk Mr. Shrikant ₹ 280.
22 Paid for printing ₹ 162.
25 Paid for advertisement ₹ 274.
27 Paid cleaning and washing charges ₹ 46.
29 Gave donation for Diwali celebration ₹ 150.
31 Purchased revenue stamps ₹ 65.
Solution:
Analytical Petty Cash Book of ___________

Question 7.
Enter the following transaction in an analytical petty cash book under the imprest system for the month of February 2019.
2019 Feb.
1 Received cheque from Head Cashier ₹ 2,700.
4 Purchased pencils for ₹ 45, inkpot ₹ 25, and papers ₹ 55.
6 Paid for repairs ₹ 190.
10 Paid taxi fare to Manager ₹ 168.
13 Paid subscription for newspaper ₹ 212.
17 Paid for refreshments to customers ₹ 175.
19 Paid to Ranjeet in settlement of his account ₹ 230.
21 Paid wages to casual labour ₹ 220.
24 Paid electricity bill ₹ 525.
26 Paid for carriage ₹ 105.
27 Purchase a wooden chair for ₹ 275.
Solution:
Analytical Petty Cash Book of ___________

Question 8.
Mr. Shrinath maintains a columnar petty cash book on the Imprest system. The Imprest amount is ₹ 850. The following information, show how his Petty Cash Book would appear for the week ended 7th September. 2019.
2019 Sept.
1 Balance in hand ₹ 125.
Postage ₹ 28.
Stationery ₹ 35.
Refreshment expenses ₹ 20.
3 Travelling and Conveyance ₹ 34.
Miscellaneous Expenses ₹ 5.
Refreshment expenses ₹ 25.
4 Repairs charges ₹ 170.
Paid for Postage ₹ 21.
5 Refreshment expenses ₹ 22.
Travelling expenses ₹ 32.
Stationery ₹ 47.
6 Refreshment expenses ₹ 19.
7 Miscellaneous Expenses ₹ 12.
Paid for Postage ₹ 7.
Repair charges ₹ 75.
Solution:
Analytical Petty Cash Book of Mr. Shrinath

Purchase Book, Sales Book, Purchase Return Book, and Sales Return Book

Question 9.
Prepare Purchase Book and Sales Book, M/s. S.K. Traders dealer in ready-made garments, Mumbai.
2019 April
1 Purchased goods from M/s. Jalna Traders, Delhi for ₹ 10,000 at 10% T.D.
4 Sold goods to M/s Kalpana Stores, Ahmedabad for ₹ 15,000 at 5% T.D.
10 Bought goods from M/s. Raj Enterprise, Patna for ₹ 5,000.
15 Sold goods to M/s Sagar Traders, Kanpur for ₹ 20,000 at 20% T.D.
25 Bought goods from M/s. Rakesh Stores, Banglore for ₹ 12,000 @ 5% T.D.
Rate of GST applicable on ready made garments are CGST @ 9%, SGST @ 9% and IGST @ 18%.
Solution:
In the books of M/s. S.K. Traders, Mumbai
Purchase Book (Analytical)

Sales Book (Analytical)

Note: In the case of inter-state sales IGST is used.

Journal Proper

Question 10.
Journalise the following transactions in the Journal Proper of Mr. Mahadev.
1 Pass opening entries: Debtors ₹ 20,000; Creditors ₹ 30,000; Computer ₹ 50,000 and Capital ₹ 40,000.
2 Pass closing entries: Purchases ₹ 10,000; Sales ₹ 25,000; Salaries ₹ 20,000 and Carriage ₹ 2,000.
3 Pass adjustment entries: Outstanding Wages ₹ 1,000 and Prepaid Insurance ₹ 500.
4 Pass transfer entries: Gross Profit ₹ 15,000; Net loss ₹ 2,000 and Drawings ₹ 4,000
5 Bought Machinery on credit from M/s. Tech ltd. for ₹ 50,000.
6 Allowed cash discount of ₹ 1,000 to Ajay.
Solution:
In the books of Mr. Mahadev
Journal Proper