# Chapter 3 – Chemical Calculations

 Book Name : Physical Science and Environment Subject : Physical Science Class : 10 (Madhyamik) Publisher : Oriental Book Company Private Limited Chapter Name : Chemical Calculations

## Multiple Choice Question

Question 1

Which of the following contains the least number of molecules?

1. 1.12 L SO2 at STP
2. 1 gm-mole SO2
3. 32 g SO2
4. 4 × 1023 molecules of SO2

(c) 32 g SO2

Explanation

• 12 L SO2 at STP ≈ 3.22 × 1023 molecules
• 1 gm-mole SO2 = 6.022 × 1023 molecules
• 32 g SO2 ≈ 3.011 × 1023 molecules
• 4 × 1023 molecules of SO2 = 4 × 1023 molecules

Among the given options, “32 g SO2” contains the least number of molecules.

Question 2

The number of atoms present in 3.7 g of nitrogen is

1. 4.45 × 1024
2. 6.023 × 1024
3. 1.59 × 1023
4. 3.023 × 1023

(c) 1.59 × 1023

Explanation

Number of moles of nitrogen

= 3.7 \over 14

= 0.264 mol

Number of atoms of nitrogen

= 0.264 mol × 6.022 × 1023

= 1.59 × 1023 atoms

Question 3

The volume of CO2 Produced at STP from 1 mole of calcium carbonate is

1. 22.4 L
2. 11.2 L
3. 5.6 L
4. 11 L

(a) 22.4 L

Explanation:

CaCO3(s) → CaO(s) + CO2(g)

From this equation, you can see that 1 mole of calcium carbonate produces 1 mole of carbon dioxide gas. At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, 1 mole of CO2 gas produced from 1 mole of calcium carbonate at STP will occupy 22.4 litres.

So, the correct answer is 22.4 litres.

Question 4

Who put forward the law of conservation of mass?

1. Cannizarro
2. Lavoisier
3. Dalton
4. Gay Lussay

(b) Lavoisier

## Very Short Answer type Questions (VSA) :

Question 1

Write the equation relating energy with mass.

E = mc2

(Where E is energy, m is mass and c is the velocity of light)

Question 2

Write the relation between molecular mass and vapour density.

Molecular mass = 2 x Vapour density.

Question 3

Calculate the vapour density of oxygen relative to hydrogen.

Vapour density of oxygen = Molar mass of oxygen / Molar mass of hydrogen

= 32\over 2 = 16

∴ The vapour density of oxygen relative to hydrogen is 16.

Question 4

6 g carbon on complete combustion produces how much volume of CO2 at STP?

6 g carbon on complete combustion produces 11.2 L CO2 at STP

Explanation:

Balanced chemical equation

C (12 g) + O2 = CO2 (22.4 L)

Since 12 g carbon on complete combustion produces 22.4 L CO2 at STP

Therefore, 6 g carbon on complete combustion produces 22.4×6\over 12 = 11.2 L CO2 at STP.

Question 5

One mole of zinc reacts with excess H2SO4 to produce how many moles of hydrogen?

1 mole of hydrogen.

Explanation :

Zn + H2SO→ ZnSO+ H2

One mole of zinc reacts with excess H2SO4 to produce one mole of hydrogen.

## Short Answer Type Questions :

Question 1

State the conservation of mass and energy.

The total mass and energy of the reaction system before the reaction will be equal to the total mass and energy of the system after the reaction.

Question 2

Why the mass loss due to the release of heat in a chemical reaction is neglected?

In ordinary chemical reactions, the amount of energy released is very low. So the corresponding mass loss is very less, which cannot be detected by a very sensitive chemical balance. Hence the mass loss due to the release of heat in a chemical reaction is neglected.

Question 3

What is vapour density? Mention its relation with molecular weight.

Vapour density: Vapour density is the ratio of the weight of a certain volume of a gas or vapour to the weight of the same volume of hydrogen under similar conditions of temperature and pressure.

Relation: Molecular mass = 2 × Vapour density.

Question 4

Why the law of conservation of mass is not applicable to nuclear reactions?

By Einstein’s equation, E=mc2, we can see that mass can be converted to energy. So, the law of conservation of mass is not applicable in nuclear reactions because some mass gets converted to energy.

Question 5

The weight of 1 litre of a gas at STP is 2.324 g. Determine the gram molecular weight of the gas.

Weight of 1 litre of gas = 2.324 g

∴ weight of 22.4 litres of gas

= 2.324 × 22.4

= 52.0576 g

Hence the gram molecular weight of the gas at STP is 52.0576 g.

Question 6

What is the volume of 1 g of hydrogen at STP?

The volume of 2 g of hydrogen = 22.4 L

∴ The volume of 1 g of hydrogen

= 22.4\over2

= 11.2 L

Question 7

Does the vapour density depend on temperature?

Yes, the vapour density of a substance does depend on temperature. The vapour density of a substance is directly proportional to its molecular weight.

Question 8

Distinguish between the vapour density and the normal density of a gas.

[Hints: Normal density of a gas is the weight of 1 litre of gas at STP, it has unit gram/litre at STP, depending on temperature and pressure. Vapour density is a ratio, has no unit, and does not depend on temperature and pressure]

Vapour density and normal density are two different concepts related to the properties of gases.

1. Vapour Density:

Vapour density is the density of a gas or vapour in relation to the density of a reference gas, which is usually dry air or hydrogen, at the same temperature and pressure.

1. Normal Density:

Normal density is the density of a gas at standard temperature and pressure (STP), which is defined as 0°C and 1 atmosphere (atm) of pressure.

Question 9

What is the number of molecules of sulphur (S) present in 24 g of solid sulphur?

The atomic mass of sulphur (S) = 32 g

Molar mass of S​​​​​​8 = 32×8 = 256 g

256 g of S​​​​​​8 = 1 mole

∴ 24 g of S​​​​​​8 = 24\over256 = 0.09375 mole

## Long Answer (LA) type questions :

Question 1

Establish the relation between molecular weight and vapour density.

From the definition of vapour density, we know,

Vapour density (VD) = (Mass of (v) volume of gas)/Mass of (v) volume of hydrogen

(At the same temperature and pressure)

According to Avogadro’s hypothesis at the same temperature and pressure equal volume of any gas contains an equal number of molecules.

∴ Vapour density (VD) = n-molecules of gas/n-molecules of hydrogen

= Weight of 1 – molecule of gas/Weight of 1-molecules of hydrogen

= molecular weight of gas (M)/molecular weight of hydrogen

VD = M/2

or, M = 2D

Question 2

How many grams of CO2 would be required to be passed through a tank of lime water to produce 100 grams of CaCO3? (Ca = 40, C = 12, 0 = 16)

Balanced Chemical Equation

CO2 + CaO = CaCO3

The molecular weight of CO2 = 12 + 16 × 2 = 44

The molecular weight of CaCO3 = 40 + 12 + 3 × 16 = 100 g

From the above-balanced chemical equation, it is clear that 44 g of CO2 is required to produce 100 grams of CaCO3.

Question 3

How many grams of calcium carbonate will be needed for the preparation of 22 gm of carbon dioxide by reaction with dilute hydrochloric acid and calcium carbonate?

[Ca = 40, C = 12, 0 =16]

Balanced Chemical Equation:

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

The molecular weight of CaCO3 = 40 + 12 + 48 = 100 g

The molecular weight of CO2 = 44 g

It is seen from the equation that to produce 44 g of CO2, 100 g of CaCO3 is necessary.

Hence, to produce 22 of CO2, CaCO3 is necessary

= 100×22\over 44 = 50 g

Question 4

Find out the amount of water that will be formed when 3.975 gm. of pure cupric oxide is reduced by pure hydrogen. Also, find the loss of weight of the cupric oxide. [Cu = 63.5, H = 1, 0 = 16]

Balanced Chemical Equation :

CuO + H2 = H2O + Cu

The molecular weight of CuO = 63.5 + 16 = 79.5 g

The molecular weight of H2O = 1×2 + 16 = 18 g

Clearly, 79.5 g of CuO produces 18 g of water

Hence, 3.975 g of CuO produces 18×3.975\over 79.5

= 0.9 g of water

Now, from the above reaction

79.5 g of CuO is reduced to 63.5 g of Cu

∴ 3.975 g of CuO is reduced to 63×3.975\over 79.5

= 3.175 g of Cu

∴ Loss of wt = 3.975 – 3.175 = 0.8 g.

Question 5

Calculate the percentage of composition of the constituents of calcium carbonate. [Ca = 40, C = 12, O = 16]

The molecular weight of CaCO3 = 40+12+16×3 = 100 g

We know, % Composition = {mass\ of\ element\over moleclar\ mass}×100

% Composition of Ca = {40\over 100}×100 = 40

% Composition of C = {12\over 100}×100 = 12

% Composition of Oxygen = {48\over 100}×100 = 48

Question 6

How many moles of potassium chlorate are required to produce 4.8 gm of oxygen?

Balanced chemical equation:

2KClO3 = 2KCl + 3O2

2 moles              3 moles

32 gm of oxygen = 1 mole

∴ 4.8 gm of oxygen = 1×4.8\over32 = 0.15 mole

It is seen from the equation that 3 moles of oxygen are produced from 2 moles of potassium chlorate.

∴ 0.15 moles of oxygen is produced from 2×0.15\over3 = 0.10 moles of potassium chlorate.

Question 7

What amount of ferrous sulphide will be required to prepare 1.7 gm of H2S by reaction of Ferrous sulphide with excess dilute H2SO4 [Fe = 56, S = 32, H = 1, 0 = 16]

Balanced chemical equation:

FeS + dil H2SO4 = FeSO4 + H2S

Molecular Mass of FeS = 56 + 32 = 88 g

Molecular Mass of H2S = 1×2+32 = 34 g

It can be seen from the equation that 34 gm of H2S is produced from 88 gm of FeS

∴ To prepare 1.7 gm of H2S, the amount of FeS needed

= 88×1.7\over 34

= 4.4 gm

Question 8

How much NH4Cl is needed to prepare 10 litres of ammonia at STP?

Balanced chemical equation:

NH4Cl = NH3 + HCl

Molecular Mass of NH4Cl = 53.5 g

The volume of NH3 = 22.4 L

Clearly, 22.4 L of ammonia is produced from 53.5 g of ammonium chloride.

∴ 10 L of ammonia can be produced from (53.5×10)/22.4 = 23.884 g of ammonium chloride.

Question 9

How many litres of oxygen is needed to burn completely 2.2 g propone (C3H8)?

Balanced chemical equation:

C3H8 + 5O2 = 3CO2 + 4H2O

Molecular Mass of C3H8 = 12 × 3 + 1 × 8 = 44 g

Volume of 5O2 = 5 × 22.4 = 112 L

Clearly, 44 g of propane burns completely in 112 L of oxygen

∴ To burn 2.2g of propane completely, the amount of oxygen needed is (112×2.2)/44 = 5.6 L

Also checkout: Class 10 Physical Science Formula List

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