Book Name |
: Physical Science and Environment |

Subject |
: Physical Science |

Class |
: 10 (Madhyamik) |

Publisher |
: Oriental Book Company Private Limited |

Chapter Name |
: Chemical Calculations |

Table of Contents

Toggle## Multiple Choice Question

**Question 1**

**Which of the following contains the least number of molecules?**

**1.12 L SO**_{2 }at STP**1 gm-mole SO**_{2}**32 g SO**_{2}**4 × 10**^{23}molecules of SO_{2}

**Answer**

(c) 32 g SO_{2}

**Explanation**

- 12 L SO
_{2}at STP ≈ 3.22 × 10^{23}molecules - 1 gm-mole SO2 = 6.022 × 10
^{23}molecules - 32 g SO
_{2}≈ 3.011 × 10^{23}molecules - 4 × 10
^{23}molecules of SO2 = 4 × 10^{23}molecules

Among the given options, “32 g SO_{2}” contains the least number of molecules.

**Question 2**

**The number of atoms present in 3.7 g of nitrogen is**

**4.45 × 10**^{24}**6.023 × 10**^{24}**1.59 × 10**^{23}**3.023 × 10**^{23}

**Answer**

(c) 1.59 × 10^{23}

**Explanation**

Number of moles of nitrogen

= 3.7 \over 14

= 0.264 mol

Number of atoms of nitrogen

= 0.264 mol × 6.022 × 10^{23}

= 1.59 × 10^{23} atoms

**Question 3**

**The volume of CO _{2} Produced at STP from 1 mole of calcium carbonate is**

**22.4 L****11.2 L****5.6 L****11 L**

**Answer**

(a) 22.4 L

**Explanation:**

CaCO_{3}(s) → CaO(s) + CO_{2}(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?**

**Cannizarro****Lavoisier****Dalton****Gay Lussay**

**Answer**

(b) Lavoisier

## Very Short Answer type Questions (VSA) :

**Question 1**

**Write the equation relating energy with mass.**

**Answer**

E = mc^{2}

(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.**

**Answer**

Molecular mass = 2 x Vapour density.

**Question 3**

**Calculate the vapour density of oxygen relative to hydrogen.**

**Answer**

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 CO _{2} at STP?**

**Answer**

6 g carbon on complete combustion produces 11.2 L CO_{2} at STP

**Explanation:**

**Balanced chemical equation**

C (12 g) + O_{2} = CO_{2} (22.4 L)

Since 12 g carbon on complete combustion produces 22.4 L CO_{2} at STP

Therefore, 6 g carbon on complete combustion produces 22.4×6\over 12 = 11.2 L CO_{2} at STP.

**Question 5**

**One mole of zinc reacts with excess H _{2}SO_{4} to produce how many moles of hydrogen?**

**Answer**

1 mole of hydrogen.

**Explanation :**

Zn + H_{2}SO_{4 }→ ZnSO_{4 }+ H_{2}

One mole of zinc reacts with excess H_{2}SO_{4} to produce one mole of hydrogen.

## Short Answer Type Questions :

**Question 1**

**State the conservation of mass and energy.**

**Answer**

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?**

**Answer**

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.**

**Answer**

**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?**

**Answer**

By Einstein’s equation, E=mc^{2}, 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.**

**Answer**

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?**

**Answer**

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?**

**Answer**

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]**

**Answer**

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

**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.

**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?**

**Answer**

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.**

**Answer**

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 CO _{2} would be required to be passed through a tank of lime water to produce 100 grams of CaCO_{3}? (Ca = 40, C = 12, 0 = 16)**

**Answer**

**Balanced Chemical Equation**

CO_{2} + CaO = CaCO_{3}

The molecular weight of CO_{2} = 12 + 16 × 2 = 44

The molecular weight of CaCO_{3} = 40 + 12 + 3 × 16 = 100 g

From the above-balanced chemical equation, it is clear that 44 g of CO_{2} is required to produce 100 grams of CaCO_{3}.

**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]**

**Answer**

**Balanced Chemical Equation:**

CaCO_{3} + 2HCl = CaCl_{2} + CO_{2} + H_{2}O

The molecular weight of CaCO_{3} = 40 + 12 + 48 = 100 g

The molecular weight of CO_{2} = 44 g

It is seen from the equation that to produce 44 g of CO_{2}, 100 g of CaCO_{3} is necessary.

Hence, to produce 22 of CO_{2}, CaCO_{3} 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]**

**Answer**

**Balanced Chemical Equation :**

CuO + H_{2} = H_{2}O + Cu

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

The molecular weight of H_{2}O = 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]**

**Answer**

The molecular weight of CaCO_{3} = 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?**

**Answer**

**Balanced chemical equation:**

2KClO_{3} = 2KCl + 3O_{2}

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 H _{2}S by reaction of Ferrous sulphide with excess dilute H2SO_{4} [Fe = 56, S = 32, H = 1, 0 = 16]**

**Answer**

**Balanced chemical equation:**

FeS + dil H_{2}SO_{4} = FeSO_{4} + H_{2}S

Molecular Mass of FeS = 56 + 32 = 88 g

Molecular Mass of H_{2}S = 1×2+32 = 34 g

It can be seen from the equation that 34 gm of H_{2}S is produced from 88 gm of FeS

∴ To prepare 1.7 gm of H_{2}S, the amount of FeS needed

= 88×1.7\over 34

= 4.4 gm

**Question 8**

**How much NH _{4}Cl is needed to prepare 10 litres of ammonia at STP?**

**Answer**

**Balanced chemical equation:**

NH_{4}Cl = NH_{3} + HCl

Molecular Mass of NH_{4}Cl = 53.5 g

The volume of NH_{3} = 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 (C3H _{8})?**

**Answer**

**Balanced chemical equation:**

C_{3}H_{8} + 5O_{2} = 3CO_{2} + 4H_{2}O

Molecular Mass of C_{3}H_{8 }= 12 × 3 + 1 × 8 = 44 g

Volume of 5O_{2} = 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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