# Class 10 FRANK Solutions Chemistry Chapter 5 - Mole Concept And Stoichiometry

Practise Chemistry concepts with Frank Solutions for ICSE Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry. Revise Gay-Lussac’s law of combining volumes and Avogadro’s law. Through our Frank Chemistry solutions, learn to calculate the relative molecular masses of chloroform, sodium acetate, ammonium sulphate and potassium chlorate.

In addition, TopperLearning’s ICSE Class 10 Chemistry Frank Solutions will also help you learn the usage of the empirical formula and the molecular formula. You can also practise these chapter concepts using our online Selina Solutions and other Chemistry learning resources.

## Mole Concept And Stoichiometry Exercise 115

### Solution 14

### Solution 1

(a) Gay-Lussac's law: It states that 'when gases react, they do so in volumes which bear a simple ratio to one another, and also to the volume of the gaseous product, provided all the volumes are measured at the same temperature and pressure'.

(b) Avogadro's law : It states that 'Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules'.

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

(a) Gram atom: "The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element".

For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram atom of hydrogen.

(b) Gram mole: "A sample of substance with its mass equal to its gram molecular mass is called one gram molecule of this substance or one gram mole".

For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 32g.

### Solution 8

### Solution 9

Molecular formula: "Molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the compound".

### Solution 10

_{6}H

_{6}is: CH

(b) The empirical formula of C

_{6}H

_{12}O

_{6}is: CH

_{2}O.

(c) The empirical formula of C

_{2}H

_{2}is: CH

(d) The empirical formula of CH

_{3}COOH is: CH

_{2}O.

### Solution 11

Three pieces of information conveyed by the formula H_{2}O is that:

It shows that there are 2hydrogen atoms and 1oxygen atoms present in H_{2}O.

The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.

It represents one molecule of compound water.

### Solution 12

### Solution 13

### Solution 15

_{2}SO

_{4}.10H

_{2}O.

(b) C

_{6}H

_{12}O

_{6}.

## Mole Concept And Stoichiometry Exercise 116

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## Mole Concept And Stoichiometry Exercise 117

### Solution 1996-1

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### Solution 1998-1

## Mole Concept And Stoichiometry Exercise 118

### Solution 1998-2

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## Mole Concept And Stoichiometry Exercise 119

### Solution 2001-1

### Solution 2001-2

According to Gay-Lussac's law:

2 vol. of C

_{2}H

_{6}requires= 7 vol. of oxygen

Vol. of C

_{2}H

_{6}= 2 vol. = 100 L

Vol. of oxygen required = 7 vol. =350 L

### Solution 2001-3

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### Solution 2001-5

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## Mole Concept And Stoichiometry Exercise 120

### Solution 2003-1

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## Mole Concept And Stoichiometry Exercise 121

### Solution 2006-1

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### Solution 2007-1

## Mole Concept And Stoichiometry Exercise 122

### Solution 2007-2

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## Mole Concept And Stoichiometry Exercise 123

### Solution 2009-1

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### Solution 2009-3

### Solution 2009-4

### Solution 2010-1

(ii) Molecular mass of NH_{4}(NO_{3}) = 80

H = 1, N = 14, O = 16

% of nitrogen

As 80 g of NH_{4} (NO_{3})
contains 28 g of nitrogen,

% of nitrogen

As80 g of NH_{4}(NO_{3})
contains48 g of oxygen

### Solution 2010-2

(i) Equation for the reaction of calcium carbonate with dilute hydrochloric acid:

(ii) Relative molecular mass of calcium carbonate=100

Mass of 4.5 moles of calcium carbonate

= No. of moles× Relative molecular mass

= 4.5×100

= 450g

(iii)

As100g of
calcium carbonate gives 22.4dm^{3} of CO_{2},_{}

(iv) Molecular mass of calcium carbonate =100

Relative molecular mass of calcium chloride =111

As 100 g of calcium carbonate gives 111g of calcium chloride,

(v) Molecular mass of HCl=36.5

Molecular mass of calcium carbonate =100

As 100 g of calcium carbonate gives (2×36.5)= 73g of HCl,

## Mole Concept And Stoichiometry Exercise 124

### Solution 2011-1

(i) Atomic mass: S = 32 and O = 16

Molecular
mass of SO_{2}=32+(2×16)=64g

As 64 g of
SO2 = 22.4dm^{3},

(ii) Gay-Lussac's law: When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure.

(iii) C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O

Molar mass of propane = 44

44 g of propane requires 5 × 22.4 litres of oxygen at STP.

### Solution 2011-2

Element |
Relative atomic mass |
% Compound |
Atomic ratio |
Simplest ratio |

H |
1 |
2.13 |
2.13/1 = 2.13 |
2 |

C |
12 |
12.67 |
12.67/12 = 1.055 |
2 |

Br |
80 |
85.11 |
85.11/80 = 1 |
1 |

Empirical
formula = CH_{2}Br

n(Empirical
formula mass of CH_{2}Br) = Molecular mass (2 × VD)

n(12 + 2 + 80) = 94 × 2

n = 2

Molecular formula = Empirical formula × 2

=
(CH_{2}Br) × 2

=
C_{2}H_{4}Br_{2}

### Solution 2011-3

(i) 10^{22} atoms of sulphur

6.022 × 10^{23}
atoms of sulphur will have mass = 32 g

(ii) 0.1 mole of carbon dioxide

1 mole of carbon dioxide will have mass = 44 g

0.1 mole of carbon dioxide will have mass = 4.4 g

### Solution 2013-1

### Solution 2013-2

### Solution 2014-1

## Mole Concept And Stoichiometry Exercise 125

### Solution 2014-2

(i) Vapour density

(ii) Ionisation

### Solution 2014-3

(i) Avogadro's law:Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

(ii)

### Solution 2015-1

6.02 × 10^{23} atoms of carbon

### Solution 2015-2

(i) 3.2 g of S has number of atoms = 6.023 × 10^{23} x 3.2 /32

=
0.6023 × 10^{23}

So, 0.6023 × 10^{23} atoms of Ca has mass=40 × 0.6023 ×10^{23}/6.023 × 10^{23} = 4g

(ii) 6 litres of hydrogen and 4 litres of chlorine when mixed result in the formation of 8 litres of HCl gas.

When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leaving behind only 2 litres of hydrogen gas.

Therefore, the volume of the residual gas will be 2 litres.

(iii)

### Solution 2015-3

(i)

(ii) 252 g of ammonium dichromate gives 22.4 dm^{3}of
N_{2}

(iii)

(iv) 252 g of ammonium dichromate gives 152 g of CrO_{3}

### Solution 2016-1

### Solution 2016-2

% of carbon = 82.76%

% of hydrogen = 100 - 82.76 = 17.24%

Element |
% Weight |
Atomic Weight |
Relative No. of moles |
Simplest Ratio |

C |
82.76 |
12 |
82.76/12 = 6.89 |
6.89/6.8 = 1 × 2 = 2 |

H |
17.24 |
1 |
17.24/1 = 17.24 |
17.24/6.89 = 2.5 × 2 = 5 |

Empirical formula = C_{2}H_{5}

Empirical formula weight = 2 × 12 + 1 × 5 = 24 + 5 = 29

Vapour density = 29

Relative molecular mass = 29 × 2 = 58

## Mole Concept And Stoichiometry Exercise 126

### Solution 2016-3

### Solution 2016-4

Mass of gas = 32 g

Volume occupied by 32g of gas = 20litres

### Solution 2016-5

2Ca(NO_{3})_{2}→ 2CaO + 4NO_{2} + O_{2}

Molecular
weight of 2Ca(NO_{3})_{2} = 2[40+2(14+48) = 32g

Molecular weight of CaO = 2(40 + 16) = 112g

328 g of
Ca(No_{3}]_{2} liberates 4 moles of NO_{2}

328 g of
Ca (NO_{3})_{2} liberates 4 × 22.4 L of
NO_{2}

### Solution 2016-6

### Solution 2017-1

(a)

(b)

(c)

### Solution 2017-2

(a)

Given:

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

Volume
of air = 1000 cm^{3}

Percentage of oxygen in air = 20%

From the given information,

According to Gay-Lussac's law,

1 vol. of propane consumes 5 vol. of oxygen.

Volume of oxygen = 1000 cm^{3} × 20% = 200
cm^{3}

Therefore,

Volume of propane burnt for every 200 cm^{3}
of oxygen,

40 cm^{3}of propane is burnt.

(b)

(i) Given:

Volume of gas at STP = 11.2 litres

Mass of gas at STP = 24 g

Gram molecular mass = ?

Mass of 22.4 L of a gas at STP is equal to its gram

molecular mass.

11.2 L of the gas at STP weighs 24 g

So,

22.4 L of the gas will weigh

Gram molecular mass = 48 g

### Solution 2017-3

(a)

Given:

Mass of hydrogen = 1 kg at 298 K and 1 atm pressure

Moles of hydrogen =?

(b) Molecular mass of CO_{2} = 12 + 2 × 16 = 44g

So, vapour density (VD) = mol. Mass/2 = 44/2 = 22

(c) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

(d) So, number of molecules of carbon dioxide in the cylinder = number of molecules of hydrogen in the cylinder = X