Mole Concept And Stoichiometry ICSE Class-10 Concise Selina Solutions

Mole Concept And Stoichiometry ICSE Class-10 Concise Selina Solutions Chapter-5. We Provide Step by  Step Answer of  Exercise – 5 (A) ,  Numericals – 5 (A),      Exercise – 5 (B),     Numerical – 5(B),  Exercise-5(C),     Exercise – 5(D),   Miscellaneous Exercise  and   Previous Year  Questions of Exercise-5 Mole Concept And Stoichiometry ICSE Class-10. Visit official Website CISCE  for detail information about ICSE Board Class-10.

Board	ICSE
Publications	Selina Publishers PVT LTD
Subject	Concise Chemistry
Class	10th
writer	Dr SP Singh
Chapter-5	Mole Concept And Stoichiometry
Topics	Solutions of Exe – 5 (A) ,  Num – 5 (A),      Exe – 5 (B),     Num – 5(B),  Exe-5(C),     Exe – 5(D),   Miscellaneous Exe  and   Previous Year Questions
Edition	2021-2022

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How to solve Mole Concept And Stoichiometry

Note:– Before viewing Solution of Mole Concept And Stoichiometry of Concise Selina Chapter 5 . Read the Mole Concept And Stoichiometry  Chapter 5 Carefully to understand the concept in better way .Therefor After reading the Mole Concept And Stoichiometry Chapter 5 solve all example of your text book. For more practice on Mole Concept And Stoichiometry try to solve exercise of other publications such as Goyal Brothers Prakshan / new Simplified Dr Dalal Chemistry.

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Exercise – 5 (A) Mole Concept And Stoichiometry

Page-75

Question 1

State :

(a) Gay-Lussac’s Law of combining volumes.

(b) Avogadro’s law

Answer 1

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

(b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Question 2

(a) What do you mean by stoichiometry?

(b) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.

(c) Differentiate between N₂ and 2N.

Answer 2

(a) stoichiometry measures quantitative relationships and is used to determine the amount of products/reactants that are produced/needed in a given reaction.

(b) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.

(c) N₂ means 1 molecule of nitrogen and 2N means two atoms of nitrogen.

N₂ can exist independently but 2N cannot exist independently.

Question 3

Explain Why?

(a) “The number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium at the same temperature and pressure.”

(b) “When stating the volume of a gas, the pressure and temperature should also be given.”

(c) Inflating a balloon seems to violate Boyle’s law.

Answer 3

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Now volume of hydrogen gas =volume of helium gas

n molecules of hydrogen =n molecules of helium gas

nH₂=nHe

1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium

Therefore 2H=He

Therefore atoms in hydrogen is double the atoms of helium.

(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.

(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.

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Numerical Problems:- 5 (A) Mole Concept And Stoichiometry Selina Solutions

Question 4

(a) Calculate the volume of oxygen at S.T.P required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.

2CO + O₂   2CO₂

(b) 200 cm³ of hydrogen and 150 cm³ of oxygen are mixed and ignited, as per the following reaction,

2H₂ + O₂   2H₂O

What volume of oxygen remains unreacted?

Answer 4

(a) 2CO + O₂   2CO₂

     2 V    1 V     2 V

2 V of CO requires = 1V of O₂

so, 100 litres of CO requires = 50 litres of O₂

(b) 2H₂ + O₂     2H₂O

       2 V         1V            2V

 From the equation, 2V of hydrogen reacts with 1V of oxygen

so 200cm³ of Hydrogen reacts with = 200/2= 100 cm³

Hence, the unreacted oxygen is 150 – 100 = 50cm³ of oxygen.

Question 5

24 cc Marsh gas (CH₄) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculation.

Answer 5

This experiment supports Gay lussac’s law of combining volumes.

Since the unchanged or remaining O₂ is 58 cc so, used oxygen 106 – 58 = 48cc

According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.

CH₄+2O₂    CO₂ + 2H₂O

1 V2 V

24 cc48 cc

i.e. methane and oxygen react in a 1:2 ratio.

Question 6

What volume of oxygen would be required to burn completely 400 ml of acetylene [C₂H₂]? Also calculate the volume of carbon dioxide formed.

2C₂H₂ + 5O₂   4CO₂ + 2H₂O (l)

Answer 6

2C₂H₂ + 5O₂    4CO₂ + 2H₂O (l)

2 V        5 V           4 V

From equation, 2 V of C₂H₂ requires = 5 V of O₂

So, for 400ml C₂H₂ , O₂ required = 400  5/2 =1000 ml

Similarly, 2 V of C₂H₂ gives = 4 V of CO₂

So, 400ml of C₂H₂ gives CO₂ = 400   4/2 = 800ml

Question 7

112 cm³ of H₂S(g) is mixed with 120 cm³ of Cl₂(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate (i) the volume of gaseous product formed (ii) composition of the resulting mixture.

Answer 7

Balanced chemical equation:

(i) At STP, 1 mole gas occupies 22.4 L.

As 1 mole H₂S gas produces 2 moles HCl gas,

22.4 L H₂S gas produces 22.4 × 2 = 44.8 L HCl gas.

Hence, 112 cm³ H₂S gas will produce 112 × 2 = 224 cm³ HCl gas.

(ii) 1 mole H₂S gas consumes 1 mole Cl₂ gas.

This means 22.4 L H₂S gas consumes 22.4 L Cl₂ gas at STP.

Hence, 112 cm³ H₂S gas consumes 112 cm³ Cl₂ gas.

120 cm³ – 112 cm³ = 8 cm³ Cl₂ gas remains unreacted.

Thus, the composition of the resulting mixture is 224 cm³HCl gas + 8 cm³ Cl₂ gas.

Question 8

1250cc of oxygen was burnt with 300cc of ethane [C₂H₆]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed;

Answer 8

From the equation, 2V of ethane reacts with 7V oxygen.

So, 300 cc of ethane reacts with   

Hence, unused O₂ = 1250 – 1050 = 200 cc

From 2V of ethane, 4V of CO₂ is produced.

So, 300 cc of ethane will produce   

Question 9

What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C₂H₄] at 273^o C and 380 mm of Hg pressure?

C₂H₄+3O₂    2CO₂ + 2H₂O

Answer 9

  C₂H₄+3O₂ 2CO₂ + 2H₂O

  1V      3V

11litre   33 litre

Question 10

Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at S.T.P.

CH₄ + 2Cl₂  CH₂Cl₂+2HCl

Answer 10

CH₄ + 2Cl₂    CH₂Cl₂ +2HCl

1 V      2 V            1 V       2 V

From equation, 1V of CH₄ gives = 2 V HCl

so, 40 ml of methane gives = 80 ml HCl

For 1V of methane = 2V of Cl₂ required

So, for 40ml of methane = 40  2 = 80 ml of Cl₂

Question 11

What volume of propane is burnt for every 500 cm³ of air used in the reaction under the same conditions? (assuming oxygen is 1/5^th of air)

C₃H₈ + 5O₂    3CO₂ + 4H₂O

Answer 11

C₃H₈ + 5O₂  3CO₂ + 4H₂O

1 V        5 V        3 V

From equation, 5 V of O₂ required = 1V of propane

_{so, 100 cm³ of O2 will require = 20 cm³ of propane}

Question 12

450 cm³ of nitrogen monoxide and 200 cm³ of oxygen are mixed together and ignited. Caclulate the composition of resulting mixture.

2NO + O₂ 2NO₂

Answer 12

2NO + O₂    2NO₂

2 V       1 V         2 V

From equation, 1V of O₂ reacts with = 2 V of NO

200cm³ oxygen will react with = 200 2 =400 cm³ NO

Hence, remaining NO is 450 – 400 = 50 cm³

NO₂ produced = 400cm³ because 1V oxygen gives 2 V NO₂

Total mixture = 400 + 50 = 450 cm³

Question 13

If 6 liters of hydrogen and 4 liters of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.

Answer 13

(i) 6 litres of hydrogen and 4 litres of chlorine when mixed, results in the formation of 8 litres of HCl gas.

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

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

Question 14

Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.

4NH₃ + 5O₂   4NO + 6H₂O

If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

Solution 14

4NH₃ + 5O₂  4NO + 6H₂O

4 V        5 V         4 V

9 litres of reactants gives 4 litres of NO

So, 27 litres of reactants will give = 27  4/9 = 12 litres of NO

Question 15

A mixture of hydrogen and chlorine occupying 36 cm³ was exploded. On shaking it with water, 4cm³ of hydrogen was left behind. Find the composition of the mixture.

Answer 15

H₂ + Cl₂     2HCl

1V    1V            2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm³ of H₂ remained behind so the mixture had composition: 16 cm³ hydrogen and 16 cm³ chlorine.

Therefore Resulting mixture is H₂ =4cm³,HCl=32cm³

Question 16

What volume of air (containing 20% O₂ by volume) will be required to burn completely 10 cm³ each of methane and acetylene?

CH₄ + 2O₂   CO₂ + 2H₂O

2C₂H₂ + 5O₂   4CO₂ + 2H₂O

Answer 16

CH₄ + 2O₂     CO₂ + 2H₂O

1 V       2 V          1 V

2C₂H₂ + 5O₂     4CO₂ + 2H₂O

2 V          5 V            4 V

From the equations, we can see that

1V CH₄ requires oxygen = 2 V O₂

So, 10cm³ CH₄ will require =20 cm³ O₂

Similarly 2 V C₂H₂ requires = 5 V O₂

So, 10 cm³ C₂H₂ will require = 25 cm³ O₂

Now, 20 V O₂ will be present in 100 V air and 25 V O₂ will be present in 125 V air ,so the volume of air required is 225cm³

Question 17

LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.

C₃H₈ + 5O₂   3CO₂ + 4H₂O

2C₄H₁₀ + 13O₂    8CO₂ + 10H₂O

Answer 17

C₃H₈ + 5O₂    3CO₂ + 4H₂O

2C₄H₁₀ + 13O₂     8CO₂ + 10H₂O

60 ml of propane (C₃H₈) gives 3 60 = 180 ml CO₂

40 ml of butane (C₄H₁₀) gives = 8  40/2 = 160 ml of CO₂

Total carbon dioxide produced = 340 ml

So, when 10 litres of the mixture is burnt = 34 litres of CO₂ is produced.

Question 18

200 cm³ of CO₂ is collected at S.T.P when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at S.T.P. in original mixture.

2C₂H₂(g) + 5O₂(g)    4CO₂(g)+ 2H₂O(g)

Answer 18

2C₂H₂(g) + 5O₂(g) 4CO₂(g)+ 2H₂O(g)

4 V CO₂ is collected with 2 V C₂H₂

So, 200cm³ CO₂ will be collected with = 100cm³ C₂H₂

Similarly, 4V of CO₂ is produced by 5 V of O₂

So, 200cm³ CO² will be produced by = 250 ml of O₂

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Page-76

Question 19

You have collected (a) 2 litres of CO₂ (b) 3 litres of chlorine (c) 5 litres of hydrogen (d) 4 litres of nitrogen and (e) 1 litres of SO₂, under similar conditions of temperature and pressure. Which gas sample will have :

(a) the greatest number of molecules, and

(b) The least number of molecules?

Justify your answers.

Answer 19

According to Avogadro’s law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules. So,

(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.

(b) 1 litre of SO₂ contains the least number of molecules since it has the smallest volume.

Question 20

The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.

Gas	Volume (in litres)	Number of molecules
Chlorine Nitrogen Ammonia Sulphur dioxide	10 20 20 5	x

Answer 20

Gas	Volume (in litres)	Number of molecules
Chlorine	10	x/2
Nitrogen	20	x
Ammonia	20	X
Sulphur dioxide	5	x/4

Question 21

(i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B?

The gases A and B are under the same conditions of temperature and pressure.

(ii) Name the law on which the above problem is based.

Answer 21

(i) According to Avogadro’s law, under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

As 150 cc of gas A contains X molecules, 150 cc of gas B also contains X molecules.

So, 75 cc of B will contain X/2 molecules.

(ii) The problem is based on Avogadro’s law.

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Exercise – 5(B) Mole Concept And Stoichiometry Concise Selina ICSE Class-10

(Relative Atomic Mass ,Relative Molecular Mass and Mole Concept)

Page-83

Question 1

(a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.

(b) What is the value of Avogadro’s number?

(c) What is the value of molar volume of a gas at S.T.P?

Answer 1

(a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.

(b)The value of Avogadro’s number is 6.023 10²³

(c) The molar volume of a gas at STP is 22.4 dm³at STP

Question 2

Define or explain the terms

(a) Vapour density

(b) Molar volume

(c) Relative atomic mass

(d) Relative molecular mass

(e) Avogadro’s number

(f) Gram atom

(g) Mole

Answer 2

(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm³.

(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

(d) The relative molecular mass of an compound is the number that represents how many times one molecular of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

(e)  Avogadro’s number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x10²³ atoms.

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

(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

Question 3

(a) What are the main applications of Avogadro’s Law?

(b) How dose Avogadro’s Law explain Gay-Lussac’s Law of combining volumes?

Answer 3

(a) Applications of Avogadro’s Law :

(1) It explains Gay-Lussac’s law.

(2) It determines atomicity of the gases.

(3) It determines the molecular formula of a gas.

(4) It determines the relation between molecular mass and vapour density.

(5) It gives the relationship between gram molecular mass and gram molecular volume.

(b) According to Avogadro’s law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac’s Law says.

H₂ + Cl₂ ? 2HCl

1V 1V 2V(By Gay-Lussacs law)

n molecules n molecules 2n molecules (By Avogadros law)

Numericals of  – 5 (B)  Exe 

Question 4

Calculate the relative molecular masses of:

(a) Ammonium chloroplatinate, (NH₄)₂ PtCl₆

(b) Potassium chlorate

(c) CuSO₄. 5H₂O

(d) (NH₄)₂SO₄

(e) CH₃COONa

(f) CHCl₃

(g) (NH₄)₂ Cr₂O₇

Answer 4

(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444

(b) KClO₃ = (K)39 + (Cl)35.5 + (3O)48 = 122.5

(c) (Cu)63.5 + (S)32 + (4O)64 + (5H₂O)5 x 18 = 249.5

(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132

(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82

(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5

(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252

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Page-84

Question 5

Find the

(a) number of molecules in 73 g of HCl,

(b) weight of 0.5 mole of O₂,

(c) number of molecules in 1.8 g of H₂O

(d) number of moles in 10 g of CaCO₃

(e) Weight of 0.2 mole of H₂ gas,

(f) Number of molecules in 3.2 g of SO_2.

Answer 5

(a) No. of molecules in 73 g HCl = 6.023 x10²³ x 73/36.5(mol.

mass of HCl)

= 12.04 x 10²³

(b) Weight of 0.5 mole of O₂ is = 32(mol. Mass of O₂) x 0.5=16 g

(c) No. of molecules in 1.8 g H₂O = 6.023 x 10²³ x 1.8/18

= 6.023 x 10²²

(d) No. of moles in 10g of CaCO₃ = 10/100(mol. Mass CaCO₃)

= 0.1 mole

(e) Weight of 0.2 mole H₂ gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO₂ = 6.023 x 10²³ x 3.2/64

= 3.023 x 10²²

Question 6

Which of the following would weigh most?

(a) 1 mole of H₂O

(b) 1 mole of CO₂

(c) 1 mole of NH₃

(d) 1 mole of CO

Answer 6

Molecular mass of H₂O is 18, CO₂ is 44, NH₃ is 17 and CO is 28

So, the weight of 1 mole of CO₂ is more than the other three.

Question 7

Which of the following contains maximum number of molecules?

(a) 4 g of O₂

(b) 4 g of NH₃

(c) 4 g of CO₂

(d) 4 g of SO₂

Answer 7

4g of NH₃ having minimum molecular mass contain maximum molecules.

Question 8

Calculate the number of

(a) Particles in 0.1 mole of any substance.

(b) Hydrogen atoms in 0.1 mole of H₂SO₄.

(c) Molecules in one Kg of calcium chloride.

Answer 8

(a) No. of particles in s1 mole = 6.023 x 10²³

So, particles in 0.1 mole = 6.023 x 10 ²³ x 0.1 = 6.023 x 10²²

(b)1 mole of H₂SO₄ contains =2 x 6.023 x 10²³

So, 0.1 mole of H₂SO₄ contains =2 x 6.023 x 10²³ x0.1

= 1.2×10²³ atoms of hydrogen

(c) 111g CaCl₂ contains = 6.023 x 10²³ molecules

So, 1000 g contains = 5.42 x 10²⁴ molecules

Question 9

How many grams of

(a) Al are present in 0.2 mole of it?

(b) HCl are present in 0.1 mole of it?

(c) H₂O are present in 0.2 mole of it?

(d) CO₂ is present in 0.1 mole of it?

Answer 9

(a) 1 mole of aluminium has mass = 27 g

So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g

(b) 0.1 mole of HCl has mass = 0.1 x 36.5(mass of 1 mole)

= 3.65 g

(c) 0.2 mole of H₂O has mass = 0.2 x 18 = 3.6 g

(d) 0.1 mole of CO₂ has mass = 0.1 x 44 = 4.4 g

Question 10

(a) The mass of 5.6 litres of a certain gas at S.T.P. is 12 g. What is the relative molecular mass or molar mass of the gas? (2017)

(b) Calculate the volume occupied at S.T.P. by 2 moles of SO₂.

Answer 10

(a) 5.6 litres of gas at STP has mass = 12 g

So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6

= 48g(molar mass)

(b)1 mole of SO₂ has volume = 22.4 litres

So, 2 moles will have = 22.4 x 2 = 44.8 litre

Question 11

Calculate the number of moles of

(a) CO₂ which contain 8.00 g of O₂

(b) Methane in 0.80 g of methane.

Answer 11

(a) 1 mole of CO₂ contains O₂ = 32g

So, CO₂ having 8 gm of O₂ has no. of moles = 8/32 = 0.25 moles

(b) 16 g of methane has no. of moles = 1

So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

Question 12

Calculate the actual mass of

(a) An atom of oxygen

(b) an atom of hydrogen

(c) a molecule of NH₃

(d) the atom of silver

(e) the molecule of oxygen

(f) 0.25 gram atom of calcium

Answer 12

(a) 6.023 x 10 ²³ atoms of oxygen has mass = 16 g

So, 1 atom has mass = 16/6.023 x 10²³ = 2.656 x 10^-23 g

(b) 1 atom of Hydrogen has mass = 1/6.023 x 10²³ = 1.666 x 10^-24

(c) 1 molecule of NH₃ has mass = 17/6.023 x10²³ = 2.82 x 10^-23 g

(d) 1 atom of silver has mass = 108/6.023 x 10²³ =1.701 x 10^-22

(e) 1 molecule of O₂ has mass = 32/6.023 x 10²³ = 5.314 x 10^-23 g

(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

Question 13

Calculate the mass of 0.1 mole of each of the following

(a) CaCO₃

(b) Na₂SO₄.10H₂O

(c) CaCl₂

(d) Mg

(Ca = 40, Na=23, Mg =24, S=32, C = 12, Cl = 35.5, O=16, H=1)

Answer 13

(a) 0.1 mole of CaCO₃ has mass =100(molar mass) x 0.1=10 g

(b) 0.1 mole of Na₂SO₄.10H₂O has mass = 322 x 0.1 = 32.2 g

(c) 0.1 mole of CaCl₂ has mass = 111 x 0.1 = 11.1g

(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

Question 14

Calculate the number of

(a) oxygen atoms in 0.10 mole of Na₂CO₃.10H₂O. (2017)

(b) gram atoms in 4.6 gram of sodium (2019)

(c) Mole in 12 g of oxygen

Answer 14

(a) 1 molecule of Na₂CO₃.10H₂O contains oxygen atoms = 13

So, 6.023 x10²³ molecules (1mole) has atoms=13 x 6.023 x 10²³

So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10²³ =7.8×10²³

(b) Given Na = 4.6 gm

At. mass = 23

No. of gram atoms of Na=    Mass of Na

At. mass of Na

= 4.6

23

= 0.2 gms

(c) 32 g of oxygen gas = 1 mole

1 gram of oxygen gas = 1/32 mole

Given that 12 g of oxygen gas

No: of moles = given mass / molar mass

= 12 / 32  = 0.375 mole

Question 15

What mass of Ca will contain the same number of atoms as are present in 3.2 g of S? (2015)

Answer 15

3.2 g of S has number of atoms = 6.023 x10²³ x 3.2 /32

= 0.6023 x 10²³

So, 0.6023 x 10²³ atoms of Ca has mass=40 x0.6023×10²³/6.023

x10²³

= 4g

Question 16

Calculate the number of atoms in each of the following:

(a) 52 moles of He

(b) 52 amu of He

(c) 52 g of He

Answer 16

(a) No. of atoms = 52 x 6.023 x10²³ = 3.131 x 10²⁵

(b) 4 amu = 1 atom of He

so, 52 amu = 13 atoms of He

(c) 4 g of He has atoms = 6.023 x10²³

So, 52 g will have = 6.023 x 10²³ x 52/4 = 7.828 x10²⁴ atoms

Question 17

Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

Answer 17

Molecular mass of Na₂CO₃ = 106 g

106 g has 2 x 6.023 x10²³ atoms of Na

So, 5.3g will have = 2 x 6.023 x10²³x 5.3/106=6.022 x10²² atoms

Number of atoms of C = 6.023 x10²³ x 5.3/106 = 3.01 x 10²² atoms

And atoms of O = 3 x 6.023 x 10²³ x 5.3/106= 9.03 x10²² atoms

Question 18

(a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH₂)₂] [O = 16; N = 14; C = 12 ; H = 1 ]

(b) Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P. [S = 32; O = 16]

Answer 18

(a) 60 g urea has mass of nitrogen(N₂) = 28 g

So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg

(b) 64 g has volume = 22.4 litre

So, 320 g will have volume = 22.4 x 320/64=112 litres

Question 19

(a) What do you understand by the statement that ‘vapour density of carbon dioxide is 22’?

(b) Atomic mass of Chlorine is 35.5.What is its vapour density?

Answer 19

(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.

(b) Vapour density of Chlorine atom is 35.5.

Question 20

What is the mass of 56 cm³ of carbon monoxide at STP?

(C=12 ,O=16)

Answer 20

22400 cm³ of CO has mass = 28 g

So, 56 cm³ will have mass = 56 x 28/22400 = 0.07 g

Question 21

Determine the number of molecules in a drop of water which weighs 0.09g.

Answer 21

18 g of water has number of molecules = 6.023 x 10²³

So, 0.09 g of water will have no. of molecules = 6.023 x 10²³ x 0.09/18 = 3.01 x 10²¹ molecules

Question 22

The molecular formula for elemental sulphur is S₈.In sample of 5.12 g of sulphur

(a) How many moles of sulphur are present?

(b) How many molecules and atoms are present?

Answer 22

(a) No. of moles in 256 g S₈ = 1 mole

So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 x 6.023 x 10²³ = 1.2 x 10²² molecules

No. of atoms in 1 molecule of S = 8

So, no. of atoms in 1.2 x 10²² molecules = 1.2 x 10²² x 8

= 9.635x 10²² molecules

Question 23

If phosphorus is considered to contain P₄ molecules, then calculate the number of moles in 100g of phosphorus?

Answer 23

Atomic mass of phosphorus P = 30.97 g

Hence, molar mass of P₄ = 123.88 g

If phosphorus is considered as P₄ molecules,

then 1 mole P₄ ≡ 123.88 g

Therefore, 100 g of P₄ = 0.807 g

Question 24

Calculate:

(a) The gram molecular mass of chlorine if 308cm³ of it at STP weighs 0.979 g

(b) The volume of 4g of H₂ at 4 atmospheres.

(c) The mass of oxygen in 2.2 litres of CO₂ at STP.

Answer 24

(a) 308 cm³ of chlorine weighs = 0.979 g

So, 22400 cm³ will weigh = gram molecular mass

= 0.979 x 22400/308 =71.2 g

(b) 2 g(molar mass) H₂ at 1 atm has volume = 22.4 litres

So, 4 g H₂ at 1 atm will have volume = 44.8 litres

Now, at 1 atm(P₁) 4 g H₂ has volume (V₁) = 44.8 litres

So, at 4 atm(P₂) the volume(V₂) will be =

(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)

So, mass of oxygen in 2.2 litres = 2.2 x 32/22.4=3.14 g

Question 25

A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^-12 g, calculate the number of carbon atoms in the signature.

Answer 25

No. of atoms in 12 g C = 6.023 x10²³

So, no. of carbon atoms in 10^-12 g = 10^-12 x 6.023 x10²³/12

= 5.019 x 10¹⁰ atoms

Question 26

An unknown gas shows a density of 3 g per litre at 273⁰C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?

Answer 26

Given:

P= 1140 mm Hg

Density = D = 2.4 g / L

T = 273 ⁰C = 273+273 =   546 K

M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L

Hence we have to find out the volume of the unknown gas at STP.

First, apply Charle’s law.

We have to find out the volume of one litre of unknown gas at standard temperature 273 K.

V₁= 1 L  T₁ = 546 K

V₂=?   T₂ = 273 K

V₁/T₁ = V₂/ T₂

V₂ = (V₁ x T₂)/T₁

= (1 L x 273 K)/546 K

= 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.

P ₁ = 1140 mm Hg  V₁ = 0.5 L

P₂ = 760 mm Hg  V₂ = ?

P₁ x V₁ = P₂ x V₂

V₂ = (P₁ x V₁)/P₂

= (1140 mm Hg x 0.5 L)/760 mm Hg

= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP

X moles = 0.75 L / 22.4 L

=  0.0335 moles

The original mass is 2.4 g

n = m / M

0.0335 moles    = 2.4 g / M

M = 2.4 g / 0.0335 moles

M= 71.6 g / mole

Hence, the gram molecular mass of the unknown gas is 71.6 g

Question 27

Cost of Sugar (C₁₂H₂₂ O₁₁) is Rs 40 per kg; calculate its cost per mole.

Answer 27

1000 g of sugar costs = Rs. 40

So, 342g(molar mass) of sugar will cost=342×40/1000 = Rs. 13.68

Question 28

Which of the following weighs the least?

(a) 2 g atom of N

(b) 3 x10²⁵ atoms of carbon

(c)1mole of sulphur

(d) 7 g of silver

Answer 28

(a) Weight of 1 g atom N = 14 g

So, weight of 2 g atom of N = 28 g

(b) 6.023 x10²³ atoms of C weigh = 12 g

So, 3 x10²⁵ atoms will weigh =

(c) 1 mole of sulphur weighs = 32 g

(d) 7 g of silver

So, 7 grams of silver weighs least.

Question 29

Four grams of caustic soda contains:

(a) 6.02 x 10²³ atoms of it

(b) 4 g atom of sodium

(c) 6.02 x10²² molecules

(d) 4 moles of NaOH

Answer 29

Option C is correct.

40 g of NaOH contains 6.023 x 10²³ molecules

So, 4 g of NaOH contains = 6.02 x10²³ x 4/40

= 6.02 x10²² molecules

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Page-85

Question 30

The number of molecules in 4.25 g of ammonia is:

(a) 1.0 x 10²³

(b) 1.5 x 10²³ 

(c) 2.0 x 10²³ 

(d) 3.5x 10²³

Answer 30

The number of molecules in 18 g of ammonia= 6.02 x10²³

So, no. of molecules in 4.25 g of ammonia = 6.02 x10²³ x 4.25/18

= 1.5 x 10²³

Question 31

Correct the statements, if required

(a) One mole of chlorine contains 6.023 x 10¹⁰ atoms of chlorine.

(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour.

(c) Relative atomic mass of an element is the number of times one molecule of an element is heavier than 1/12 the mass of an atom of carbon-12.

(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.

Answer 31

(a) One mole of chlorine contains 6.023 x 10²³ atoms of chlorine.

(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.

(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.

(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

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Exercise – 5 (C)   Selina Solutions of ICSE Class-10 Chemistry 

(Percentage Composition, Empirical and Molecular Formula)

Page-90

Question 1

Give three kinds of information conveyed by the formula H₂O.

Answer 1

Information conveyed by H₂O

(1)That H₂O contains 2 volumes of hydrogen and 1 volume of oxygen.

(2)That ratio by weight of hydrogen and oxygen is 1:8.

(3)That molecular weight of H₂O is 18g.

Question 2

Explain the terms empirical formula and molecular formula.

Answer 2

The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.

The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

Question 3

Give the empirical formula of:

(a) C₆H₆

(b) C₆H₁₂O₆

(c) C₂H₂

(d) CH₃COOH

Answer 3

(a) CH

(b) CH₂O

(c) CH

(d) CH₂O

Question 4

Find the percentage of water of crystallisation in CuSO₄.5H₂O. (At. Mass Cu = 64, H = 1, O = 16, S = 32) (2017)

Answer 4

Question 5

Calculate the percentage of phosphorus in

(a) Calcium hydrogen phosphate Ca(H₂PO₄)₂

(b) Calcium phosphate Ca₃(PO₄)₂

Answer 5

(a) Molecular mass of Ca(H₂PO₄)₂ = 234

So, % of P = 2 31 100/234 = 26.5%

(b) Molecular mass of Ca₃(PO₄)₂ = 310

% of P = 2 31 100/310 = 20%

Question 6

Calculate the percent composition of Potassium chlorate KClO₃.

Answer 6

Molecular mass of KClO₃ = 122.5 g

And , % of K = 39 /122.5 = 31.8%

So % of Cl = 35.5/122.5 = 28.98%

Hence % of O = 3 16/122.5 = 39.18%

Question 7

Find the empirical formula of the compounds with the following percentage composition:

Pb = 62.5%, N = 8.5%, O = 29.0%

Answer 7

Element % At. mass Atomic ratio Simple ratio

Pb 62.5 207  1

N 8.5 14   2

O 29.0 16     6

So, Pb(NO₃)₂ is the empirical formula.

Question 8

Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.

Answer 8

In Fe₂O₃ , Fe = 56 and O = 16

Molecular mass of Fe₂O₃ = 2 56 + 3 16 = 160 g

Iron present in 80% of Fe₂O₃ =

So, mass of iron in 100 g of ore = 56 g

mass of Fe in 10000 g of ore = 56 10000/100

= 5.6 kg

Question 9

If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula. (2015)

Answer 9

For acetylene , molecular mass = 2 V.D = 2 13 = 26 g

The empirical mass = 12(C) + 1(H) = 13 g

n =

Molecular formula of acetylene= 2 Empirical formula =C₂H₂

Similarly, for benzene molecular mass= 2 V.D = 2 39 = 78

n = 78/13=6

So, the molecular formula = C₆H₆

Question 10

Find the empirical formula of a compound containing 17.7% hydrogen and 82.3% nitrogen. (2018)

Answer 10

Element % At. mass Atomic ratio Simple ratio

H 17.7 1

N 82.3 14

So, the empirical formula = NH₃

Question 11

On analysis, a substance was found to contain

C = 54.54%, H = 9.09%, O = 36.36%

The vapour density of the substance is 44,calculate;

(a) its empirical formula, and

(b) its molecular formula

Answer 11

Element % at. mass atomic ratio simple ratio

C 54.54 12 2

H 9.09 1  4

O 36.36 16 1

(a) So, its empirical formula = C₂H₄O

(b) empirical formula mass = 44

Since, vapour density = 44

So, molecular mass = 2 V.D = 88

Or n = 2

so, molecular formula = (C₂H₄O)₂ = C₄H₈O₂

Question 12

An organic compound ,whose vapour density is 45, has the following percentage composition

H=2.22%, O = 71.19% and remaining carbon. Calculate ,

(a) its empirical formula, and

(b) its molecular formula

Answer 12

Element % at. mass atomic ratio simple ratio

C 26.59 12 1

H 2.22 1 1

O 71.19 16 2

(a) its empirical formula = CHO₂

(b) empirical formula mass = 45

Vapour density = 45

So, molecular mass = V.D 2 = 90

so, molecular formula = C₂H₂O₄

Question 13

An organic compound contains H = 4.07%, Cl = 71.65% chlorine and remaining carbon. Its molar mass = 98.96. Find,

(a) Empirical formula, and

(b) Molecular formula

Answer 13

Element % at. mass atomic ratio simple ratio

Cl71.6535.5   1

H4.071 2

C24.2812  1

(a) its empirical formula = CH₂Cl

(b) empirical formula mass = 49.5

Since, molecular mass = 98.96

so, molecular formula = (CH₂Cl)₂ = C₂H₄Cl₂

Question 14

A hydrocarbon contains 4.8g of carbon per gram of hydrogen. Calculate

(a) the g atom of each

(b) find the empirical formula

(c) Find molecular formula, if its vapour density is 29.

Answer 14

(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1

(b) Element Given mass At. mass Gram atom Ratio

C 4.8 12 0.4 1 2

H 1 1 1 2.5 5

So, the empirical formula = C₂H₅

(c) Empirical formula mass = 29

Molecular mass = V.D 2 = 29 2 = 58

So, molecular formula = C₄H₁₀

Question 15

0.2 g atom of silicon Combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.

Answer 15

Since, g atom of Si = given mass/mol. Mass

so, given mass = 0.2 28 = 5.6 g

Element mass At. mass Gram atom Ratio

Si5.6280.21

Cl21.335.5 3

Empirical formula = SiCl₃

Question 16

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula. (2016)

Answer 16

% 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	6.89/6.89 = 1 x 2 = 2
H	17.24	1	17.24/1 = 17.24	17.24/6.89 = 2.5 x 2 = 5

 

Empirical formula = C₂H₅

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

Vapour Density = 29

Relative molecular mass = 29 x 2 = 58

N = 

Molecular formula = n x empirical formula

= 2 x C₂H₅

= C₄H₁₀

Question 17

In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions.

(a) How many gram- atoms of magnesium are equal to 18g?

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Page-91

(b) How many gram- atoms of nitrogen are equal to 7g of nitrogen?

(c) Calculate simple ratio of gram- atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.

Answer 17

(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg

(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N

(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3

So, the formula is Mg₃ N₂

Question 18

Barium chloride crystals contain 14.8% water of crystallization. Find the number of molecules of water of crystallization per molecule.

Answer 18

Barium chloride = BaCl₂.x H₂O

Ba + 2Cl + x[H₂ + O]

=137+ 235.5 + x [2+16]

=[208 + 18x] contains water = 14.8% water in BaCl₂.x H₂O

=[208 + 18 x] 14.8/100 = 18x

=[104 + 9x] 2148=18000x

=[104+9x] 37=250x

=3848 + 333x =2250x

1917x =3848

x = 2 molecules of water

Question 19

Urea is very important nitrogenous fertilizer. Its formula is CON₂H₄.Calculate the percentage of nitrogen in urea. (C=12,O=16 ,N=14 and H=1).

Answer 19

Molar mass of urea; CON₂H₄ = 60 g

So, % of Nitrogen = 28 100/60 = 46.66%

Question 20

Determine the formula of the organic compound if its molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.

Answer 20

Element % At. mass Atomic ratio Simple ratio

C 42.1 12 3.5 1

H 6.48 1 6.48 2

O 51.42 16 3.2 1

The empirical formula is CH₂O

Since the compound has 12 atoms of carbon, so the formula is

C₁₂ H₂₄ O_12.

Question 21

(a) A compound with empirical formula AB₂, has the vapour density equal to its empirical formula weight. Find its molecular formula.

(b) A compound with empirical formula AB has vapour density 3 times its empirical formula weight. Find the molecular formula.

(c)10.47 g of a compound contained 6.25 g of metal A and rest non-metal B. Calculate the empirical formula of the compound [At. wt of A = 207, B = 35.5]

Answer 21

(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A₂B₄.

(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D

Empirical formula weight = V.D/3

So, n = molecular mass/ Empirical formula weight = 6

Hence, the molecular formula is A₆B₆

(c) 

Given:

Wt. of the compound: 10.47g

Wt. of metal A: 6.25g

Wt. of non-metal B: 10.47 – 6.25 = 4.22g

Element	mass	At. Wt.	Relative no. of atoms	Simplest ratio
A	6.25g	207	6.25/207=0.03	0.03/0.03=1
B	4.22g	35.5	4.26/35.5=0.12	0.12/0.03=4

 

Hence, the empirical formula is AB₄

Question 22

A hydride of nitrogen contains 87.5% per cent by mass of nitrogen. Determine the empirical formula of this compound.

Answer 22

Atomic ratio of N = 87.5/14 =6.25

Atomic ratio of H= 12.5/1 = 12.5

This gives us the simplest ratio as 1:2

So, the molecular formula is NH₂

Question 23

A compound has O=61.32%, S= 11.15%, H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 amu. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallization.

Answer 23

Element % at. mass atomic ratio simple ratio

Zn 22.65 65 0.348 1

H 4.88 1 4.88 14

S 11.15 32 0.348 1

O 61.32 16 3.83 11

Empirical formula of the given compound =ZnSH₁₄O₁₁

Empiricala formula mass = 65.37+32+141+11+16=287.37

Molecular mass = 287

n = Molecular mass/Empirical formula mass = 287/287=1

Molecular formula = ZnSO₁₁H₁₄

=ZnSO₄.7H₂O

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 Exercise – 5(D)  Mole Concept And Stoichiometry

Page-94

Question 1

Complete the following blanks in the equation as indicated.

CaH₂ (s) + 2H₂O (aq)  Ca(OH)₂ (s) + 2H₂ (g)

(a) Moles: 1 mole + ——-   ——– + ————–

(b) Grams: 42g + ——-  ——– + —————-

(c) Molecules: 6.02 x 10²³ + ——-   ——– + ———–

Answer 1

(a) Moles:1 mole + 2 mole 1 mole + 2 mole

(b) Grams: 42g + 36g 74g + 4 g

(c) Molecules = 6.02 10²³ + 12.046 10²³     6.02 10²³+ 12.046  10²³

Question 2

The reaction between 15 g of marble and nitric acid is given by the following equation:

CaCO₃ + 2HNO₃  Ca(NO₃)₂+ H₂O + CO₂

Calculate: (a) the mass of anhydrous calcium nitrate formed, (b) the volume of carbon dioxide evolved at S.T.P.

Answer 2

(a) 100 g of CaCO₃ produces = 164 g of Ca(NO₃)₂

So, 15 g CaCO₃ will produce = 164 15/100 = 24.6 g Ca(NO₃)₂

(b) 1 V of CaCO₃ produces 1 V of CO₂

100 g of CaCO₃ has volume = 22.4 litres

So, 15 g will have volume = 22.4 15/100 = 3.36 litres CO₂

Question 3

66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate:

(a) Mass of ammonia required.

(b) The volume of the gas used at the S.T.P.

(c) The mass of acid required.

Answer 3

2NH₃ + H₂SO₄ (NH₄)₂SO₄

66 g

(a) 2NH₃ + H₂SO₄ (NH₄)₂SO₄

34 g98 g132 g

For 132 g (NH₄)₂SO₄ = 34 g of NH₃ is required

So, for 66 g (NH₄)₂SO₄ = 66 32/132 = 17 g of NH₃ is required

(b) 17g of NH₃ requires volume = 22.4 litres

(c) Mass of acid required, for producing 132g (NH₄)₂SO₄ = 98g

So, Mass of acid required, for 66g (NH₄)₂SO₄ = 66 98/132 = 49g

Question 4

The reaction between red lead and hydrochloric acid is given below:

Pb₃O₄ + 8HCl  3PbCl₂ + 4H₂O + Cl₂

Calculate:

(a) the mass of lead chloride formed by the action of the 6.85 g of red lead,

(b) the mass of the chlorine and

(c) the volume of the chlorine evolved at S.T.P.

Answer 4

(a) Molecular mass of Pb₃O₄ = 3 207.2 + 4 16 = 685 g

685 g of Pb₃O₄ gives = 834 g of PbCl₂

Hence, 6.85 g of Pb₃O₄ will give = 6.85 834/685 = 8.34 g

(b) 685g of Pb₃O₄ gives = 71g of Cl₂

Hence, 6.85 g of Pb₃O₄ will give = 6.85 71/685 = 0.71 g Cl₂

(c) 1 V Pb₃O₄ produces 1 V Cl₂

685g of Pb₃O₄has volume = 22.4 litres = volume of Cl₂ produced

So, 6.85 Pb₃O₄ will produce = 6.85 22.4/685 = 0.224 litres of Cl₂

Question 5

Find the mass of KNO₃ required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO₃ is required for the same purpose.

KNO₃+ H₂SO₄    KHSO₄ + HNO₃

NaNO₃ + H₂SO₄   NaHSO₄ + HNO₃

Answer 5

Molecular mass of KNO₃ = 101 g

63 g of HNO₃ is formed by = 101 g of KNO₃

So, 126000 g of HNO₃ is formed by = 126000 101/63 = 202 kg

Similarly,126 g of HNO₃ is formed by 170 kg of NaNO₃

So, smaller mass of NaNO₃ is required.

Question 6

Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27^oC and normal pressure.

Calculate :

(a) The mass of salt required.

(b) The mass of the acid required to prepare the 2 litres of CO₂ at 27 C and normal pressure.

CaCO₃ + 2HCl   CaCl₂ + H₂O + CO₂

Answer 6

CaCO₃ + 2HCl   CaCl₂ + H₂O + CO₂

100g73g22.4L

(a) V₁ =2 litresV₂ =?

T₁ = (273+27)=300KT₂=273K

V₁/T₁=V₂/T₂

V₂=V₁T₂/T₁=

Now at STP 22.4 litres of CO₂ are produced using CaCO₃ =100g

So,    litres are produced by =100/22.4 2274/300 =.125g

(b)  22.4 litres are CO₂ are prepared from acid =73g

  litres are prepared from = 73/22.4 2273/300=5.9g

Question 7

Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water

Answer 7

2H₂O  2H₂ + O₂

2 V2 V1 V

2 moles of H₂O gives = 1 mole of O₂

So, 1 mole of H₂O will give = 0.5 moles of O₂

so, mass of O₂ = no. of moles x molecular mass

= 0.5 32 = 16 g of O₂

and 1 mole of O₂ occupies volume =22.4 litre

so, 0.5 moles will occupy = 22.4   0.5 = 11.2 litres at S.T.P.

Question 8

1.56 g of sodium peroxide reacts with water according to the following equation:

2Na₂O₂ + 2H₂O     4NaOH + O₂

Calculate:

(a) mass of sodium hydroxide formed,

(b) Volume of oxygen liberated at S.T.P.

(c) Mass of oxygen liberated.

Answer 8

2Na₂O₂ + 2H₂O   4NaOH + O₂

2 V4 V1 V

(a) Mol. Mass of Na₂O₂ = 2 23 + 2 16 = 78 g

Mass of 2Na₂O₂= 156 g

156 g Na₂O₂ gives = 160 g of NaOH (4 40 g)

So, 1.56 Na₂O₂ will give = 160 1.56/156 = 1.6 g

(b) 156 g Na₂O₂ gives = 22.4 litres of oxygen

So, 1.56 g will give = 22.4 1.56/156 = 0.224 litres

= 224 cm³

(c)156 g Na₂O₂ gives = 32 g O₂

So, 1.56 g Na₂O₂ will give = 32 1.56/156

= 32/100 = 0.32 g

Question 9

(a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH₄Cl by the reaction:

2NH₄Cl + Ca(OH)₂     CaCl₂ +2H₂O + 2NH₃

(b) What will be the volume of ammonia when measured at S.T.P?

The molar volume of a gas = 22.4 litres at STP.

Answer 9

2NH₄Cl + Ca(OH)₂  CaCl₂+2H₂O + 2NH₃

2 V1 V1 V2 V

Mol. Mass of 2NH₄Cl = 2[14 + (1 4) + 35.5] = 2[53.5] = 107 g

(a) 107 g NH₄Cl gives = 34 g NH₃

So, 21.4 g NH₄Cl will give = 21.4 34/107 = 6.8 g NH₃

(b) The volume of 17 g NH₃ is 22.4 litre

So, volume of 6.8 g will be = 6.8 22.4/17 = 8.96 litre

Question 10

Aluminium carbide reacts with water according to the following equation.

Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄

(a) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?

(b) What volume of methane s.t.p. is obtained from 12g of aluminium carbide? (2018)

Answer 10

Question 11

MnO₂ + 4HCl   MnCl₂ + 2H₂O +Cl₂

0.02 moles of pure MnO₂is heated strongly with conc. HCl. Calculate:

(a) mass of MnO₂ used

(b) moles of salt formed,

(c) mass of salt formed,

(d) moles of chlorine gas formed,

(e) mass of chlorine gas formed,

(f) volume of chlorine gas formed at S.T.P.,

(g) moles of acid required,

(h) Mass of acid required.

Answer 11

MnO₂ + 4HCl   MnCl₂ + 2H₂O +Cl₂

1 V4 V1 V1 V

(a) 1 mole of MnO₂ weighs = 87 g (mol. Mass)

So, 0.02 mole will weigh = 87 0.02 = 1.74 g MnO₂

(b) 1 mole MnO₂ gives = 1 mole of MnCl₂

So, 0.02 mole MnO₂will give =0.02 mole of MnCl₂

(c) 1 mole MnCl₂ weighs = 126 g(mol mass)

So, 0.02 mole MnCl₂ will weigh = 126 0.02 g = 2.52 g

(d) 0.02 mole MnO₂will form =0.02 mole of Cl₂

(e) 1 mole of Cl₂ weighs = 35.5 g

So, 0.02 mole will weigh = 71 0.02 = 1.42 g of Cl₂

(f) 1 mole of chlorine gas has volume = 22.4 litres

So, 0.02 mole will have volume = 22.4  0.02 = 0.448 litre

(g) 1 mole MnO₂requires HCl = 4 mole

So, 0.02 mole MnO₂ will require =4 0.02 = 0.08 mole

(h) For 1 mole MnO₂ , acid required = 4 mole of HCl

So, for 0.02 mole, acid required = 4 0.02 =0.08 mole

Mass of HCl = 0.08 x 36.5 = 2.92 g

Question 12

Nitrogen and hydrogen react to form ammonia.

N₂ (g) + 3H₂ (g)  2NH₃ (g)

If 1000g H₂ react with 2000g of N₂:

(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?

(b) Calculate the mass of ammonia(NH₃) that will be formed?

Answer 12

N₂ + 3H₂   2NH₃

28g6g34g

28g of nitrogen requires hydrogen = 6g

2000g of nitrogen requires hydrogen = 6/28    2000=3000/7g

So mass of hydrogen left unreacted =1000-3000/7=571.4g of H₂

(b) 28g of nitrogen forms NH₃ = 34g

2000g of N₂ forms NH₃

= 34/28 2000

=2428.6g

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Misc.- Exercise  of Mole Concept And Stoichiometry

Page-95

Question 1

From the equation for burning of hydrogen and oxygen

2H₂ + O₂   2H₂O (Steam)

Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

Solution 1

From equation: 2H₂ + O₂ 2H₂O

1 mole of Oxygen gives = 2 moles of steam

so, 0.5 mole oxygen will give = 2 0.5 = 1mole of steam

Question 2

From the equation

3Cu + 8HNO₃ 3Cu  (NO₃)₂+ 4H₂O + 2NO

(At.mass Cu=64, H=1, N=14,O=16)

Calculate:

(a) Mass of copper needed to react with 63g of HNO₃

(b) Volume of nitric oxide at S.T.P. that can be collected.

Answer 2

3Cu + 8HNO₃    3Cu (NO₃)₂+ 4H₂O + 2NO

1 V8         V3              V2 V

Mol. Mass of 8HNO₃ = 8 63 = 504 g

(a) For 504 g HNO₃, Cu required is = 192 g

So, for 63g HNO₃Cu required = 192 63/504 = 24g

(b) 504 g of HNO₃ gives = 2 22.4 litre volume of NO

So, 63g of HNO₃ gives =2 22.4 63/504 =5.6 litre of NO

Question 3

(a) Calculate the number of moles in 7g of nitrogen.

(b) What is the volume at S.T.P. of 7.1 g of chlorine?

(c) What is the mass of 56 cm³ of carbon monoxide at S.T.P?

Answer 3

(a) 28g of nitrogen = 1mole

So, 7g of nitrogen = 1/28 7= 0.25 moles

(b) Volume of 71 g of Cl2 at STP =22.4 litres

Volume of 7.1 g chlorine =22.4 7.1/71=2.24 litre

(c) 22400cm³ volume have mass =28 g of CO(molar mass)

So, 56cm³ volume will have mass =28 56/22400= 0.07 g

Question 4

Some of the fertilizers are sodium nitrate NaNO₃, ammonium sulphate (NH₄)₂SO₄ and urea CO(NH₂)₂. Which of these contains the highest percentage of nitrogen?

Answer 4

% of N in NaNO₃=

% of N in (NH₄)₂SO₄ =

% of N in CO(NH₂)₂ =

So, highest percentage of N is in urea.

Question 5

Water decomposes to O₂ and H₂ under suitable conditions as represented by the equation below:

2H₂O      2H₂+O₂

(a) If 2500 cm³ of H₂ is produced, what volume of O₂ is liberated at the same time and under the same conditions of temperature and pressure?

(b) The 2500 cm³ of H₂ is subjected to 2 times increase in pressure (temp. remaining constant). What volume of H₂ will now occupy?

(c) Taking the value of H₂ calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm³ pressure remaining constant.

Answer 5

2H₂O 2H₂+O₂

2 V2      V1 V

(a) From equation, 2 V of water gives 2 V of H₂ and 1 V of O₂

where 2 V = 2500 cm³

so, volume of O₂ liberated = 2V/V = 1250 cm³

(b)

(c)

i.e. temperature should be increased by 3.5 times.

Question 6

Urea [CO(NH₂)₂] is an important nitrogeneous fertilizer. Urea is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Answer 6

Molecular mass of urea=12 + 16+2(14+2) =60g

60g of urea contains nitrogen =28g

So, in 50g of urea, nitrogen present =23.33 g

50 kg of urea contains nitrogen=23.33kg

Question 7

Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Answer 7

% of hydrogen = 20%

% of carbon = 100 – 20 = 80%

	% Weight	Atomic Weight	Relative No. of Moles	Simplest Ratio
C	80	12	80/12 = 6.667	6.667/6.667 = 1
H	20	1	20/1 = 20	20/6.667 = 2.99 ≈ 3

Empirical formula = CH₃

Empirical formula weight = 1 x 12 + 1 x 3 = 12 + 3 = 15

Vapour Density = 15

Relative molecular mass = 15 x 2 = 30

N =

Molecular formula = n x empirical formula

= 2 x CH₃

= C₂H₆

Question 8

The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.

0.145 g of X was heated with dry copper (II) oxide and 224 cm³ of carbon dioxide was collected at S.T.P.

(a) Which elements does X contain?

(b) What was the purpose of copper (II) oxide?

(c) Calculate the empirical formula of X by the following steps:

(i) Calculate the number of moles of carbon dioxide gas.

(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.

(iii) Calculate the mass of hydrogen in sample X.

(iv) Deduce the ratio of atoms of each element in X (empirical formula).

Answer 8

22400cm³ CO₂ has mass = 44g

so, 224 cm³ CO₂ will have mass= 0.44 g

Now since CO₂ is being formed and X is a hydrocarbon so it contains C and H.

In 0.44g CO₂, mass of carbon=0.44-0.32=0.12g=0.01g atom

So, mass of Hydrogen in X = 0.145-0.12 = 0.025g

= 0.025g atom

Now the ratio of C:H is C=1: H=2.5 or C=2 : H=5

i.e. the formula of hydrocarbon is C₂H₅

(a) C and H

(b) Copper (II) oxide was used for reduction of the hydrocarbon.

(c)

(i) no. of moles of CO₂= 0.44/44 = 0.01 moles

(ii) mass of C = 0.12 g

(iii) mass of H = 0.025 g

(iv) The empirical formula of X = C₂H₅

Question 9

A compound is formed by 24g of X and 64g of oxygen. If atomic mass of X=12 and O=16, calculate the simplest formula of compound.

Answer 9

Mass of X in the given compound =24g

Mass of oxygen in the given compound =64g

So total mass of the compound =24+64=88g

% of X in the compound = 24/88 100 = 27.3%

% of oxygen in the compound=64/88 100 =72.7%

Element % At. Mass Atomic ratio Simplest ratio

X 27.3 12 27.3/12=2.27 1

O 72.7 16 72.2/16=4.54 2

So simplest formula = XO₂

Question 10

A gas cylinder filled with hydrogen holds 5g of the gas. The same cylinder holds 85 g of gas X under same temperature and pressure. Calculate :

(a) Vapour density of gas X.

(b) Molecular weight of gas X.

Answer 10

(a) V.D =

(b) Molecular mass = 17(V.D) x 2= 34g

Question 11

(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :

CO₂ + C     2CO

What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?

(b) 60 cm³ of oxygen was added to 24 cm³ of carbon monoxide and mixture ignited. Calculate:

(i) volume of oxygen used up and

(ii) Volume of carbon dioxide formed.

Answer 11

(a) CO₂ + C 2CO

1 V    1 V      2 V

12 g of C gives = 44.8 litre volume of CO

So, 3 g of C will give = 11.2 litre of CO

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Page-96

(b) 2CO + O₂   2CO₂

2 V    1 V     2 V

(i) 2 V CO requires oxygen = 1 V

so, 24 cm³ CO will require = 24/2 =12 cm³

(ii) 2 x 22400 cm³ CO gives = 2 x 22400 cm³ CO₂

so, 24cm³ CO will give = 24 cm³ CO₂

Question 12

How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO₂ evolved: (2016)

Answer 12

Molecular weight of  ==

 =328g

Molecular weight of CaO =2(40+16)

=112g

(a) 328g of Ca(NO₃)₂liberates 4 moles of NO₂

328g of Ca(NO₃)₂ liberates   L of NO₂  82g will liberate

=22.4dm³ of NO₂

(b) 328 g of calcium nitrate gives 112g ofCaO

82 g will give

=28 g of CaO

Question 13

The equation for the burning of octane is:

2C₈H₁₈ + 25O₂      16CO₂ + 18H₂O

(i) How many moles of carbon dioxide are produced when one mole of octane burns?

(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?

(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?

(iv) What is the empirical formula of octane?

Answer 13

2C₈H₁₈ + 25O₂      16CO₂ + 18H₂O

25 V         16 V            18 V

(i) 2 moles of octane gives = 16 moles of CO₂

so, 1 mole octane will give = 8 moles of CO₂

(ii) 1 mole CO₂ occupies volume = 22.4 litre

so, 8 moles will occupy volume = 8 22.4 = 179.2 litre

(iii) 1 mole CO₂ has mass = 44 g

so, 16 moles will have mass = 44 16 = 704 g

(iv) Empirical formula is C₄H₉.

Question 14

Ordinary chlorine gas has two isotopes ³⁵₁₇Cl and ³⁷₁₇Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.

Answer 14

The relative atomic mass of Cl = (35 3 + 1 37)/4=35.5 amu

Question 15

Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.

Answer 15

Mass of silicon in the given compound =5.6g

Mass of the chlorine in the given compound=21.3g

Total mass of the compound=5.6g+21.3g=26.9g

% of silicon in the compound = 56/26.9 100 = 20.82%

% of chlorine in the compound = 21.2/26.9 100 = 79.18%

Element % At. Mass At. Ratio Simplest ratio

Si 20.82 28 20.82/28=0.74 1

Cl 79.18 35.5 79.18/35.5=2.23 3

So the empirical formula of the given compound =SiCl₃

Question 16

An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47 %; oxygen = 59.26 %. Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is 162.

Answer 16

% composition Atomic ratio Simple ratio

P = 38.27% 38.27/31 =1.23 1

H = 2.47% 2.47/1 = 2.47 2

O = 59.26% 59.26/16 = 3.70 3

So, empirical formula is PH₂O₃ or H₂PO₃

Empirical formula mass = 31+ 2 1 + 3 16 = 81

The molecular formula is = H₄P₂O₆, because n = 162/81=2

Question 17

Calculate the mass of substance ‘A’ which in gaseous form occupies 10 litres at 27⁰C and 700 mm pressure. The molecular mass of ‘A’ is 60.

Answer 17

V₁ = 10 litres V₂=?

T₁= 27+ 273 = 300K T₂=273K

P₁=700 mm P₂ = 760 mm

Using the gas equation

Question 18

A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.

(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?

(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result

Answer 18

(a) Molecular mass of CO₂ = 12+ 2×16 = 44 g

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

V.D =

So, mass of CO₂ = 22 kg

(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

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

Question 19

Following questions refer to one mole of chlorine gas.

(a) What is the volume occupied by this gas at S.T.P.?

(b) What will happen to volume of gas, if pressure is doubled?

(c) What volume will it occupy at 273⁰C?

(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?

Solution 19

(a) The volume occupied by 1 mole of chlorine = 22.4 litre

(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.

(c) V₁/V₂ = T₁/T₂

22.4/V₂ =273/546

V₂ = 44.8 litres

(d) Mass of 1 mole Cl₂ gas =35.5 x 2 =71 g

Question 20

(a) A hydrate of calcium sulphate CaSO₄.xH₂O contains 21% water of crystallisation. Find the value of x.

(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.

(c) How much volume will be occupied by 2g of dry oxygen at 27⁰C and 740 mm pressure?

(d) What would be the mass of CO₂ occupying a volume of 44 litres at 25⁰ C and 750 mm pressure?

(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.

AgNO₃ (aq) + NaCl (aq)  AgCl (s) + NaNO₃

Calculate the percentage of NaCl in the mixture.

Answer 20

(a) Total molar mass of hydrated CaSO₄.xH₂O = 136+18x

Since 21% is water of crystallization, so

So, x = 2 i.e. water of crystallization is 2.

(b) For 18 g water, vol. of hydrogen needed = 22.4 litre

So, for 1.8 g, vol. of H₂ needed= 1.8 x 22.4/18 = 2.24 litre

Now 2 vols. of water = 1 vol. of oxygen

1 vol. of water =1/2 vol. of O₂ =22.4/2=11.2 lit.

18 g of water = 11.2 lit. of O₂

1.8 g of water = 11.2/18 18/10=1.12 lit.

(c) 32g of dry oxygen at STP = 22400cc

2g will occupy = 224002/32=1400cc

P₁=760mm P₂ =740mm

V₁=1400cc V₂ =?

T₁ =273 K, T₂ = 27 +73 = 300K

(d) P₁= 750mm P₂=760mm

V₁= 44lit. V₂=?

T₁= 298K T₂=273K

(e) Since 143.5g of AgCl is produced from =58.5 g of NaCl

so, 1.435 g of AgCl is formed by =0.585 g of NaCl

% of NaCl =0.585 x100 = 58.5%

Question 21

(a) From the equation :

Calculate :

(i) The mass of carbon oxidized by 49 g of sulphuric acid.

(ii) The volume of sulphur dioxide measured at STP,  liberated at the same time.

(b) 

(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)

(ii) The relative molecular mass of this compound is 168, so what is its molecular formula?

Answer 21

(a)

For CO₂ 12+32

(i) Molecular mass of sulphuric acid = 2(2+32+64)

= 196

196 g of suphuric acid oxidized 12g of Carbon

49 g of suphuric acid will

=3 g

(ii) 196 g of sulphuric acid gives 2(22.4)

=44.8L

49 g og sulphuric acid will give

=11.2 L of SO₂

_(b) 

_(i) 

Element	% Weight	Atomic Weight	Atomic Ratio	Simplest Ratio
C	14.4	12	14.4/12 = 1.2	1.2/1.2=1
H	1.2	1	1.2/1 =1.2	1.2/1.2=1
Cl	84.5	35.5	84.5/35.5=2.3	2.3/1.2=1.9=2

Empirical formula = CHCl₂

_(ii) 

Empirical formula = CHCl₂

Empirical formula weight = 1 x 12 + 1 x 1+(2 x35.5)

= 12 + 1+70

= 83

Relative molecular mass = 168

N =      2.02≈2

Molecular formula = n x empirical formula

= 2 x CHCl₂

= C₂H₂Cl₄

Question 22

Find the percentage of

(a) oxygen in magnesium nitrate crystals [Mg (NO₃) 6H₂O].

(b) boron in Na₂B₄O₇.10H₂O. [H = 1, B = 11, O =16, Na = 23].

(c) phosphorus in the fertilizer superphosphate Ca(H₂PO₄)₂

Answer 22

(a) Relative molecular mass of [Mg (NO₃) 6H₂O]

=24+14+(3 x16)+(6 x18)=194

Since, 194g of [Mg (NO₃) 6H₂O] contains 144g of oxygen

100g of [Mg (NO₃) 6H₂O] contains    of oxygen = 74.22%

(b) Relative molecular mass of Na₂B₄O₇.10H₂O

(23 × 2) + (4 × 11) + (7 × 16) + 10(18) = 382

Since 382g of Na₂B₄O₇.10H₂O contains 44g of boron

100g Na₂B₄O₇.10H₂O of contains       of boron

=11.5%

(c) Relative molecular mass of Ca(H₂PO₄)₂

= 40 + 2(2 + 31 + 64) = 234

Since,234g of Ca(H₂PO₄)₂ contains 62g of phosphorus

100g of Ca(H₂PO₄)₂ contains

=26.5%

Question 23

A gas occupied 360 cm³ at 87⁰C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find its relative molecular mass.

Answer 23

Question 24

Solid ammonium dichromate decomposes as:

If 63 g of ammonium dichromate decomposes. Calculate

(a) the quantity in moles of (NH₄)₂Cr₂O₇

(b) the quantity in moles of nitrogen formed

(c) the volume of N₂ evolved at STP.

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Page-97

(d) what will be the loss of mass ?

(e) calculate the mass of chromium (III) oxide formed at the same time.

Answer 24

Solid ammonium dichromate decomposes as:

(a)Molecular mass of ammonium dichromate

= 2(14+4)+104+112

= 252 g

Number of moles=

=

=0.25moles

(b)

252 g of ammonium dichromate gives 22.4 dm³ of N₂

63 g of ammonium dichromate gives

=5.6 L

= 0.25 moles

(c)

252 g of ammonium dichromate gives 22.4 dm³ of N₂

63 g of ammonium dichromate gives

= 5.6 L

(d)

Number of moles=

=

=0.25 moles

0.25 moles of ammonium dichromate gives

0.25 moles of N₂=7 g

1 mole of H₂O =18 g

Therefore, total loss of mass=7+18

=25 g

(e)

252 g of ammonium dichromate gives 152 g of CrO₃

63 g of ammonium dichromate gives

=38 g

Question 25

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under :

2H₂S + 3O₂    2H₂O + 2SO₂

Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at S.T.P. which will complete the combustion of hydrogen sulphide determined in (litres).

Answer 25

2H₂S + 3O₂    2H₂O + 2SO₂

2 V          3 V         2 V

128 g of SO₂ gives = 2 22.4 litres volume

So, 12.8 g of SO₂ gives = 2 22.4 12.8/128

= 4.48 litre volume

Or one can say 4.48 litres of hydrogen sulphide.

2 22.4 litre H₂S requires oxygen = 3 22.4 litre

So, 4.48 litres H₂S will require = 6.72 litre of oxygen

Question 26

Ammonia burns in oxygen and the combustion, in the presence of a catalyst. May be represented by 2NH₃ + 2O_2 ? 2NO + 3H₂O

[H= 1, N= 14, O=16]

What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

Answer 26

From equation, 2NH₃ + 2   O₂ 2NO + 3H₂O

When 60 g NO is formed, mass of steam produced = 54 g

So, 1.5 g NO is formed, mass of steam produced = 541.5/60

=1.35 g

Question 27

If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate Ca(NO₃)₂ would be required to replace the nitrogen in a 10 hectare field ?

Answer 27

In 1 hectare of soil, N₂ removed = 20 kg

So, in 10 hectare N₂ removed = 200 kg

The molecular mass of Ca(NO₃)₂ =164

Now, 28 g N₂ present in fertilizer = 164 g Ca(NO₃)₂

So, 200000 g of N₂ is present in = 164 200000/28

= 1171.42 kg

Question 28

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P+ 5HNO₃   H₃PO₄+ 5NO₂ + H₂O

If 6.2g of phosphorus was used in the reaction calculate:

(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.

(b) mass of nitric acid will be consumed at the same time?

(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and 273⁰C?

Answer 28

(a) 1 mole of phosphorus atom = 31 g of phosphorus

31 g of P =1 mole of P

6.2g of P =       =0.2 mole of P

(b) 31 g P reacts with HNO₃ = 315 g

so, 6.2 g P will react with HNO₃ = 315 6.2/31 = 63 g

(c)

Moles of steam formed from 31g phosphorus = 18g/18g = 1mol

Moles of steam formed from 6.2 g phosphorus = 1mol/31g6.2=0.2 mol

Volume of steam produced at STP =0.2 22.4 l/MOL=4.48 litre

Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled

So, Volume of steam produced at 760mm Hg and 273⁰C=4.48 2=8.96litre

Question 29

112 cm³ of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [ F=19, P=31].

Answer 29

112cm³ of gaseous fluoride has mass = 0.63 g

so, 22400cm³ will have mass = 0.63 22400/112

= 126 g

The molecular mass = At mass P + At. mass of F

126= 31 + At. Mass of F

So, At. Mass of F = 95 g

But, at. mass of F = 19 so 95/19 = 5

Hence, there are 5 atoms of F so the molecular formula = PF₅

Question 30

Washing soda has formula Na₂CO₃.10H₂O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda?

Answer 30

Na₂CO₃.10H₂O  Na₂CO₃ + 10H₂O

286 g 106 g

So, for 57.2 g Na₂CO₃.10H₂O = 106 57.2/286 = 21.2 g Na₂CO₃

Question 31

A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M = 56).

Answer 31

Simple ratio of M = 34.5/56 = 0.616 = 1

Simple ratio of Cl = 65.5/35.5 = 1.845 = 3

Empirical formula = MCl₃

Empirical formula mass = 162.5, Molecular mass = 2V.D = 325

So, n = 2

So, molecular formula = M₂Cl₆

Question 32

A compound X consists of 4.8% carbon and 95.2% bromine by mass.

(i) Determine the empirical formula of this compound working correct to one decimal place (C = 12 ; Br = 80)

(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?

Answer 32

(i) Element % atomic mass atomic ratio simple ratio

C4.812 1

Br95.280      3

So, empirical formula is CBr₃

(ii) Empirical formula mass = 12 + 3 80 = 252 g

molecular formula mass = 2 252(V.D) = 504 g

n= 504/252 = 2

so, molecular formula = C₂Br₆

Question 33

The reaction: 4N₂O + CH₄       CO₂ + 2H₂O + 4N₂ takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of di nitrogen oxide (N₂O) required to give 150 cm³ of steam.

Answer 33

4N₂O + CH₄   CO₂ + 2H₂O + 4N₂

4 V  1 V 1 V 2 V 4 V

2 x 22400 litre steam is produced by N₂O = 4 x 22400 cm³

So, 150 cm³ steam will be produced by= 4 22400 150/2 x 22400

= 300 cm³ N₂O

Question 34

Samples of the gases O₂, N₂, CO₂ and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same of temperature and pressure

What is the volume occupied by:

(a) x molecules of N₂

(b) 3x molecules of CO

(c) What is the mass of CO₂ in grams?

(d) In answering the above questions, which law have you used?

Solution 34

(a) Volume of O₂ = V

Since O₂ and N₂ have same no. of molecules = x

so, the volume of N₂ = V

(b) 3x molecules means 3V volume of CO

(c) 32 g oxygen is contained in = 44 g of CO₂

So, 8 g oxygen is contained in = 44 x 8/32 = 11 g

(d) Avogadro’s law is used in the above questions.

Question 35

The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound?

Answer 35

simple ratio of Na = 42.1/23 = 1.83 = 3

and simple ratio of P = 18.9/31 = 0.609 = 1

Hence simple ratio of O = 39/16 = 2.43 = 4

So, the empirical formula is Na₃PO₄

Question 36

What volume of oxygen is required to burn completely a mixture of 22.4 dm³ of methane and 11.2 dm³ of hydrogen into carbon dioxide and steam?

CH₄ + 2O₂ CO₂ + 2H₂O

2H₂ + O₂  2H₂O

Answer 36

CH₄ + 2O₂    CO₂ + 2H₂O

1 V 2 V 1 V 2 V

From equation:

22.4 litres of methane requires oxygen = 44.8 litres O₂

2H₂ + O₂ 2H₂O

2 V 1 V 2 V

From equation,

44.8 litres hydrogen requires oxygen = 22.4 litres O₂

So, 11.2 litres will require = 22.4 x 11.2/44.8 = 5.6 litres

Total volume = 44.8 + 5.6 = 50.4 litres

Question 37

The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?

Answer 37

According to Avogadros law:

Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.

So, 1 mole of each gas contains = 6.02 10²³ molecules

Mol. Mass of H₂ (2),O₂(32) ,CO₂(44),SO₂(64),Cl₂(71)

(1) Now 2 g of hydrogen contains molecules =6.02 10²³

So, 8g of hydrogen contains molecules = 8/2 6.02 10²³

=4 6.02 10²³ = 4 M molecules

(2) 32g of oxygen contains molecules = 8/32 6.02 10²³=M/4

(3) 44g of carbon dioxide contains molecules = 8/44 6.02 10²³=2M/11

(4) 64g of sulphur dioxide contains molecules =6.02 10²³

So, 8g of sulphur dioxide molecules = 8/64 6.02 10²³= M/8

(5) 71 g of chlorine contains molecules =6.02 10²³

So, 8g of chlorine molecules = 8/72 6.02 10²³ = 8M/71

Since 8M/71<M/8<2M/11<M/4<4M

Thus Cl₂<SO₂<CO₂<O₂<H₂

(i) Least number of molecules in Cl₂

(ii) Most number of molecules in H₂

Question 38

10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na₂SO₄ + BaCl₂     BaSO₄ + 2NaCl

Calculate the percentage of sodium sulphate in the original mixture.

Answer 38

Na₂SO₄ + BaCl₂    BaSO₄ + 2NaCl

Molecular mass of BaSO₄ = 233 g

Now, 233 g of BaSO₄ is produced by Na₂SO₄ = 142 g

So, 6.99 g BaSO₄ will be produced by = 6.99 142/233 = 4.26

The percentage of Na₂SO₄ in original mixture = 4.26 100/10

= 42.6%

Question 39

When heated, potassium permanganate decomposes according to the following equation :

(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.

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(b) Given that the molecular mass of potassium permanganate is 158. What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres)

Answer 39

(a) 1 litre of oxygen has mass = 1.32 g

So, 24 litres (molar vol. at room temp.) will have mass = 1.32 x 24

= 31.6 or 32 g

(b) 2KMnO₄ K₂MnO₄ + MnO₂ + O₂

316 g of KMnO₄ gives oxygen = 24 litres

So, 15.8 g of KMnO₄ will give = 24 316/15.8 = 1.2 litres

Question 40

(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:

(i) The moles of sulphur dioxide present in the flask.

(ii) The number of molecules of sulphur dioxide present in the flask.

(iii) The volume occupied by 3.2 g of sulphur dioxide at S.T.P.

(S= 32, O= 16)

(b) An Experiment showed that in a lead chloride solution, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chlorine? (Pb = 207; Cl = 35.5)

Answer 40

(a)

(i) The no. of moles of SO₂ = 3.2/64 = 0.05 moles

(ii) In 1 mole of SO₂, no. of molecules present = 6.02 10²³

So, in 0.05 moles, no. of molecules = 6.02 10²³ 0.05

= 3.0 10²²

(iii) The volume occupied by 64 g of SO₂ = 22.4 dm³

3.2 g of SO₂ will be occupied by volume = 22.4 3.2/64 =1.12 dm³

(b) Gram atoms of Pb = 6.21/207=0.03 = 1

Gram atoms of Cl = 4.26/35.5 = 0.12 = 4

So, the empirical formula = PbCl₄

Question 41

The volume of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure .

(i) Which sample of gas contains the maximum number of molecules?

(ii) If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?

(iii) If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?

(iv) If the volume of A is actually 5.6 dm³ at S.T.P., calculate the number of molecules in the actual Volume of D at S.T.P. (Avogadro’s number is 6 10²³).

(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N₂O)

Answer 41

(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.

V₁/V₂ = n₁/n₂

V₁/V₂ = n₁/2n₁

So, V₂ = 2V₁

(iii) Gay lussac’s law of combining volume is being observed.

(iv) The volume of D = 5.6 4 = 22.4 dm³, so the number of molecules = 6 x 10²³ because according to mole concept 22.4 litre volume at STP has = 6 x 10 ²³ molecules

(v) No. of moles of D = 1 because volume is 22.4 litre

so, mass of N₂O = 1 44 = 44 g

Question 42

The equation given below relate the manufacture of sodium carbonate (molecular weight of Na₂CO₃ = 106).

1. NaCl+NH₃+ CO₂+ H₂O    NaHCO₃+NH₄Cl
2. 2NaHCO₃     Na₂CO₃+H₂O + CO₂

Equations (1) and (2) are based on the production of 21.2 g of sodium carbonate.

(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate?

(b) To produce the mass of sodium hydrogen carbonate calculate in (a), what volume of carbon dioxide, measured at S.T.P. would be required?

Answer 42

(a) NaCl+NH₃+ CO₂ + H₂O   NaHCO₃+NH₄Cl

2NaHCO₃  Na₂CO₃+H₂O + CO₂

From equation:

106 g of Na₂CO₃ is produced by = 168 g of NaHCO₃

So, 21.2 g of Na₂CO₃ will be produced by = 168 21.2/106

= 33.6 g of NaHCO₃

(b) For 84 g of NaHCO₃, requiredvolume of CO₂ = 22.4 litre

So, for 33.6 g of NaHCO₃, required volume of CO₂ = 22.4 x 33.6/84

= 8.96 litre

Question 43

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measure at S.T.P.)

NH₄NO₃  N₂O + 2H₂O

(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?

(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam? (Relative molecular mass of ammonium nitrate is 80)

(iii) Determine the percentage of oxygen in ammonium nitrate (O = 16).

Answer 43

(a) NH₄NO₃  N₂O+2H₂O

1mole 1mole 2 mole

1 V       1 V      2 V

44.8 litres of water produced by = 22.4 litres of NH₄NO₃

So, 8.96 litres will be produced by = 22.4 x 8.96/44.8

= 4.48 litres of NH₄NO₃

So, 4.48 litres of N₂O is produced.

(i) 44.8 litre H₂O is produced by = 80 g of NH₄NO₃

So, 8.96 litre H₂O will be produced by = 80 x 8.96/44.8

= 16g NH₄NO₃

(iii) % of O in NH₄NO₃ = 3×16/80 = 60%

Question 44

Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120g of copper oxide? The equation for the reaction is :

Answer 44

Molecular mass of      =(240) g

Molecular mass of  = 2 × 22.4 = 44.8dm³

Molecular mass of  = (192)g

240 g of CuO requires 44.8 dm³ of NH₃

120g of CuO will require

=22.4dm³

Question 45

(a) Calculate the number of moles and the number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?

(b) What is the vapour density of ethylene?

Answer 45

(a) The molecular mass of ethylene(C₂H₄) is 28 g

No. of moles = 1.4/28 = 0.05 moles

No. of molecules = 6.023 x10²³ x 0.05 = 3 x 10²² molecules

Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D

S0, V.D = 28/2 = 14

Question 46

(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na₃AlF₆) correct to the nearest whole number.

(F = 19; Na =23; Al = 27)

(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

2CO + O₂   2CO₂

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Answer 46

(a) Molecular mass of Na₃AlF₆ = 210

So, Percentage of Na = 3x23x100/210 = 32.85%

(b) 2CO + O₂   2CO₂

2 V           1 V           2 V

1 mole of O₂ has volume = 22400 ml

Volume of oxygen used by 2 x 22400 ml CO = 22400 ml

So, Vol. of O₂ used by 560 ml CO =22400 x 560/(2 x 22400)

= 280 ml

So, Volume of CO₂ formed is 560 ml.

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 Previous Year Question of Mole Concept And Stoichiometry

Page-98

 Question 47 (2009)

(a) A gas cylinder of capacity of 20 dm³is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:

(i) 5

(ii) 10

(iii).15

(iv) 20

(b) 

(i) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm³ of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.

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(ii) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find molecular formula of the compound if its relative molecular mass is 37. [N = 14, H = 1].

 (c)

(i) A gas cylinder contains 24 x 10²⁴ molecules of nitrogen gas. If Avogadro’s number is 6 x 10²³ and the relative atomic mass of nitrogen is 14, calculate :

1 Mass of nitrogen gas in the cylinder

2. Volume of nitrogen at STP in dm³.

(ii) Commercial sodium hydroxide weighing 30g has some sodium chloride in it. The Mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is

[Relative molecular mass of NaCl = 58; AgCl = 143]

(iii) A certain gas ‘X’ occupies a volume of 100 cm³ at S.T.P. and weighs 0.5 g. Find its relative molecular mass

Answer 47  (2009)

(a) Mass of gas X =10g

Mass of hydrogen gas= 2

Relative vapour density

= =5

Relative molecular mass of the gas= 2×relative vapour

density = 2×5

=10

(b) 

(i) The combustion reaction 

According to Gay-Lussac’s law,

2 volume of acetylene requires 5 volume of oxygen to burn it

 1 volume of acetylene requires 2.5 volume of oxygen to burn it

  200cm³ requires 2.5×200=500 cm³ of oxygen

2 volume of acetylene on combustion gives 4CO₂ 

1 volume of acetylene on combustion gives 2CO₂ 

200 cc of acetylene on combustion will give  200×2=400cc of CO₂

(iii) Hydrogen = 12.5%

∴ Nitrogen= 100-12.5= 87.5%

Element	Weight	Atomic Weight	Atomic Ratio	Simplest Ratio
N	87.5	14	87.5/14=6.25	6.25/6.25=1
H	12.5	1	12.5/1=12.5	12.5/6.25=2

The Empirical formula of the compound is NH₂

 Empirical formula weight =14+2=16

Relative molecular mass =37

N = 2.3≈2

Molecular formula = n x empirical formula  = 2 x NH₂

=N₂H₄

(C)

(i) Molecules of nitrogen gas in a cylinder = 24 x 10²⁴

Avogadro’s number = 6 x 10²³

1. Mass of nitrogen in a cylinder =

=1120g

2. Volume of nitrogen at stp

and Volume of 28 g of N₂  = 22.4dm³ 

So Volume of 1120g of N₂

hence =896 dm³

Question 48 (2010) 

(a) 

(i) LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as :

(ii) Calculate the percentage of nitrogen and oxygen in ammonium nitrate. [Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16].

(b) 4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.

(i) Write the equation for the reaction.

(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)

(iii). What is the volume of carbon dioxide liberated at STP?

(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).

(v) How many moles of HCl are used in this reaction?

Answer 48 (2010)

(a) 

(i) 10 litres of LPG contains 

Propane

Butane

 

18+16=34 L

(ii) Molecular mass of NH₄(NO₃) =80

H=1, N=14, O=16

% of Nitrogen

As 80 g of NH4(NO3) contains 28 g of nitrogen

100 g of of NH₄(NO₃) will contain

= 35%

% of Oxygen

As,80 g of NH4 (NO3) contains 48 g of oxygen

100 g of of NH₄(NO₃) will contain

= 60%

(b)

(i) Equation for 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 x Relative molecular mass

and = 4.5 x 100

therefore = 450g

(iii).

As, 100g of calcium carbonate gives 22.4dm³ of CO₂

  450 g of calcium carbonate will give

=100.8 L

(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

450 g of calcium carbonate will give

=499.5 g

(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

450 g of calcium carbonate will give

=328.5g

Number of moles of HCl=

=

= 9 moles

Question 49 (2011) 

(a) 

(i) Calculate the volume of 320 g of SO₂ at STP. (Atomic mass: S = 32 and O = 16).

(ii) State Gay-Lussac’s Law of combining volumes.

(iii). Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈). (Atomic mass: C = 12, O = 16, H = I, Molar Volume = 22.4 dm³ at stp.)

(b) 

(i) An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula. [Atomic mass : C = 12, H = 1, Br = 80]

(ii) Calculate the mass of:

1. 10²²atoms of sulphur.

2. 0.1 mole of carbon dioxide.

[Atomic mass : S = 32, C = 12 and O = 16 and

Avogadro’s Number = 6 x 10²³]

Answer 49 (2011)

(a) 

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

Molecular mass of SO₂=32+(2×16)

=64g

As 64 g of SO₂ = 22.4dm³

Then, 320 g of SO₂ =

=112 L

(ii) Gay-Lussac’s law Gay-Lussac’s Law states “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). C₃H₈ + 5O₂→3CO₂ + 4H₂O

Molar mass of propane = 44

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

8.8 g of propane requires          = 22.4 litres

(b) 

(i)

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 = CH₂Br

n(Empirical formula mass of CH₂Br) = Molecular mass (2 × VD)

n(12 + 2 + 80) = 94 × 2

n = 2

Molecular formula = Empirical formula × 2

= (CH₂Br) × 2

= C₂H₄Br₂

(ii) 10²² atoms of sulphur

6.022 × 10²³ atoms of sulphur will have mass = 32 g

10²² atoms of sulphur will have mass =

= 0.533 g

(iii). 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

Question 50 (2012) 

(a) Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

If 9.3 g of phosphorus was used in the reaction, calculate :

(i) Number of moles of phosphorus taken.

(ii) The mass of phosphoric acid formed.

(iii)  The volume of nitrogen dioxide produced at STP.

(b) 

(i) 67.2 litres of hydrogen combines with 44.8 litres of nitrogen to form ammonia under specific conditions as:

Calculate the volume of ammonia produced. What is the other substance, if any, that remains in the resultant mixture?

(ii) The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

(iii). Find the total percentage of Magnesium in magnesium nitrate crystals, Mg(NO₃)₂-6H₂O.

[Mg = 24, N = 14, 0 = 16 and H = 1]

Answer 50

(a)              

(i) Number of moles of phosphorus taken =

= 0.3 mol

(ii) 1 mole of phosphorus gives 98 gm of phosphoric  acid.

So, 0.3 mole of phosphorus gives (0.3 × 98) gm of  phosphoric acid

= 29.4 gm of phosphoric acid

(iii) 1 mole of phosphorus gives 112 L of NOgas at STP.

So, 0.3 mole of phosphorus gives (112 × 0.3) L of

NOgas at STP.

= 33.6 L of NOgas at STP

(b) 

(i) According to the equation

3 volumes of hydrogen produce 2 volumes of ammonia

67.2 litres of hydrogen produce    = 44.8 L

3 volumes of hydrogen combine with 1 volume of ammonia.

67.2 litres of hydrogen combine with    =22.4L Nitrogen left = 44.8 – 22.4 = 22.4 litres

(ii) 5.6 dm³ of gas weighs 12 g

1 dm³ of gas weighs = (12/56) gm

22.4 dm³ of gas weighs = (12/56 × 22.2) gm = 48g

Therefore, the relative molecular mass of gas = 48 gm.

(iii). Molar mass of Mg (NO₃)₂.6H₂O

= 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g

Mass percent of magnesium =

Question 51 (2013) 

(a) 

(i) What volume of oxygen is required to burn completely 90 dm³ of butane under similar conditions of temperature and pressure?

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Page-100

(ii) The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?

(iii). A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

(b) O₂is evolved by heating KClO₃ using MnO₂ as a catalyst.

(i) Calculate the mass of KClO₃ required to produce 6.72 litre of O₂ at STP.

[atomic masses of K = 39, Cl = 35.5, 0 = 16].

(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

(iii). Calculate the volume occupied by 0.01 mole of CO₂ at STP.

Answer 51 (2013)

(a)

(i) 

2 vols. of butane requires O₂ = 13 vols

90 dm³ of butane will require O₂ =         × 90

= 585 dm³

(ii) Molecular mass = 2 × Vapour density

So, molecular mass of gas = 2 × 8 = 16 g

As we know, molecular mass or molar mass occupies 22.4 litres.

That is,

16 g of gas occupies volume = 22.4 litres

So, 24 g of gas will occupy volume

=

(iii). According to Avogadro’s law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

So, molecules of nitrogen gas present in the same vessel = X

(b) 

(i)

3 vols. of oxygen require KClO₃ = 2 vols.

So, 1 vol. of oxygen will require KClO₃ =

So, 6.72 litres of oxygen will require KClO₃

=

22.4 litres of KClO₃ has mass = 122.5 g

So, 4.48 litres of KClO₃ will have mass

=

22.4 litres of oxygen = 1 mole

So, 6.72 litres of oxygen =

No. of molecules present in 1 mole of O₂

= 6.023 × 10²³

 So, no. of molecules present in 0.3 mole of O₂

= 6.023 × 10²³ × 0.3

= 1.806 × 10²³

(iii). Volume occupied by 1 mole of CO₂ at STP = 22.4  litres

So, volume occupied by 0.01 mole of CO₂ at STP = 22.4 × 0.01= 0.224 litres

Question 52 (2014)

(a) 

(i) Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:

What volume of ethyne gas at s.t.p. is require to produce 8.4 dm³ of carbon dioxide at STP [H = 1, C = 12, 0 = 16]

(ii) A compound made up of two elements X and Y has an empirical formula X₂Y. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density (V.D.) 25, find its molecular formula.

(b) A cylinder contains 68 g of Ammonia gas at STP

(i) What is the volume occupied by this gas?

(ii) How many moles of ammonia are present in the cylinder?

(iii). How many molecules of ammonia are present in the cylinder?

Answer 52 (2014) 

(a) 

(i)

2 moles of the   C₂H₂=4 moles of CO₂

x dm³ of Ethine   =8.4 dm³ of CO₂

x=

=4.2 dm³ of Etheline

(ii) Empirical formula= X₂Y

Atomic weight (X)= 10

Atomic weight (Y)= 5

Empirical formula weight = (2 × 10) + 5

=25

= 2

So, the molecular formula = X₂Y×2

= X₄Y₂ 

_(b) 

(i) A cylinder contains 68 g of ammonia gas at STP.

Molecular weight of ammonia = 17 g/mole

68 g of ammonia gas at STP =?

1 mole = 22.4 dm³

∴ 4 mole = 22.4 × 4 = 89.6 dm³

(ii) 4 moles of ammonia gas is present in the cylinder.

(iii). 1 mole = 6.023 × 10²³ molecules

4 moles = 24.092 × 10²³ molecules

Return to:-  Concise Selina Solutions for ICSE Chemistry Class-10

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