New Simplified Chemistry Class 10 ICSE Solutions – Mole Concept and Stoichiometry : Gay Lussac’s Law – Avogadro’s Law

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Practice Questions

Question 1.
What volume of oxygen would he required to burn completely 400 ml of acetylene (C2H2) ? Calculate the volume of CO2 formed.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 1

Question 2.
2500 cc of oxygen was burnt with 600 cc of ethane (C2H6). Calculate the volume of unused oxygen and the volume of carbon dioxide formed, after writing the balanced equation :
Ethane + Oxygen → Carbon dioxide + Water vapour
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 2
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 3

Question 3.
80 cm3 of methane are mixed with 200 cm3 of pure oxygen at similar temperature and pressure. The mixture is then ignited. Calculate the composition of resulting mixture if it is cooled to initial room temperature and pressure.
CH4 + 2O2 → CO2 + 2H2O
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 4

Question 4.
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:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 5

Question 5.
What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions ?
C3H8 + 5O2 → 3CO2 + 4H2O
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 6

Question 6.
450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Calculate the composition of resulting mixture.
2NO + O2 → 2NO2
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 7

Question 7.
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 were unchanged oxygen. Which law does their experiment supports ? Explain with calculations.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 8
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 9
Avogadro’s Law : “Under the same conditions of temperature and pressure equal volumes of all gases contain die same number of molecules.”
If we assume that 7 litre of oxygen gas contains ‘n’ molecules of the gas then by Avogadro’s Law :

  1. 1 litre of oxygen will contain ‘n’ molecules of hydrogen
  2. 1 litre of nitrogen will contain ‘n’ molecules of nitrogen.
  3. 1 litre of any gas will contain ‘n’ molecules of that gas. e.g.

New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 10

Relative Molecular Mass : “It is the number that represents how many times one molecule of a substance is heavier than one atom of hydrogen whose weight has been taken unity or 1/12 of 6C12.
Relative Molecular Mass : “It is the number that represents how many times one molecule of a substance is heavier than one atom of hydrogen whose weight has been taken unity or 1/12 of 6C12.
Avogadro’s Number : “The number of atoms present in 12 g (gm atomic mass) of 6C12 is called Avogadro’s number.
Na or L = 6.023 × 1023

New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 11
Mole : Mole is the mass of substance containing particles equal to Avogadro’s number i.e. 6.023 × 1023.
Gram Atom : “The relative atomic mass of an element expressed in grams is called gram atom.
Gram Mole : “The relative atomic mass of a substance expressed in grams is called gram mole.
Molar Volume : Volume occupied by one mole of any gas at STP is called molar volume.

Applications Of Avogadro’S Law :

  1. Determines the atomicity of the gas.
  2. Determines the molecular formula of a gas.
  3. Determines the relation between molecular weight and vapour density.
  4. Explains Gay-Lussac’s law.
  5. Determines the relationship between gram molecular weight and gram molecular volume.

1. Determines the atomicity of a gas :
Atomicity : The number of atoms present in one molecule of that element is called atomicity.
Monoatomic : Elements which have one atom in their molecules e.g. Helium, Neon.
Diatomic : Elements which have two atoms in their molecule e.g. Hydrogen, nitrogen, oxygen.
e.g. of Determination of atomicity of a gas :
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 12
A molecules of nitrogen contains two atoms : atomicity – Diatomic

2. Determines the molecular formula of a gas.
Molecular formula : A chemical formula which gives the actual or exact number of atoms of the elements present in one molecule of a compound.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 13

3. Determines the relation between molecular weight and vapour density.
Molecular weight: It is the ratio of the weight of 1 molecule of a substance to the weight of one atom of hydrogen.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 14
Vapour Density : It is the ratio of the mass of a certain volume of gas or vapour of the mass of the same volume of hydrogen.
Mol. wt. = 2× vapour density

4. Explains Gay-Lussac’s Law :
Gay-Lussac’s Law is explained by Avogadro’s Law which states “Under similar conditions of temperature and pressure, equal volumes of different gases have same number of molecules.”
Since substances react in simple ratio of number of molecules, volumes of gaseous reactants and products will also bear a simple ratio to one another. This is what Gay-Lussac’s Law says.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 15

5. Determines relationship between gram molecular mass and gram molecular volume :
Gram molecular mass is the relative molecular mass of a substance expressed in grams. It is also called gram molecule of that element.
Gram molecular volume : The volume occupied by e.g. molecular wt. of a gas at s.t.p.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 16

Additional Problems

Q.1. Lussac’S Law

Question 1.
Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with 6 litres of hydrogen. All volumes measured at s.t.p.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 17

Question 2.
2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 18

Question 3.
20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure. What will be the volume and composition of the mixture of the gases when they are cooled to room temperature.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 19

Question 4.
224 cm3 of ammonia undergoes catalytic oxidation in presence of Pt to given nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measured at room temperature and pressure.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 20
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 21

Question 5.
Acetylene [C2H2] burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn 50 cm3 of acetylene. [Assume air contains 20% oxygen].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 22

Question 6.
On igniting a mixture of acetylene [C2H2] and oxygen, 200 cm3 of CO2 is collected at s.t.p. Calculate the volume of acetylene & O2 at s.t.p. in the original mixture.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 23

Question 7.
Ammonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitable conditions. Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia formed if only 10% conversion has taken place.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 24

Question 8.
100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temp. Calculate the composition of the resultant mixture. [Water gas contains CO and H2 in equal ratio]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 25

Q.2. Mole Concept – Avogadro’S Law – Avogadro’S Number
Calculate the following : [all measurement at s.t.p. or as stated in the problem]

Question 1.
The mass of 2.8 litres of C02. [C = 12, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 27
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 28

Question 2.
The volume occupied by 53.5g of Cl2. [Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 29

Question 3.
The number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 30

Question 4.
The number of

  1. molecules [S = 32]
  2. atoms in 192 g. of sulphur. [S8]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 31

Question 5.
The mass of (Na) sodium which will contain 6.023 × 1023 atoms. [Na = 23]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 32

Question 6.
The no. of atoms of potassium present in 117 g. of K. [K = 39]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 33

Question 7.
The number of moles and molecules in 19.S6 g. of Pb (NO3)2. [Pb = 207, N = 14, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 34

Question 8.
The mass of an atom of lead [Pb = 202].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 35

Question 9.
The number of molecules in 1 1/2 litres of water. [density of water 1.0 g./cm3. — ∴ mass of water = volume × density]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 36
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 37

Question 10.
The gram-atoms in 88.75 g of chlorine [Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 38

Question 11.
The number of hydrogen atoms in 0.25 mole of H2SO4.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 39

Question 12.
The gram molecules in 21 g of nitrogen [N = 14]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 40

Question 13.
The number of atoms in 10 litres of ammonia [N = 14, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 41
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 42

Question 14.
The number of atoms in 60 g of neon [Ne = 20]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 43

Question 15.
The number of moles of ‘X’ atoms in 93 g of ‘X’ [X is phosphorus = 31]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 44

Question 16.
The Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm presure. [O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 45
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 46

Question 17.
The moles of sodium hydroxide contained in 160 g of it. [Na = 23, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 47

Question 18.
The weight in g. of 2.5 moles of ethane [C2H6]. [C = 12, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 48

Question 19.
The molecular weight of 2.6 g of a gas which occupies 2.24 lits. at 0°C and 760 mm press.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 49

Question 20.
The gram atoms in 46 g of sodium [Na = 23]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 50

Question 21.
The number of moles of KCl03 that will be required to give 6 moles of oxygen.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 51

Question 22.
The weight of the substance of its molecular weight is 70 and in the gaseous form occupies 10 lits. at 27°C and 700 mm pressure.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 52

Question 23.
Has higher number of moles : 5 g. of N2O or 5 g. of NO [N = 14, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 53

Question 24.
Has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 54

Question 25.
Has higher no. of atoms : 1 g of O2 or 1 g of Cl2 [O = 16, Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 55

Q.3. Vapour Density And Molecular Weight

Question 1.
500 ml. of gas ‘X’ at s.t.p. weighs 0.50 g. Calculate the vapour density and molecular weight of the gas. [1 lit. of H2 at s.t.p. weighs 0.09 g].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 56

Question 2.
A gas cylinder holds 85 g of a gas ‘X’. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure Calculate the molecular weight of ‘X’.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 57

Question 3.
Calculate the relative molecular mass [molecular weight] of 290 ml. of a gas ‘A’ at 17°C and 1520 mm pressure which weighs 2.73 g at s.t.p. [1 litre of hydrogen at s.t.p. weighs 0.09 g.]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 58
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 59

Question 4.
State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if its V.D. is 8.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 60

Question 5.
Calculate the atomcity of a gas X [at. no. 35.5] whose vapour density is equal to its relative atomic mass.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 61

Question 6.
Calculate the relative molcular mass and vapour density of methyl alcohol [CH3OH] if 160 g. of the alcohol on vaporization has a volume of 112 litres at s.t.p.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 62

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