Mole Concept Dalal Simplified ICSE Chemistry Class-10

Mole Concept Dalal Simplified ICSE Chemistry Class-10 Solutions Chapter 4:-  Solutions of Dr Dalal Simplified ICSE Chemistry by Dr Viraf and J Dalal for Class 10. Step by step Solutions of Dr Dalal Simplified ICSE Chemistry by Dr Viraf and J Dalal  with topics -Gay Lussac ‘Law ,Avogadro numbers Mole Concept Vapour Density and Molecular Weight Percentage Composition Empirical and Molecular Formula Chemical Equations. Numericals of Mole Concept of Simplified Dalal Chemistry.

Mole Concept Dalal Simplified ICSE Chemistry Class-10

–: Select Topics :-

Part A (Mole Concept)

Gay Lussac’ s Law  

Mole Concept -Avogadro Law – Avogadro numbers  

Vapour Density & Molecular Weight

 Part B (Stoichiometry)

Percentage Composition 

Empirical and Molecular Formula 

Chemical Equations 

Get Other Chapter  Dalal Simplified ICSE Chemistry Class-10 Solutions


How to Solve Mole Concept ICSE Chemistry Class-10

Note:– Before viewing Solutions of Mole Concept And Stoichiometry by Dr Viraf and J Dalal Simplified ICSE Chemistry Solutions of Chapter-4 .Read the Chapter-4 Mole Concept And Stoichiometry Carefully to understand the concept in better way .After reading the Chapter-4 Mole Concept And Stoichiometry solve all example of your text book with ICSE Specimen Sample Paper for Class-10 Exam of Council. Focus on Numerical’s Problems. Mole Concept is the Most important Chapter in ICSE Class 10 Chemistry.Previous Year Solved Question Paper for ICSE Board


Part-A  Additional Problems

Gay Lussac’ s Law — Mole Concept Dr Viraf and J Dalal Simplified ICSE Chemistry for Class-10–

 

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:

Ans 1 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:
Ans 2 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:
Ans 3 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:

Ans 4 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:
Ans 5 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:
Ans 6 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:
Ans 7 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry

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:
Ans 8 Gay Lussac Law Mole Concept Dalal Simplified ICSE Chemistry


Mole Concept – Avogadro’S Law – Avogadro’S Number

Question 1.

The mass of 2.8 litres of C02. [C = 12, O = 16]
Answer:
Ans 1 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 2.

The volume occupied by 53.5g of Cl2. [Cl = 35.5]
Answer:

Ans 2 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 3.

The number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]
Answer:

Ans 3 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 4.

The number of

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

Answer:

Ans 4 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 5.

The mass of (Na) sodium which will contain 6.023 × 1023 atoms. [Na = 23]
Answer:

Ans 5 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 6.

The no. of atoms of potassium present in 117 g. of K. [K = 39]
Answer:

Ans 5 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 7.

The number of moles and molecules in 19.S6 g. of Pb (NO3)2. [Pb = 207, N = 14, O = 16]
Answer:

Ans 7 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 8.

The mass of an atom of lead [Pb = 202].
Answer:

Ans 8 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

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:

Ans 9 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 10.

The gram-atoms in 88.75 g of chlorine [Cl = 35.5]
Answer:

Ans 10 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 11.

The number of hydrogen atoms in 0.25 mole of H2SO4.
Answer:

Ans 11 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 12.

The gram molecules in 21 g of nitrogen [N = 14]
Answer:

Ans 12 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 13.

The number of atoms in 10 litres of ammonia [N = 14, H = 1]
Answer:

Ans 13 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 14.

The number of atoms in 60 g of neon [Ne = 20]
Answer:

Ans 14 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 15.

The number of moles of ‘X’ atoms in 93 g of ‘X’ [X is phosphorus = 31]
Answer:

Ans 15 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 16.

The Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm presure. [O = 16]
Answer:

Ans 16 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 17.

The moles of sodium hydroxide contained in 160 g of it. [Na = 23, O = 16, H = 1]
Answer:

Ans 17 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 18.

The weight in g. of 2.5 moles of ethane [C2H6]. [C = 12, H = 1]
Answer:

Ans 18 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

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:

Ans 19 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 20.

The gram atoms in 46 g of sodium [Na = 23]
Answer:

Ans 20 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 21.

The number of moles of KCl03 that will be required to give 6 moles of oxygen.
Answer:

Ans 21 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

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:

Ans 22 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 23.

Has higher number of moles : 5 g. of N2O or 5 g. of NO [N = 14, O = 16]
Answer:

Ans 23 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 24.

Has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]
Answer:

Ans 24 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

Question 25.

Has higher no. of atoms : 1 g of O2 or 1 g of Cl2 [O = 16, Cl = 35.5]
Answer:

Ans 25 Avogadro' Law Mole Concept Dalal Simplified ICSE Chemistry

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:

Ans 1 Vapour Density And Molecular Weight Mole Concept Dalal Simplified ICSE Chemistry

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:

Ans 2 Vapour Density And Molecular Weight Mole Concept Dalal Simplified ICSE Chemistry

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:

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