NCERT Solutions Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry

This foundational chapter covers the mole concept, atomic and molecular masses, percentage composition, empirical and molecular formulas, and stoichiometry. Mastering these concepts is essential for all chemistry calculations.

Exercise Questions and Solutions

Q1. Calculate the molecular mass of the following: (i) H₂O (ii) CO₂ (iii) CH₄

Solution:

Atomic masses: H = 1, C = 12, O = 16

• (i) H₂O = 2(1) + 16 = 18 u
• (ii) CO₂ = 12 + 2(16) = 44 u
• (iii) CH₄ = 12 + 4(1) = 16 u

Q2. Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

Solution:

Molar mass of Na₂SO₄ = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 g/mol

• Mass % of Na = (46/142) × 100 = 32.39%
• Mass % of S = (32/142) × 100 = 22.54%
• Mass % of O = (64/142) × 100 = 45.07%

Q3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.

Solution:

Empirical formula = Fe₂O₃

Q4. Calculate the amount of carbon dioxide that could be produced when 1 mole of carbon is burnt in air.

Solution:

C + O₂ → CO₂

From the balanced equation: 1 mole C produces 1 mole CO₂

Mass of CO₂ = 1 × 44 = 44 g

Q5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution.

Solution:

Molarity = moles/volume(L)

Moles = 0.375 × 0.5 = 0.1875 mol

Molar mass of CH₃COONa = 12 + 3(1) + 12 + 2(16) + 23 = 82 g/mol

Mass = 0.1875 × 82 = 15.375 g

Important Formulas

Key Takeaways

• 1 mole contains 6.022 × 10²³ particles (Avogadro’s number)
• Molar mass = mass of 1 mole of substance in grams
• At STP, 1 mole of gas occupies 22.4 L
• Stoichiometry uses balanced equations to calculate amounts
• Limiting reagent determines the amount of product formed
• Empirical formula shows simplest whole number ratio of atoms