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Some Basic Concepts of Chemistry

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199 questions

A solution of HNO₃ has a molarity of 1.5 M and a density of 1.05 g/mL. What is its molality? (Molar mass: HNO₃ = 63 g/mo

Mass of 1 L solution = 1000 × 1.05 = 1050 g. Mass of HNO₃ = 1.5 × 63 = 94.5 g. Mass of water = 955.5 g = 0.9555 kg. Molality = 1.5 / 0.9555 ≈ 1.57 m.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

What volume of CO₂ at STP is produced when 10 g of C₄H₁₀ is burned completely? (Molar mass: C₄H₁₀ = 58 g/mol)

Reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Moles of C₄H₁₀ ≈ 0.1724 mol. 2 mol produce 8 mol CO₂; 0.1724 mol produce ≈ 0.6896 mol. Volume = 0.6896 × 22.4 ≈ 15.45 L.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

What is the mass of SO₃ produced when 16 g of SO₂ reacts with 8 g of O₂? (Molar masses: SO₂ = 64 g/mol, SO₃ = 80 g/mol,

Reaction: 2SO₂ + O₂ → 2SO₃. Moles: SO₂ = 0.25 mol, O₂ = 0.25 mol. 0.25 mol SO₂ produces 0.25 mol SO₃ = 0.25 × 80 = 20 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent

What mass of Fe can be obtained from 16 g of Fe₂O₃ using excess CO? (Molar masses: Fe₂O₃ = 160 g/mol, Fe = 56 g/mol)

Reaction: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Moles of Fe₂O₃ = 16/160 = 0.1 mol. 0.1 mol Fe₂O₃ produces 0.2 mol Fe = 0.2 × 56 = 11.2 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent