Some question of Class 11 Chemistry (NCERT) Chapter 1: Some Basic Concepts of Chemistry with answers topic wise are given below..
1. Laws of Chemical Combination
Q1. State the Law of Conservation of Mass.
Ans: Mass can neither be created nor destroyed in a chemical reaction (Total mass of reactants = Total mass of products).
Q2. 2.5 g of calcium carbonate decomposes to give 1.4 g of calcium oxide and 0.7 g of carbon dioxide. Verify the law of conservation of mass.
Ans:
Mass of reactants = 2.5 g
Mass of products = 1.4 g (CaO) + 0.7 g (CO₂) = 2.1 g
Discrepancy due to experimental error. The law holds theoretically.
Q3. Copper oxide (CuO) has 79.9% Cu and 20.1% O. A sample of CuO from Mars has 77.5% Cu. Which law does this violate?
Ans: Law of Definite Proportions (composition must be fixed).
2. Mole Concept & Molar Mass
Q4. Calculate the molar mass of:
(i) H₂SO₄
(ii) C₂H₅OH
Ans:
(i) 2(1) + 32 + 4(16) = 98 g/mol
(ii) 2(12) + 6(1) + 16 = 46 g/mol
Q5. How many moles are in 11 g of CO₂?
Ans: Molar mass of CO₂ = 44 g/mol
Moles = 11/44 = 0.25 mol
Q6. Find the number of molecules in 360 g of glucose (C₆H₁₂O₆).
Ans: Molar mass = 180 g/mol
Moles = 360/180 = 2 mol
Molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴
3. Percentage Composition & Empirical Formula
Q7. Find the percentage of nitrogen in urea (NH₂CONH₂).
Ans: Molar mass = 60 g/mol
Mass of N = 28 g
% N = 28/60×100% = 46.67%
Q8. A compound has 68.4% Cr and 31.6% O. Find its empirical formula.
Ans:
| Element | % | Atomic Mass | Moles | Ratio |
|---|---|---|---|---|
| Cr | 68.4 | 52 | 68.4/52 ≈ 1.315 | 1.315/1.315 = 1 |
| O | 31.6 | 16 | 31.6/16 ≈ 1.975 | 1.975/1.315 ≈ 1.5 |
| Multiply by 2: Cr:2, O:3 → Cr₂O₃ |
4. Stoichiometry
Q9. What mass of oxygen is required to burn 24 g of methane (CH₄)?
Ans:
CH₄ + 2O₂ → CO₂ + 2H₂O
Moles of CH₄ = 24161624 = 1.5 mol
Moles of O₂ required = 3 mol (since 1 mol CH₄ : 2 mol O₂)
Mass of O₂ = 3 × 32 = 96 g
Q10. How much CO₂ is produced when 6.0 g of carbon is burnt?
Ans:
C + O₂ → CO₂
Moles of C = 612126 = 0.5 mol
Mass of CO₂ = 0.5 × 44 = 22 g
5. Concentration Terms
Q11. Define molarity and molality.
Ans:
- Molarity (M): Moles of solute per liter of solution.
- Molality (m): Moles of solute per kg of solvent.
Q12. Calculate molarity of 4.9 g H₂SO₄ in 500 mL solution.
Ans: Moles of H₂SO₄ = 4.998984.9 = 0.05 mol
Volume = 0.5 L
Molarity = 0.050.50.50.05 = 0.1 M
Q13. Find molality of 20% (w/w) glucose (C₆H₁₂O₆) solution.
Ans: In 100 g solution: 20 g glucose, 80 g solvent (water)
Moles of glucose = 20180=1918020=91 mol
Molality = 1/90.080.081/9 = 1.389 m
6. Limiting Reagent & Yield
Q14. 2.0 g H₂ reacts with 8.0 g O₂ to form water. Identify the limiting reagent.
Ans:
2H₂ + O₂ → 2H₂O
Moles of H₂ = 22=122=1 mol
Moles of O₂ = 832=0.25328=0.25 mol
O₂ required for 1 mol H₂ = 0.5 mol (since 2 mol H₂ : 1 mol O₂)
Available O₂ (0.25 mol) < required (0.5 mol) → O₂ is limiting.
Q15. If 4.0 g CaCO₃ decomposes to give 1.792 g CaO, calculate % yield.
Ans:
CaCO₃ → CaO + CO₂
Theoretical yield: 56 g CaO from 100 g CaCO₃
For 4 g CaCO₃ → 56100×4=2.2410056×4=2.24 g CaO
% yield = 1.7922.24×1002.241.792×100 = 80%
7. Significant Figures & Scientific Notation
Q16. Express 0.00456 in scientific notation with 3 significant figures.
Ans: 4.56 × 10⁻³
Q17. Calculate: (3.24×10−3)(1.2×102)(1.2×102)(3.24×10−3) with correct significant figures.
Ans: 3.24×10−31.2×102=2.7×10−51.2×1023.24×10−3=2.7×10−5 (2 significant figures, as per least precise term 1.2).
8. Mixed Problems
Q18. How many significant figures are in:
(i) 0.050
(ii) 126,000
Ans: (i) 2 (5 and 0 after decimal) (ii) 3 (1,2,6)
Q19. Find the empirical formula of a compound with C=40%, H=6.67%, O=53.33%.
Ans:
| Element | % | Moles | Simplest Ratio |
|---|---|---|---|
| C | 40 | 40/12 ≈ 3.33 | 1 |
| H | 6.67 | 6.67/1 ≈ 6.67 | 2 |
| O | 53.33 | 53.33/16 ≈ 3.33 | 1 |
| → CH₂O |
Q20. What volume of 0.5 M HCl is needed to neutralize 100 mL of 0.2 M NaOH?
Ans:
HCl + NaOH → NaCl + H₂O
Moles of NaOH = 0.2 × 0.1 = 0.02 mol
Moles of HCl required = 0.02 mol
Volume of HCl = 0.02/0.5=0.04 L = 40 mL