More practice questions

4. Q4. Chlorine is made by 4HCl(aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g). How many grams of HCl react with 5.0 g of MnO₂ (MnO₂ = 86.94, HCl = 36.46 g mol^⁻¹; NCERT Exercise 1.36, page 28)?

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   Moles of MnO₂ = 5.0/86.94 = 0.0575 mol. HCl needed = 4 × 0.0575 = 0.230 mol, that is, 0.230 × 36.46 ≈ 8.4 g.

5. Q5. The density of methanol is 0.793 kg L^⁻¹. What volume of it is needed to make 2.5 L of its 0.25 M solution (CH₃OH = 32.04 g mol^⁻¹; NCERT Exercise 1.12, page 26)?

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   Moles = 0.25 × 2.5 = 0.625 mol, that is, 0.625 × 32.04 = 20.03 g. 0.793 kg L^⁻¹ = 0.793 g mL^⁻¹, so volume = 20.03 ÷ 0.793 ≈ 25.3 mL, reported as 25 mL to two significant figures from 0.25 M and 2.5 L.

6. Q6. Drinking water contains 15 ppm (by mass) of chloroform, CHCl₃. (i) Express this in per cent by mass. (ii) Find the molality of chloroform (CHCl₃ = 119.4 g mol^⁻¹; NCERT Exercise 1.17, page 26).

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   (i) 15 ppm = 15 g in 10^⁶ g = 15/10^⁶ × 100 = 1.5 × 10^⁻³ %. (ii) 15 g is 15/119.4 = 0.126 mol, in about 10^⁶ g = 1000 kg of water, so molality ≈ 0.126/1000 ≈ 1.26 × 10^⁻⁴ m, or 1.3 × 10^⁻⁴ m to two significant figures from 15 ppm.

7. Q7. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. Assume the density of water to be one (NCERT Exercise 1.29, page 28).

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   Take 1000 g (1 L) of water: 1000/18.02 ≈ 55.5 mol. With n mol of ethanol, n/(n + 55.5) = 0.040, so n ≈ 2.31 mol. The density of water gives the solvent volume, not the final solution volume. If the intended additional approximation is V(solution) ≈ 1 L, the result is approximately 2.31 M (2.3 M to two significant figures). The stated data alone do not determine exact molarity: use the measured solution volume or its density. Ethanol and water volumes are not strictly additive, so an exact correction cannot be inferred by simply adding pure-liquid volumes.

8. Q8. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different (NCERT Exercise 1.25, page 27)?

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   0.50 mol Na₂CO₃ is an amount: 0.50 × 106.0 = 53 g of the solid. 0.50 M Na₂CO₃ is a concentration: 0.50 mol dissolved in every litre of solution, so the amount depends on the volume taken. It is also 1.0 M in Na^⁺ ions, since each Na₂CO₃ gives two.

9. Q9. What is the concentration of sugar (C₁₂H₂₂O₁₁, 342.3 g mol^⁻¹) in mol L^⁻¹ if 20 g of it is dissolved in enough water to make 2 L of solution (NCERT Exercise 1.11, page 26)?

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   Moles = 20/342.3 = 0.0584 mol. Molarity = 0.0584/2 ≈ 0.0292 mol L^⁻¹.

10. Q10. Calculate the concentration of nitric acid in moles per litre in a sample with density 1.41 g mL^⁻¹ that is 69% nitric acid by mass, HNO₃ = 63.01 g mol^⁻¹ (NCERT Exercise 1.6, page 26).

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    Mass of 1 L = 1000 mL × 1.41 g mL^⁻¹ = 1410 g. Mass of HNO₃ in it = 0.69 × 1410 = 972.9 g. Moles = 972.9 ÷ 63.01 ≈ 15.4 mol, so about 15.4 mol L^⁻¹ (15 M to the two figures of 69%).