Question

The normality of a solution of a mixture containing HCl and H₂SO₄ is N/5. Twenty millilitres of this solution reacts with excess of AgNO₃ solution to give 0.287 g of silver chloride. The percentage of HCl in the mixture by mass is (Ag = 108)

• A. 42.69%
• B. 57.31%
• C. 40%
• D. 33.18%

Answer 1

Answer: (A)

42.69%

Answer 2

1. Calculate moles of AgCl: $n_{AgCl} = \frac{0.287 \text{ g}}{143.5 \text{ g/mol}} = 0.002$ mol.
2. From the reaction HCl + AgNO₃ → AgCl + HNO₃, moles of HCl = moles of AgCl = 0.002 mol.
3. Calculate the mass of HCl in 20 mL: Mass$_{HCl} = 0.002 \text{ mol} \times 36.5 \text{ g/mol} = 0.073$ g.
4. Calculate the normality of HCl in the 20 mL solution: $N_{HCl} = \frac{\text{moles of HCl}}{\text{Volume (L)}} = \frac{0.002 \text{ mol}}{0.020 \text{ L}} = 0.1$ N.
5. The total normality of the mixture is given as N/5, which is $0.2$ N.
6. The normality of H₂SO₄ in the 20 mL solution is $N_{H_2SO_4} = \text{Total Normality} - N_{HCl} = 0.2 \text{ N} - 0.1 \text{ N} = 0.1$ N.
7. Calculate the number of equivalents of H₂SO₄ in 20 mL: Equivalents$_{H_2SO_4} = N_{H_2SO_4} \times \text{Volume (L)} = 0.1 \text{ N} \times 0.020 \text{ L} = 0.002$ equivalents.
8. Calculate the mass of H₂SO₄ in 20 mL: Mass$_{H_2SO_4} = \text{Equivalents}_{H_2SO_4} \times \text{Equivalent weight of H}_2SO_4 = 0.002 \text{ eq} \times 49 \text{ g/eq} = 0.098$ g. (Equivalent weight of H₂SO₄ = Molar mass / 2 = 98 / 2 = 49 g/eq).
9. Calculate the total mass of the mixture in 20 mL: Total Mass = Mass$_{HCl}$ + Mass$_{H_2SO_4} = 0.073 \text{ g} + 0.098 \text{ g} = 0.171$ g.
10. Calculate the percentage of HCl by mass: Percentage$_{HCl} = \frac{\text{Mass}_{HCl}}{\text{Total Mass}} \times 100 = \frac{0.073 \text{ g}}{0.171 \text{ g}} \times 100 = 42.69\%$.