Ncert Exercises & Solutions

2.30 Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.

0.15 M solution of benzoic acid in methanol means,

1000 mL of solution contains 0.15 mol of benzoic acid

Therefore, 250 mL of solution contains = $\frac{0.15 \times 250}{1000} mol of benzoic acid$ = 0.0375 mol of benzoic acid

Molar mass of benzoic acid (C₆H₅COOH) = 7 × 12 + 6 × 1 + 2 × 16 = 122 g mol⁻¹

Hence, required benzoic acid = 0.0375 mol × 122 g mol⁻¹ = 4.575 g