Question

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

Answer 1

0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L 

i.e. 1000 mL of the solution. Molar mass of benzoic acid (C₆H₅COOH) = 72 + 5 + 12 + 32 + 1 = 122 g mol^-1 

∴ 0.15 mole of benzoic acid = 0.15 × 122g = 18.39 

Thus, 1000 mL of the solution contain benzoic acid = 18.39 

∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250 = 4.575g