Question

Calculate the amount of benzoic acid $(C_6H_5COOH)$ required for preparing 250 mL of 0.15 M solution in methanol.

Answer 1

The correct answer is: 4.575 g
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 $=\frac{0.15\times250}{1000} mol$ of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid $(C_6H_5COOH) = 7\times12+6\times1+2\times16$
$= 122 g mol^{-1}$
Hence, required benzoic acid $= 0.0375 mol \times 122 g mol$
= 4.575 g