Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing 250 mL of 0.15 M solution in methanol.

Explanation

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
`= (0.15xx250)/(1000)` mol of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid `(C_(6)H_(5)COOH)`
`= 7xx12+6xx1+2xx16`
`= 122 g mol^(-1)`
Hence, required benzoic acid
= 0.0375 mol `xx 122 g mol^(-1)`
= 4.575 g.