Question -
Answer -
0.15 M solution of benzoic acidin methanol means,
1000 mL of solution contains 0.15mol of benzoic acid
Therefore,250 mL of solution contains = 
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