Question:
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
0.15 M solution of benzoic acid in methanol means,
1000 mL of solution contains 0.15 mol of benzoic acid
Therefore, $250 \mathrm{~mL}$ 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
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