Calculate the amount of benzoic acid $(C_{6}H_{5}COOH)$ required for preparing $250 \ mL$ of $0.15 \ M$ solution in methanol.
(N/A) $0.15 \ M$ solution of benzoic acid in methanol means that $1000 \ mL$ of solution contains $0.15 \ mol$ of benzoic acid.
Therefore,$250 \ mL$ of solution contains $= \frac{0.15 \times 250}{1000} \ mol = 0.0375 \ mol$ of benzoic acid.
Molar mass of benzoic acid $(C_{6}H_{5}COOH) = (7 \times 12) + (6 \times 1) + (2 \times 16) = 122 \ g \ mol^{-1}$.
Hence,the required mass of benzoic acid $= 0.0375 \ mol \times 122 \ g \ mol^{-1} = 4.575 \ g$.
Therefore,$250 \ mL$ of solution contains $= \frac{0.15 \times 250}{1000} \ mol = 0.0375 \ mol$ of benzoic acid.
Molar mass of benzoic acid $(C_{6}H_{5}COOH) = (7 \times 12) + (6 \times 1) + (2 \times 16) = 122 \ g \ mol^{-1}$.
Hence,the required mass of benzoic acid $= 0.0375 \ mol \times 122 \ g \ mol^{-1} = 4.575 \ g$.
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