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Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

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Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

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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 = \(\frac{0.15 \times 250}{1000}\) 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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