# Calculate the amount of benzoic acid (C6HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.

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Calculate the amount of benzoic acid (C6HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.

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0.15 M solution means that 0.15 moles of C6H5COOH is present in 1L = 1000 mL of the solution

Molar mass of C6H6COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol-1

Thus, 1000 mL of solution contains benzoic acid = 18.3 g

250 mL of solution will contain benzoic acid = $\frac{18.3}{1000}$ x 250 = 4.575 g