Calculate the amount of benzoic acid (C6H5COOH) required

Monjur Alahi

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


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Salim Ahmed Kawsar

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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 =0.15x250/1000      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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