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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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Number of moles of benzoic acid required = 0.15 × 250/ 1000 = 0.0375 moles

Molar mass of benzoic acid = 7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122 g/mol

Mass of benzoic acid required = 122 × 0.0375 = 4.575 g

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