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

No.of moles of benzoic acid required = 0.15 x 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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