Question

what will be the amount of benzoic acid (C6H6COOH) required for preparing 200 ml of 0.1M solution in methanol(CH3OH)​

Answer 1

Number of moles of benzoic acid required =0.15 × $\frac{250}{1000}$

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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