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Calculate the amount of benzoic acid (C(...

Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing `250mL` of `0.15 M` solution in methanol.

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0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L, i.e., 1000 mL of the solution.
Molar mass of benzoic acid `(C_(6)H_(5)COOH)=72+5+12+32+1="122 g mol"^(-1)`
`therefore" 0.15 mole of benzoic acid "=0.15xx122 g=18.3g`
`therefore" 250 mL of the solution will contain benzoic acid "=(18.3)/(1000)xx250="4.575 g."`
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