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

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Correct Answer - `4.575g`
Molecular mass of `(C_(6)H_(5)COOH)`
`=12xx6+5+12+32+1=122g mol^(-1)`
mmoles of benzoic aci`=MxxV` ltbr. `=250xx0.15=37.5`
`=37.5xx10^(-3)mol`
Amount of benzoic acid `=` Moles `xx Mw_(2)` ltbr. `=37.5xx10^(-3)xx122`
`=4.575g`

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