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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 than 0.15 mole of benzoic acid is dissolved in 1L of solution.
Molar mass of `C_(6)H_(5)COOH`
`=12xx6+5+12+2xx16+1=122 " g mol"^(-1)`
`therefore 0.15 " mol of " C_(6)H_(5)COOH=0.15xx122=18.3 g`
Thus, 1L or 1000 mL of solution contain
`=18.3 " g of " C_(6)H_(5)COOH`
` therefore 250 mL` of the solution will contain
`(18.3)/(1000)xx250=4.575 " g of " C_(6)H_(5)COOH`.

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