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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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Calculate the amound of benzoic acid (C6H5COOH) required for preparing 250mL of 0.15M solution in methanol. Select the correct answer from above options

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Correct Answer - 4.575g Molecular mass of (C6H5COOH) =12×6+5+12+32+1=122gmol-1 mmoles of benzoic aci=M×V ltbr. =250×0.15=37.5 =37.5×10-3mol Amount of benzoic acid = Moles ×Mw2 ltbr. =37.5×10-3×122 =4.575g

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