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Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing 250 ml of 0.15 M solution in methanol. |
Answer» Given Molarity = 0.15 M Volume V = 250 ml Molecular weight of benzoic acid `(C_(6)H_(5)COOH)=122` Molarity (M) `=("Weight")/(GMw)xx(1000)/(V(ml))` `0.15=(W)/(122)xx(1000)/(250)` `W=(122xx0.15)/(4)=4.575" gms."` |
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