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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. |
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Answer» Solution :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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