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Calculate the amount of benzoic acid (C_(6) H_(5) COOH) required for preparing 250 mL of 0.15 Msolution in methanol. |
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Answer» Solution :`0.15` M solution means that `0.15 ` moles of `C_(6) H_(5) COOH` is present in 1L `=1000 ML ` of the solution Molar mass of `C_(6) H_(5) COOH = 72 + 5 + 12 + 32 + 1 = 122 G mol ^(-1)` `therefore 0.15` mole of benzoic acid `=0.15 xx 122 g =18.3 g` Thus, 1000 mL of solution contains benzoic acid `=18.3g` `therefore 250mL` of solution will contain benzoic acid `= (18.3)/(1000) xx 250=4.575g` |
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