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Calculate molality of 1 litre solution of 93% `H_(2) SO_(4)` by volume. The density of solution is `1.84 g mL^(-1)`. |
Answer» Given `H_(2) SO_(4)` is 93% by volume `:.` Weight of `H_(2) SO_(4) = 93 g` Volume of solution `= 100 mL` Weight of solution `= 100 xx 1.84 = 184 g` Weight of water `= 184 - 93 = 91 g` Molality `(m) = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (93 xx 1000)/(98 xx 91) = 10.42` |
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