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Calcualate the molarity and molality of 20% aqueous ehtanol `(C_(5) H_(5) OH)` solution by volume. (density of solution `= 0.96 g mL^(-1)`) |
Answer» Correct Answer - C::D `C_(2) H_(5) OH` is 20% by volume, i.e., `20 mL of C_(2) H_(5) OH` is dissolved in `100 mL` of solution Volume of `H_(2) O = 100 - 20 = 80 mL` Weight of `H_(2) O = 80 xx 1 = 80 g (d_(H_(2)O) = 1)` Weight of solution `= 100 xx 0.96 = 96 g` weight of `C_(2) H_(5) OH = 96 - 80 = 16 g` `(Mw of C_(2) H_(5) OH = 12 xx 2 + 5 + 16 + 1 = 46)` `:. M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)) = (16 xx 1000)/(46 xx 100) = 3.48` `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (16 xx 1000)/(46 xx 80) = 4.35` |
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