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What is would be the molality of a solution obtained by mixing equal volumes of 30% by weight `H_(2) SO_(4) (d = 1.218 g mL^(-1))` and 70% by weight `H_(2) SO_(4) (d = 1.610 g mL^(-1))`? If the resulting solution has density `1.425 g mL^(-1)`, calculate its molality. |
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Answer» Let `V ml` of each are mixed. For solution I `H_(2) SO_(4)` is 30 % by wegith. `:.` Weight of `H_(2) SO_(4) = 30 g` and weight of solution `= 100 g` `:.` Volume of solution `= (100)/(1.218) mL` contains `30 g H_(2) SO_(4)` `:. V mL` contains `(30 xx V xx 1.218)/(100) g H_(2) SO_(4)`. For solution II `H_(2) SO_(4)` is 70% by weight. `:.` weight of `H_(2) SO_(4) = 70 g` Weight of solution `= 100 g` `:.` Volume of solution `= (100)/(1.610) mL` i.e., `(100)/(1.610) mL` contains `70 g H_(2) SO_(4)` `:. V mL` contains `= (70 xx V xx 1.610)/(100) H_(2) SO_(4)` `= [ (30 xx 1.218)/(100) + (70 xx 1.610)/(100)] V g = 1.492 V g` ltbRgt Total volume of solution `= 2 V mL` `:.` Molarity `(M)` of solution `= (1.4924 V)/(98 xx (2 V)/(1000)) = 7.61` Now, Weight of total solution `= 2 V xx 1.425 g = 2.85 V g` `:.` Weight of water `= (2.85 V - 1.4924 V) g` `= 1.376 V g` `:.` Molality `(m)` of solution `= (1.7924 V)/(98 xx (1.3576 V)/(1000)) = 11.22` Alternatively : (use the formula) `d_(sol) = M ((Mw_(2))/(1000) + (1)/(m))` `1.425 = 7.61 ((98)/(1000) + (1)/(m))` `:. M = 11.22` |
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