Question

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.

Answer 1

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`