What 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)`.

Answer

7.6M,11.4 M

Explanation

Suppose 100 mL of each solution is mixed .
Total volume of the solution = 200 mL
Wt. of `H_(2)SO_(4)` in 30% `H_(2)SO_(4) = (100 xx 1.218 xx 30)/(100)`
` = 36.54 g`
Wt. Of `H_(2)SO_(4) ` in 70 %`H_(2)SO_(4) = (100 xx 1.610 xx 70)/(100)`
` = 112.7 g`
Total wt.of `H_(2)SO_(4)= 36.54 + 112.7 = 149.24 g`
Wt. of water ` = (100 xx 1.218 + 100 xx 1.610) - 149.24`
` = 133.56 g`
Molarity ` = (149.24//98)/(200) xx 1000 = 7.61 M`
Molality `= (149.24//98)/(133.56) xx 1000 = 11.4m`