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)`.

Explanation

Let the solution contains 100 ml of each varety of `H_(2)SO_(4)`.
Hence, total volume `=200 ml`
Wt. of 100 ml of `H_(2)SO_(4)` solution (70%) `=100xx1.218`
`=121.8g`
Wt of 100 ml of `H_(2)SO_(4)` solution (70%) `=100xx1.610` `:.` Wt of `H_(2)SO_(4)(30%)=121.8 g xx(300)/(100)=36.54g`
Wt . of `H_(2)SO_(4)(70%)=161xx(70)/(100)=112.7=149.24g`
`:.` Wt of `H_(2)SO_(4)` (Solvent) =Wt. solution -wt. of solute
`=(121.8+161)-(149.24)`
`=133.56g`
Moles of `H_(2)SO_(4)=(149.24)/(98)=1.5228`
Mpolality `=(1.5228)/(0.2)=7.6M` M
Molality `=(1.5228)/(13.3.56)xx1000=11.40M`