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 100 mL of one solution be mixed with 100 mL of the other solution
Mass of 100 mL of 30% `H_(2)SO_(4) = 100 xx 1.218 = 121.8g`
Mass of `H_(2)SO_(4)` in 121.8 g of 30% `H_(2)SO_(4)`
`= (30)/(100)xx121.8g=36.54g`
Mass of water `=(121.8-36.54)=85.26g`
Mass of 100mL of 70% `H_(2)SO_(4) = 100 xx 1.61 = 161.0g`
Mass of `H_(2)SO_(4)` in 161.0g of 70% `H_(2)SO_(4)`
`= (70)/(100) xx 161.0g =112.7g`
Mass of water `= (161.0-112.7)=48.30g`
Total `H_(2)SO_(4)` in solution `=36.54+112.7 =149.24g`
No. of moles of `H_(2)SO_(4)=(149.24)/(98)`
Total mass of water in solution `=(85.26 + 48.30)`
`=133.56g = (133.56)/(1000)kg`
Molality `= ("No. of moles "H_(2)SO_(4))/("Mass of water in kg")=(149.24)/(98)xx(1000)/(133.56)`
`=11.4m`.