Find the molality fo `H_(2)SO_(4)` solution whose specific gravity is `1.98 g mL^(-1)` and 98% (Weight/volume) `H_(2)SO_(4)`.

Answer

A, B, D

Explanation

`H_(2) SO_(4)` is by volume,
weight of `H_(2) SO_(4) = 95 g`
Volume of solution `= 100 mL`
Weight of solution `= 100 xx 1.98 = 198 g`
Weight of `H_(2) O = 198 - 95 = 103 g`
`m = (W_(2) xx 1000)/(Mw xx W_(1)) = (95 xx 1000)/(98 xx 103) = 9.412`