Find the molality of `H_(2)SO_(4)` solution whose specific gravity is `1.98 mL^(-1)` and `93%` by volume `H_(2)SO_(4)`

Answer

9

Explanation

`H_(2)SO_(4)` is `93%` by volume
`:.` weight of `H_(2)SO_(4) = 93g`
and volume of solution `= 100mL`
`:.` wt.of solution `= 1.98 xx 100 = 198 g`
`:.` wt. of water = wt. of solution -wt.of `H_(2)SO_(4)`
`193 - 93 = 105g`
Now, molarity `= ("Moles of" H_(2)SO_(4))/("Wt.of water"("in" kg))`