Find the molality of `H_(2)SO_(4)` solution whose specific gravity is `1.98 mL^(-1)` and `90%` by volume `H_(2)SO_(4)`

Explanation

`H_(2)SO_(4)` is `90%` by volume.
Volume of solution `= 100mL`
`:.` Wt. of `H_(2)SO_(4)= 90 g`
`:.` Moles of `H_(2)SO_(4)= 90//98`
and Weight of solution `=100xx1.98`
`:.` Weight of water `= 198-90`
`108 g`
`:. "Molality"(m=("moles")/("solvent"("in" kg)))=(90)/(98xx108//1000)`
`8.50`