Calculate the molality of 1 litre solution of 93% `H_(2)SO_(4) ("weight"//"volume")`. The density of solution is 1.84 g `mL^(-1)`.

Explanation

The solution is `1000mL` containing `930g` of `H_(2)SO_(4)`
The weight of the solution will be `1840g`. The weight of the solvent `(H_(2)O)` will therefore be `(1840-930)` i.e.910g
`:.` molality `=("mole of" H_(2)SO_(4))/("wt.of"H_(2)O(g))xx1000`
`=(930//98)/(910)xx1000=10.428m`