Calculate the molality of 1 litre solution of 93% ${\mathrm{H}}_2{\mathrm{SO}}_4$ (weight/volume). The density of the solution is 1.84 g/mL.

Mass of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in 100ml of 93% ${\mathrm{H}}_2{\mathrm{SO}}_4$solution = 93g

$\therefore$ Mass of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in 1000 ml of the ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution = 930g

Mass of 1000 ml ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution = 1000 × 1.84 = 1840g

Mass of water in 1000 ml of solution = 1840 – 930 = 910 g

$\text{ Moles of }{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\mathrm{Wt}\text{ . of }{\mathrm{H}}_2{\mathrm{SO}}_4}{\text{ Mol }\mathrm{Wt}\text{ . of }{\mathrm{H}}_2{\mathrm{SO}}_4}=\frac{930}{98}$

$\therefore$Moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in 1 kg of water $=\frac{930}{98}\times \frac{1000}{910}=10.43 \mathrm{mol}{ \mathrm{kg}}^{-1}$

$\therefore$Molality of 1 litre solution = 10.43