Calculate molality of 1  litre solution of 93% (w/v) ${\mathrm{H}}_2{\mathrm{SO}}_4$ (The density of solution is 1.84 g/ml)

Given ${\mathrm{H}}_2{\mathrm{SO}}_4$ is 93% by volume 

weight of ${\mathrm{H}}_2{\mathrm{SO}}_4=93\mathrm{g}$

volume of solution = 100 ml

density of the solution=1.84 g

$\therefore$ Weight of solution $=100\times 1.84\mathrm{g}=184\mathrm{g}$

weight of water = 184 - 93 = 91 g

 $\mathrm{Molality} =\frac{\text{Moles}}{\text{wt. of water in kg}}$

$=\frac{93\times 1000}{98\times 91}=10.42\mathrm{m}$