The molality of $90\mathrm{\%} {\mathrm{H}}_2{\mathrm{SO}}_4$ solution is $[\text{ density }=1.8\mathrm{gm}/\mathrm{ml}]$

The density of solution $= 1.8 gm I ml$

Weight of one litre of solution = 1800 gm

$\therefore$ Weight of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in the solution $=\frac{1800\times 90}{100}$$= 1620 gm$

$\therefore$ Weight of solvent $= 1800 -1620 = 180 gm$

$\therefore$ Molality $=\frac{1620}{98}\times \frac{1000}{180}=91.83$