$224 \mathrm{mL}$ of ${\mathrm{SO}}_{2\left(\mathrm{g}\right)}$ at $298 \mathrm{K}$ and $1 \mathrm{atm}$ is passed through $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ solution. The non-volatile solute produced is dissolved in $36 \mathrm{g}$ of water. The lowering of vapour pressure of solution (assuming the solution is dilute) ($\mathrm{P}{°}_{\left({\mathrm{H}}_2\mathrm{O}\right)}=24 \mathrm{mm}$ of $\mathrm{Hg}$) is $\mathrm{x}\times {10}^{-2} \mathrm{mm}$ of $\mathrm{Hg}$ the value of $\mathrm{x}$ is                (Integer answer)

Hence, the correct answer is 18.