$224 \mathrm{~mL}$ of $\mathrm{SO}_{2(\mathrm{~g})}$ at $298 \mathrm{~K}$ and 1 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)
$\left(\mathrm{P}_{\left(\mathrm{H}_{2} \mathrm{O}\right)}=24 \mathrm{~mm}\right.$ of $\left.\mathrm{Hg}\right)$ is $x \times 10^{-2} \mathrm{~mm}$ of $\mathrm{Hg}$, the value of $x$ is___________ (Integer answer)
$\mathrm{P}^{\mathrm{s}} \quad=\mathrm{P}^{0} . \mathrm{X}_{\text {solvent }}$
$=24 \times \frac{2}{\left(2+15 \times 10^{-3}\right)}$
$=23.82$
$\Delta \mathrm{P}=0.18$ torr $=18 \times 10^{-2}$ torr.
$=24 \times \frac{(1.6+18.4)}{2020}$
$=0.2376=23.76 \times 10^{-2}$