224mL of ${\mathrm{SO}}_{2( \mathrm{g})}$ at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36 g of water. The lowering of vapour Pressure of solution (assuming the solution is dilute) $\left({\mathrm{P}}_{\left({\mathrm{H}}_2\mathrm{O}\right)}^0=24 \mathrm{mm}\right.$ of $\left.\mathrm{Hg}\right)$ is $x\times {10}^{-2} \mathrm{mm}$ of Hg, the value of x is  (Integer answer)

  $$\begin{gathered} {\mathrm{n}}_{{\mathrm{so}}_2}=\frac{224\times {10}^3}{0.0821\times 298}=9.2 \mathrm{mmol} \\ {\mathrm{n}}_{\mathrm{NaOH}}=\mathrm{MV}\left(\mathrm{L}\right)=0.1\times 0.1=0.01\mathrm{mol}=10\mathrm{mmol} \end{gathered}$$        

$$\begin{gathered} {\mathrm{SO}}_2+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O} \\ 9.2 10 \mathrm{mmol} \\ \mathrm{Here} \mathrm{NaOH} \mathrm{is} \mathrm{the} \mathrm{Limiting} \mathrm{reagent} \end{gathered}$$

$$\begin{gathered} {\mathrm{n}}_{{\mathrm{Na}}_2{\mathrm{SO}}_3} \mathrm{formed}=\frac{10}{2}=5\mathrm{mmol}=0.005 \mathrm{mol} \\ {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}} \mathrm{formed}=\frac{36}{18}=2 \\ \mathrm{for} {\mathrm{Na}}_2{\mathrm{SO}}_3 \mathrm{solution}, \mathrm{i}=3 \end{gathered}$$    

$\Delta \mathrm{P}={\mathrm{P}}^0.{\mathrm{X}}_{\text{solute }}$
$=3\times 24\times \frac{0.005}{2.005}=0.18=18\times {10}^{-2}$

Hence, the required round-off answer is 18.