224 mL of ${\mathrm{SO}}_{2\left(\mathrm{g}\right)}$ 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) ($P{°}_{\left(H_{2}O\right)}=$24 mm of Hg) is $x\times {10}^{-2}$ mm of Hg, the value of x is ________ 

$\begin{array}{l}\begin{array}{l}\text{ no. of moles of }{\mathrm{SO}}_2\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}} =\frac{1\times 224\times {10}^{-3}}{0.0821\times 298}\\ =9\times {10}^{-3}\end{array}\\ =9\text{ milli moles }\\ \text{ mill moles }\mathrm{NaOH}=0.1\times 100=10\\ {\mathrm{SO}}_2+2{\mathrm{NaOH}}_{\left(aq\right)}\to {\mathrm{Na}}_2{\mathrm{SO}}_3+H_{2}O\end{array}$

9                   10                        0

9-5             10-10                     5

No. of moles of  ${\mathrm{H}}_2\mathrm{O}=\frac{36}{18}=2$

$\begin{array}{ll}\frac{P_o-P_s}{P_0}=\frac{3\times 5\times {10}^{-3}}{2} & \text{ Po-Ps }=\frac{15\times 24\times {10}^{-3}}{2}\\ \text{ Po }-Ps=18\times {10}^{-2}& \end{array}$