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