224 mL of $$SO_{2(g)}$$ 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) ($$P^\circ_{H_2O} = 24$$ mm of Hg) is $$x \times 10^{-2}$$ mm of Hg the value of $$x$$ is ______ (Integer answer)

First, we find the moles of $$SO_2$$ gas. At 298 K and 1 atm, using the ideal gas law:

$$n_{SO_2} = \frac{PV}{RT} = \frac{1 \times 0.224}{0.0821 \times 298} = \frac{0.224}{24.466} \approx 0.00916 \text{ mol}$$

Moles of NaOH $$= 0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ mol}$$

When $$SO_2$$ reacts with NaOH, the reaction producing the normal salt is: $$SO_2 + 2NaOH \to Na_2SO_3 + H_2O$$

This reaction requires 2 mol NaOH per mol $$SO_2$$. With 0.01 mol NaOH, the maximum $$SO_2$$ that can react by this route is $$0.01/2 = 0.005$$ mol. Since we have 0.00916 mol $$SO_2$$ available, NaOH is the limiting reagent for forming $$Na_2SO_3$$.

Moles of $$Na_2SO_3$$ formed $$= 0.005$$ mol. The excess $$SO_2 = 0.00916 - 0.005 = 0.00416$$ mol escapes as gas since it is not fully consumed.

The non-volatile solute in solution is $$Na_2SO_3$$. It dissociates as: $$Na_2SO_3 \to 2Na^+ + SO_3^{2-}$$, so the van't Hoff factor $$i = 3$$.

Moles of water $$= \frac{36}{18} = 2$$ mol.

For a dilute solution, the lowering of vapour pressure is given by Raoult's law:

$$\Delta P = P^\circ \times \frac{i \times n_{\text{solute}}}{n_{\text{solvent}}} = 24 \times \frac{3 \times 0.005}{2} = 24 \times 0.0075 = 0.18 \text{ mm Hg}$$

Therefore, $$\Delta P = 18 \times 10^{-2}$$ mm Hg, so $$x = \boxed{18}$$.