A solution containing $$10 \text{ g}$$ of an electrolyte $$AB_2$$ in $$100 \text{ g}$$ of water boils at $$100.52°C$$. The degree of ionization of the electrolyte $$(\alpha)$$ is ______ $$\times 10^{-1}$$. (nearest integer) [Given: Molar mass of $$AB_2 = 200 \text{ g mol}^{-1}$$, $$K_b$$ (molal boiling point elevation const. of water) $$= 0.52 \text{ K kg mol}^{-1}$$, boiling point of water $$= 100°C$$; $$AB_2$$ ionises as $$AB_2 \rightarrow A^{2+} + 2B^-$$]

Find the degree of ionization $$\alpha$$ for electrolyte $$AB_2$$.

Boiling point elevation is calculated as $$\Delta T_b = 100.52 - 100 = 0.52°C$$.

Since $$\Delta T_b = iK_bm$$, where $$m$$ is the molality, we first determine $$m$$.

Substituting the values gives $$m = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \text{ mol/kg}$$.

Using $$\Delta T_b = iK_bm$$ again with $$0.52 = i \times 0.52 \times 0.5$$ leads to $$i = \frac{0.52}{0.26} = 2$$.

The dissociation of $$AB_2$$ is represented by $$AB_2 \rightarrow A^{2+} + 2B^-$$, yielding $$n = 3$$ ions.

Relating the van't Hoff factor to the degree of ionization via $$i = 1 + (n-1)\alpha$$, we have $$2 = 1 + 2\alpha$$, which gives $$\alpha = 0.5$$.

Therefore, $$\alpha = 0.5 = 5 \times 10^{-1}$$.

The correct answer is 5.