$$AB_2$$ is 10% dissociated in water to $$A^{2+}$$ and $$B^-$$. The boiling point of 10.0 molal aqueous solution of $$AB_2$$ is ________ °C. (Round off to the Nearest Integer).
[Given: Molal elevation constant of water $$K_b = 0.5$$ K kg mol$$^{-1}$$, boiling point of pure water = 100°C]

We are given that $$AB_2$$ is 10% dissociated in water. It dissociates as $$AB_2 \rightarrow A^{2+} + 2B^-$$. The degree of dissociation is $$\alpha = 0.1$$.

The van't Hoff factor $$i$$ is given by $$i = 1 + (n - 1)\alpha$$, where $$n$$ is the number of ions produced per formula unit. Here $$n = 3$$ (one $$A^{2+}$$ and two $$B^-$$).

So $$i = 1 + (3 - 1)(0.1) = 1 + 0.2 = 1.2$$.

The elevation in boiling point is $$\Delta T_b = i \times K_b \times m = 1.2 \times 0.5 \times 10.0 = 6.0$$ °C.

The boiling point of the solution is $$100 + 6.0 = 106$$ °C.

The answer is $$\mathbf{106}$$ °C.