${\mathrm{AB}}_2$ is $10\%$ dissociated in water to $A^{2+}$ and $B^{-}$. The boiling point of a $10.0$ molal aqueous solution of $A{B}_2$ is ____________ $°C$. (Round off to the Nearest Integer).

[Given: Molal elevation constant of water $K_b=0.5Kkgmo{l}^{-1}$ boiling point of pure water $=100°C$].

Step 1: Calculate the elevation in boiling point:

From the dissociation of ${\mathrm{AB}}_2$

$$\begin{gathered} {\mathrm{AB}}_2\underset{}{\overset{}{\to }}{\mathrm{A}}^{2+}+2{\mathrm{B}}^{-} \\ \mathrm{t}=0\mathrm{a}00 \\ \mathrm{t}=\mathrm{t}\mathrm{a}-\mathrm{a\alpha }\mathrm{a\alpha }2\mathrm{a\alpha } \end{gathered}$$

Since, ${\mathrm{AB}}_2$ is dissociated $10\%$ hence $\mathrm{\alpha }=0.1$

$$\begin{gathered} \mathrm{n}=\mathrm{a}-\mathrm{a\alpha }+\mathrm{a\alpha }+2\mathrm{a\alpha } \\ \mathrm{n}=\mathrm{a}(1+2\mathrm{\alpha }) \\ \mathrm{i}=1+2\mathrm{\alpha } \end{gathered}$$

From the formula:

$$\begin{gathered} ∆T_b=i{K}_{b}m \\ =(1+2\alpha )\times 0.5\times 10 \\ =5+10\alpha \\ =5+(10\times 0.10) \\ =6K \end{gathered}$$

Step 2: Calculate the boiling point of the solution

Let the boiling point of the solution be $T_b$

Boiling point of water, $T_b°=100°C=373K$

$$\begin{gathered} ∆T_b=T_b-T_b° \\ 6=T_b-373 \\ \Rightarrow T_b=379K=106°C \end{gathered}$$

Thus, the boiling point of a $10.0$ molal aqueous solution of $A{B}_2$ is $\mathbf{106}°C$.