AB _ 2 is 10 % dissociated in water to A ^ 2 + and B ^ - . The...

$\begin{cases} & {AB}_{2} & \to & {A}^{2 +} & + & 2{B}^{-} \\ t = 0 & a & & 0 & & 0 \\ t = 0 & a - a \alpha & & a \alpha & & 2a \alpha\end{cases}$

${n}_{T} = a - a\alpha + a\alpha + 2a\alpha$

$= a\left(1 + 2\alpha\right)$

so $i = 1 + 2\alpha$

Now ${\Delta T}_{b} = i \times m \times {K}_{b}$

${\Delta T}_{b} = \left(1 + 2\alpha\right) \times m \times {K}_{b}$

$\alpha = 0 . 1m = 10{K}_{b} = 0 . 5$

${\Delta T}_{b} = 1 . 2 \times 10 \times 0 . 5$

$= 6$

So boiling point $= 106$