$AB_2 $ is 10% dissociated in water to $A^{2+}$ and $B^-$. The boiling point of a 10.0 molal aqueous solution of $AB_2 $ is _ _ _ _ $^ \circ 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 ^\circ C $]
Solution
At ionic equilibrium, $\mathop {A{B_2}}\limits_{a(1 - \alpha )} \rightleftharpoons \mathop {{A^{2 + }}}\limits_{a\alpha } + \mathop {2{B^ - }}\limits_{2a\alpha } $
Total moles of particles $=a(1-\alpha) + a\alpha + 2a\alpha = a(1+2\alpha ) = 1.2a $
$\therefore i=\frac {1.2a}{a} = 1.2 $
$\Delta T_b = i K_b m = 1.2 \times 0.5 \times 10 = 6 $
$\therefore B.P. = 106^\circ C $