Question

$A{B}_2$ 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 ${}^{\circ }C.$ (Round off to the nearest Integer).
[Given: Molal elevation constant of water $K_b=0.5{\text{K kg mol}}^{-1}$, boiling point of pure water $=100{}^{\circ }C$]

Answer 1

By using the relation,
$\Delta T_b=i{K}_{b}m\text{Where,}\Delta T_b=\text{elevation in boiling point}{K}_b=\text{Molal elevation constant of water}{K}_b=0.5{\text{K kg mol}}^{-1}m=\text{molality of solution}=10.0\text{molal}i=\text{Van't off factor}$
$\alpha =\frac{i-1}{n-1}\alpha =\text{degree of dissociation}n=\text{number of ions}$
$A{B}_2\rightleftharpoons A^{2+}+2B^{-}$
$\therefore n=3$
$0.1=\frac{i-1}{3-1}i=1.2\Delta T_b=1.2\times 0.5\times 10=6(T_b{)}_{solution}=\text{Boiling point of water}+\Delta T_b=100+6=106{}^{\circ }C$