0.1 molal aqueous solution of an electrolyte AB₃ is 90% ionised. The boiling point of the solution at 1 atm is :
$\left({\mathrm{K}}_{\left.{\mathrm{b}}_{\left({\mathrm{H}}_2\mathrm{O}\right.}\right)}=0.52\mathrm{Kkg}{\mathrm{mol}}^{-1}\right)$

$$\begin{gathered} \mathrm{t} = 0 \underset{1}{{\mathrm{AB}}_3}\rightleftharpoons \underset{0}{{\mathrm{A}}^{3+}}+\underset{0 \\ }{3{\mathrm{B}}^{-}} \\ \mathrm{t}={\mathrm{t}}_{\mathrm{e}} 1-\mathrm{\alpha } \mathrm{\alpha } \mathrm{\alpha } \end{gathered}$$
$$\begin{gathered} \mathrm{Total} 1+\mathrm{\alpha }+\mathrm{\alpha }+3\mathrm{\alpha }=1+3\mathrm{\alpha } \\ \therefore \mathrm{T}-100=0.1(1+3\times 0.9)\times 0.52 \\ \mathrm{T} = 373.19 \mathrm{K} \end{gathered}$$