An aqueous solution of a solute AB has b.p of 101.08°C (AB is 100% ionized in boiling point of the solution) and freezes at−1.80°C . Hence, $\mathrm{AB}({\mathrm{K}}_{\mathrm{b}}/{\mathrm{K}}_{\mathrm{f}}=0.3)$

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$

$\frac{{\mathrm{K}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{f}}}=0.3$

$∆{\mathrm{T}}_{\mathrm{b}}=\mathrm{i}(0.3{\mathrm{K}}_{\mathrm{f}})\mathrm{m}, ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}$

$\mathrm{AB}\rightleftharpoons {\mathrm{A}}^{+}+{\mathrm{B}}^{-}$

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$1-\mathrm{\alpha } \mathrm{\alpha } \mathrm{\alpha }$

$\mathrm{i}=1-\mathrm{\alpha }+2\mathrm{\alpha }$

$=1+\mathrm{\alpha }$

For 100% dissociation , i = 1 + 1 = 2

$\frac{∆{\mathrm{T}}_{\mathrm{b}}}{∆{\mathrm{T}}_{\mathrm{f}}}=\frac{2(0.3) K_f.m}{i.K_f.m}$

$\Rightarrow \frac{1.08}{1.80}=\frac{0.6}{\mathrm{i}}$

$\Rightarrow \mathrm{i}=\frac{1.08}{1.08}=1$

Hence it acts as non electrolyte at freezing point.