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(B) Given: $\Delta T_b = 101.08 - 100 = 1.08\,^{\circ}C$ and $\Delta T_f = 0 - (-1.80) = 1.80\,^{\circ}C$.For $AB$,at the boiling point,it is $100\%$

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behaves as a non-electrolyte at the freezing point of the solution

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(B) Given: $\Delta T_b = 101.08 - 100 = 1.08\,^{\circ}C$ and $\Delta T_f = 0 - (-1.80) = 1.80\,^{\circ}C$.For $AB$,at the boiling point,it is $100\%$ ionised,so the van't Hoff factor $i_b = 2$.The formulas for elevation in boiling point and depression in freezing point are $\Delta T_b = i_b K_b m$ and $\Delta T_f = i_f K_f m$.Dividing the two equations: $\frac{\Delta T_b}{\Delta T_f} = \frac{i_b K_b m}{i_f K_f m} = \frac{i_b}{i_f} 	imes \frac{K_b}{K_f}$.Substituting the values: $\frac{1.08}{1.80} = \frac{2}{i_f} 	imes 0.3$.$0.6 = \frac{0.6}{i_f}$,which gives $i_f = 1$.Since the van't Hoff factor $i_f = 1$,the solute $AB$ behaves as a non-electrolyte at the freezing point of the solution.

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.

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An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.