An aqueous solution of a solute AB has a b.p. | vedclass

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(B) For boiling point elevation: $\Delta T_b = i_b 	imes m 	imes K_b$ $101.08 - 100 = 2 	imes m 	imes K_b$ (Since $AB$ is $100\%$ ionized at $b.p.

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behaves as a non-electrolyte at the $f.p.$ of the solution

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(B) For boiling point elevation: $\Delta T_b = i_b 	imes m 	imes K_b$ $101.08 - 100 = 2 	imes m 	imes K_b$ (Since $AB$ is $100\%$ ionized at $b.p.$,$i_b = 2$) $1.08 = 2 	imes m 	imes K_b \implies m 	imes K_b = 0.54$ For freezing point depression: $\Delta T_f = i_f 	imes m 	imes K_f$ $0 - (-1.80) = i_f 	imes m 	imes (K_b / 0.3)$ (Given $K_b / K_f = 0.3 \implies K_f = K_b / 0.3$) $1.80 = i_f 	imes (0.54 / 0.3)$ $1.80 = i_f 	imes 1.8 \implies i_f = 1$ Since $i_f = 1$,$AB$ behaves as a non-electrolyte at the freezing point.

An aqueous solution of a solute $AB$ has a $b.p.$ of $101.08^\circ C$ ($AB$ is $100\%$ ionized at the boiling point of the solution) and freezes at $-1.80^\circ C$. Given $K_b / K_f = 0.3$,the solute $AB$:

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