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CGP EDU Academic Team
Published on: August 13, 2026
An aqueous solution of a solute AB has b.p. of 101.08°C (AB is 100% ionized at boiling point of the solution) and freezes at – 1.80°C. Hence, AB (K b / K f = 0.3)
Text Solution
Verified by ExpertsThe correct answer is:
B
Given 
T b = 1.08 0 C, i = 2 at boiling pt. of solution.
and 
T f = 1.80 0 C , and 
= 0.3
so 
= 
so i f = 1
i.e., AB behaves as non–electrolyte at the f.p of the solution.
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