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1.0 molal aqueous solution of an electrolyte $X_{3} Y_{2}$ is 25% ionized. The Boiling point of the solution is $(K_{b} of H_{2} O = 0.52 k Kg {mol}^{- 1})$

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375.5 k

374.04 k

377.12 k

373.25 k

answer is B.

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Detailed Solution

$α = 0.25$

i = 1 − 0.25 + (0.25)5 = 2

$T_{b} = 373 = 2 \times 0.52 \times 1$

= 1.04

$T_{b} = 34.04$

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