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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 : $(K_{b} for H_{2} O = 0.52 Kkg / mol)$

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377.12 K

375.5 K

373.25 K

374.04 K

answer is B.

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Detailed Solution

$X_{3} Y_{2} \to 3 X^{2 +} + 2 Y^{3 -}$

$1 - α 3 α 3 α$

$\Rightarrow i = 1 + 4 α = 2$

$Δ T_{b} = 2 \times 0.52 \times 1 = 1.04$

$\Rightarrow T_{b} = 373 + 1.04$ $= 374.04 K$

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