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Elevation in Boiling Point

1.0 molal aqu...

Question

$1.0$ molal aqueous solution of an electrolyte $X_{3} Y_{3}$ is $25 %$ ionized. The boiling point of the solution is:
$[k_{b}$ for $H_{2} O = 0.52 K k g / m o l]$

375.5 K

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374.04 K

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377.12 K

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373.25 K

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Solution

The correct option is B $374.04 K$
As we know,
$Δ T_{b} = i k_{b} m$
$X_{3} Y_{2} ⟶ 3 X^{+ 2} + 2 Y^{+ 3}$
$i = 1 - α + 3 α + 2 α i = 0.75 + 1.25$
$i = 2$

$T_{1} - T_{0} = 2 \times 0.52 \times 1 = 1.04$

$T_{1} = 373 + 1.04 T_{1} = 374.04 K$

Hence, the correct option is $B$

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