$0.1 molal$ aqueous solution of an electrolyte $A B_{4}$ is $75 %$ ionised. The boiling point $in K$ of the solution at $1 atm$ is : $K_{b (H_{2} O)} = 0.52 K kg {mol}^{- 1}$ . (Report your answer upto two decimal only).

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Solution

$\begin{matrix} A B_{4} & \to & A^{+} & + & 4 B^{-} 1 - α & α & 4 α \end{matrix}$
van't Hoff factor is
$∴ i = \frac{1 + 4 α}{1} = 1 + 4 \times 0.75 = 4$
Elevation in boiling point is
$Δ T_{b} = i K_{b} . m = 4 \times 0.52 \times 0.1 = 0.208 \approx 0.21$
$∴ T_{b} = T_{b}^{\circ} + 0.21 = 373 + 0.21 = 373.21$

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