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Q.

KBr is 80% ionized in solution. The freezing point of 0.4 molal solution of KBr is :
$K_{f} (H_{2} O) = 1 .86 \frac{Kkg}{mole}$

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An Intiative by Sri Chaitanya

a

257.3 K

b

-1.339 K

c

274.339K

d

- 1.339 °C

answer is D.

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Detailed Solution

${ΔT}_{f} = {iK}_{f} m$
$\underset{\underset{(1 - α)}{1}}{KBr} ⟶ \underset{\underset{α}{0}}{K^{+}} + \underset{\underset{α}{0}}{{Br}^{-}}$
$i = 1 + α$
Here $α = 0.8 (80 % ionization$ )
i = 1.8
$\begin{array}{l} {ΔT}_{f} = (1 .8) (1 .86) (0 .4) = 1 .339 K \\ T_{f} = - 1 {.339}^{\circ} C \end{array}$

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