1 molal $K_{4} Fe {(CN)}_{6}$ solution has a degree of dissociation at 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute, A. The molar mass of A is ____ u.
(Density of water = $1 .0 g {cm}^{- 3}$ )

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answer is 85.

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Detailed Solution

$K_{4} Fe {(CN)}_{6} ⇌ 4 K^{+} + Fe {(CN)}_{6}^{4 -} 1 0 0 1 - 0.4 4 (0.4) 0.4 i = 1 - 0.4 + 4 (0.4) + 0.4 = 2.6$

$\begin{array}{l} {ΔT}_{f} = {iK}_{f} m \\ {ΔT}_{f} = = 2.6 \times K_{f} \times 1 \\ 2 .6 K_{f} = 1 \times K_{f} \times \frac{18 .1}{M} \times \frac{1000}{(100 - 18 .1)} \\ M = 85 \end{array}$

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