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JEE Main levelSingle correct optionDifficulty 2 of 3so203

Assuming complete ionisation, the freezing point of a $0.0100 mol k g^{- 1}$ aqueous solution of $K X_{4} [Fe (CN) X_{6}]$ is (take $K_{f} = 1.86 K kg mo l^{- 1}$ )

A-0.0186 $^{\circ} C$
B-0.0744 $^{\circ} C$
C-0.0558 $^{\circ} C$
D-0.0930 $^{\circ} C$

Answer: D

K X_{4} [Fe (CN) X_{6}] 4 K X^{+} + [Fe (CN) X_{6}] X^{4 -}

gives 5 ions, so

i = 5

.

Δ T_{f} = i K_{f} m = 5 \times 1.86 \times 0.0100 = 0.0930 K

, and the freezing point is

- 0.093 0^{\circ} C

.

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