The freezing point ( $^{0} C$ ) of a solution containing 0.1g of $K_{3} [F e {(C N)}_{6}]$ (molecular weight 329) in 100g of water $(k_{f} = 1.86 k . k g . m o l e^{- 1})$

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$5.7 \times 10^{- 3}$

$- 1.2 \times 10^{- 2}$

$5.7 \times 10^{- 2}$

$2.3 \times 10^{- 2}$

answer is A.

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Detailed Solution

$K_{3} [F e {(C N)}_{6}] ⇌ 3 K^{+} + {[F e {(C N)}_{6}]}^{- 3} i = 4 Δ T_{f} = i K_{f} m = 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1000}{100} Δ T_{f} = 0.023 = 2.3 \times 10^{- 2}$

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