The freezing point (in °C) of a solution containing 0.1 g of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (Mol. Wt. 329) in 100 g of water $(\mathrm{K}_{\mathrm{f}}=1.86\mathrm{~K}\mathrm{~kg}\mathrm{~mol}^{-1})$ is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Solutions

The freezing point (in °C) of a solution containing 0.1 g of

K_{3} [Fe (CN)_{6}]

(Mol. Wt. 329) in 100 g of water

(K_{f} = 1.86 K kg mol^{- 1})

Answer: (A)

K_{3} [Fe (CN)_{6}]

dissociates as:

K_{3} [Fe (CN)_{6}] \to 3 K^{+} + [Fe (CN)_{6}]^{3 -}

So van't Hoff factor

i = 4

.

Depression in freezing point:

Δ T_{f} = K_{f} \times i \times \frac{m}{M} \times \frac{1000}{W}

where

m = 0.1

M = 329

g/mol,

W = 100

Δ T_{f} = 1.86 \times 4 \times \frac{0.1}{329} \times \frac{1000}{100} = 1.86 \times 4 \times \frac{0.1}{329} \times 10 = \frac{7.44}{329} \approx 0.023 = 2.3 \times 1 0^{- 2}

Freezing point =

0 - 2.3 \times 1 0^{- 2} = - 2.3 \times 1 0^{- 2}

°C

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option