Discussion Forum

The freezing point (in °C) of solution containing 0.1 g of  $K_{3}[Fe(CN)_{6}]$ (molecular weight 329) in 100 g of water $(K_{f}=1.86 K kg mol^{-1})$ is 

Answer:

Explanation:

Vant Hoff's factor $(i)= 4(3K^{+}+[Fe(CN)_{6}]^{3-})$ 

Molality= $\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329}$

$\Rightarrow$    $-\triangle T_{f}=iK_{f}.m=4 \times 1.86 \times \frac{1}{329}=2.3 \times 10^{-2} \Rightarrow T_{f}=-2.3 \times 10^{-2} {^{o}}C$

(As freezing point of water is $0^{0} C$)