Question

The freezing point (in$^{\circ}$C) of solution containing $0.1\, g$ of $K_3[Fe(CN)_6$] (mol. wt. $329)$ in $100 \,g$ of water $(K_f = 1.86\, K \,kg\, mol^{-1})$ is

• A. $-2.3 \times 10^{-2}$
• B. $-5.7 \times 10^{-2}$
• C. $-5.7 \times 10^{-3}$
• D. $-1.2 \times 10^{-2}$

Answer 1

Answer: (A)

van-t Hoff factor (i)$= 4 \{ 3K^+ [Fe(CN)_6]^{3-}$
Molality $=\frac{0.1}{329} \times \frac{1000}{100} =\frac{1}{329}$
$\Rightarrow -\Delta T_f =iK_f m$
$ =4 \times 1.86 \times \frac{1}{329} =2.3 \times 10^{-2}$
$ = T_f =-2.3 \times 1^{-2} \,{}^{\circ}C$
(As % ) freezing point of water is $0^{\circ}C$)