Question

The freezing point depression of $0.001m$ $K_x\left[Fe{\left(CN\right)}_6\right]$ is $7.4\times {10}^{-3}K$. The value of $x$ is:
(Assuming complete dissociation, $K_f=1.85K$ $kg{mol}^{-1}$ for water)

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

$\Delta T_f=K_f\times m\times i$

$\therefore i=\frac{\Delta T_f}{K_f\times m}=\frac{7.4\times {10}^{-3}}{1.85\times {10}^{-3}}=4$

$\underset{1-\alpha }{K_x\left[Fe\right(CN{)}_6]}\rightleftharpoons \underset{x\alpha }{x{K}^{+}}+\underset{\alpha }{\left[Fe\right(CN{)}_6{]}^{-}}$

$\Rightarrow i=1+x\alpha =1+x$ $[\because \alpha =1$....given$]$

$\therefore x=3$