Question

The freezing point depression of $0.001M$ of $A_x{B}_y\left[Fe(CN{)}_6\right]$ is $5.58\times 10^{-3}K$. If the oxidation state of $Fe$ is $+2$ and $K_f=1.86Kkgmo{l}^{-1}$, then the total number of possibilities for different types of $A$ and $B$ cations are

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

The oxidation state of $Fe$ is $+2$ in the complex. On ionisation, reaction occur as follows :

Total number of moles of equilibrium
$=1-\alpha +4\alpha +\alpha =1+4\alpha$
$\therefore i=\frac{\text{ Total number of moles at equili. }}{\text{ Initial moles }}=\frac{1+4\alpha }{1}$
$=1+4\times 0.5=1+2=3$
Thus, the total number of possibilities for different types of $A$ and $B$ cations are $3$ .