The ratio of the value of any colligative property for ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ solution to that of ${\mathrm{Fe}}_4{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_3$ (prussian blue), solution is nearly 

van't Hoff factor (i) for ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ is 5 (assuming complete ionization) 

$\begin{array}{lllll}\begin{array}{l}{\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right] \to \\ \downarrow \\ 1 \mathrm{molecule}\end{array}& & \begin{array}{l}4{\mathrm{K}}^{+}\\ \downarrow \\ 4 \mathrm{cations}\end{array}& +& \begin{array}{l}{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{-4}\\ \downarrow \\ 1 \mathrm{anion}\end{array}\end{array}$

Similarly, i for ${\mathrm{Fe}}_4{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_3$ is 7 

As we know, colligative properties $\propto i$

$\frac{{\mathrm{i}}_{{\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}}{{\mathrm{i}}_{{\mathrm{Fe}}_4{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_3}}$$\text{=}\frac{\mathrm{Colligative} \mathrm{properties} \mathrm{of} {\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_{\mathrm{6}}\right]}{\mathrm{Colligative} \mathrm{properties} \mathrm{of} {\mathrm{Fe}}_4{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_{\mathrm{6}}\right]}_3}=\frac{5}{7}=0.71$