Question

Van't Hoff factor of centimolal solution of $K_3\left[Fe{\left(CN\right)}_6\right]$ is $\mathrm{3.333.}$ Calculate the percent dissociation of $K_3\left[Fe{\left(CN\right)}_6\right]$(Round off to the nearest integer).

Answer 1

$K_3\left[Fe{\left(CN\right)}_6\right]\to 3K^{+}+{\left[Fe{\left(CN\right)}_6\right]}^{3-}$
For the dissociation reaction,
$i=1+(n-1)\alpha$
Van't Hoff factor (i) is given to be $3.333$.
$n=\text{Total moles formed after dissociation}$
$\alpha =\text{Degree of dissociation}$
So,
$3.333=1+\alpha (4-1)\alpha (4-1)=3.333-13\alpha =2.333\alpha =\frac{2.333}{3}=0.777$
$\text{Percent dissociation}=\alpha \times 100=0.777\times 100=77.7\%\approx 78.0\%$