Question

Van’t Hoff factor of centimolal solution of $K_3[Fe(CN{)}_6]$ is $3.333$. Calculate the percent dissociation of $K_3[Fe(CN{)}_6]$.

• A. $33.33$
• B. $0.78$
• C. $78$
• D. $23.33$

Answer 1

Answer: (C)

Given, $i=3.333$

On dissociation of $K_3\left[Fe(CN{)}_6\right]$, we get four ions.

$K_3\left[Fe(CN{)}_6\right]⟶3K^{+}+{\left[Fe(CN{)}_6\right]}^{3-}$

Total number of ions $(n)=4$

$\therefore \alpha =\frac{i-1}{n-1}=\frac{3.333-1}{4-1}$

$=\frac{2.333}{3}=0.7776\dots$

$\approx 0.78$

Hence, the per cent dissociation becomes $78\%$.