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.

$K _{3}\left[ Fe ( CN )_{6}\right] \longrightarrow 3 K ^{+}+\left[ Fe ( CN )_{6}\right]^{3-}$

Total number of ions $(n)=4$

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

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

$\approx 0.78$

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