Question

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

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

Answer 1

The correct option is C 78

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

$i=\alpha n+(1-\alpha )=1+\alpha (n-1)=1+\alpha (4-1)$

Van't Hoff factor $\left(i\right)$ is given to be 3.333.

$n=$ Total moles formed after dissociation.

Let degree of dissociation be $\alpha$
So,

$3.333=1+\alpha (4-1)$

$\alpha (4-1)=3.333-1$

$\alpha (4-1)=2.333$

$3\alpha =2.333$

$\alpha =2.333/3=0.777$

Percent dissociation $=\alpha \times 100=77.7=78\%$.