Question

$0.1M{K}_4\left[Fe{\left(CN\right)}_6\right]$ is 60% ionized. What will be its Van't Hoff factor?

• A. $1.4$
• B. $3.4$
• C. $2.4$
• D. $4.4$

Answer 1

The correct option is B $3.4$

$\alpha =\frac{i-1}{n-1}$;

$n=5$, since $K_4\left[Fe{\left(CN\right)}_6\right]$ gives 5 ions in the solution.
$K_4\left[Fe{\left(CN\right)}_6\right]\to 4K^{+}+\left[Fe\right(CN{)}_6{]}^{4-}$
Now,

$\therefore 0.6=\frac{i-1}{5-1}$

$i=3.4$