The van't Hoff factor of the compound $K_3Fe(CN)_6$ in dilute solution is

$\begin{array}{1 1} 1 \\ 2 \\ 3 \\ 4 \end{array}$

Solution :
The dissociative equation for $K_3[Fe(CN)_6$ can be written as under
$K_3[Fe(CN)_6] \to 3K^{+}+[Fe(CN)_6]^{3-}$
Therefore , number of particles after dissociation =4
So,Van't Hoff factor $= 4/1=4$
edited Oct 23 by meena.p

K3[Fe(CN)6] will break in 3 k+ and 1 [Fe(CN)6]. Hence, it's van't hoff factor is (3+1=4).