- K4[Fe(CN)6]
Explanatin:
Van't Hoff factor i = $\frac{\text{number of solute particles present in solution}}{\text{theoretical number of solute particles due to solution of non electrolyte }}$
$=\frac{\text{n(observed)}}{\text{n(theoretical)}}.$
1 molecule of Al2(SO4)3 ionizes in solution to produce 5 ions.
$\text{Al}_2(\text{SO}_4)_3\rightarrow2\text{Al}^{3+}+3\text{SO}^2_4$
Hence, i $=\frac{\text{n(observed)}}{\text{n(theoretical)}}=\frac{5}{1}=5.$
1 molecule of K4[Fe(CN)6] ionize in solution to produce 5 ions.
K4[Fe(CN)6] → 4K++[Fe(CN)6]4−
Hence, i $=\frac{\text{n(observed)}}{\text{n(theoretical)}}=\frac{5}{1}=5.$
Thus, K4[Fe(CN)6] has the same value of van't Hoff's factor (i) as that of Al2(SO4)3 (assuming 100% ionization).
Note: 1 molecule each of Al(NO3)3, K2SO4 and K3[Fe(CN)6] ionizes in solution to produce 4, 3 and 4 ions respectively.