Assuming $100\%$ ionisation, the van't Hoff factor ($i$) is equal to the number of ions produced per formula unit of the electrolyte. For $\text{Al}_2(\text{SO}_4)_3$, the dissociation is: $\text{Al}_2(\text{SO}_4)_3 \rightleftharpoons 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$ Total number of ions ($i$) = $2 + 3 = 5$. Now, let's check the given options: (A) $\text{K}_2\text{SO}_4 \rightleftharpoons 2\text{K}^+ + \text{SO}_4^{2-}$ ($i = 3$) (B) $\text{K}_3[\text{Fe(CN)}_6] \rightleftharpoons 3\text{K}^+ + [\text{Fe(CN)}_6]^{3-}$ ($i = 4$) (C) $\text{Al(NO}_3)_3 \rightleftharpoons \text{Al}^{3+} + 3\text{NO}_3^-$ ($i = 4$) (D) $\text{K}_4[\text{Fe(CN)}_6] \rightleftharpoons 4\text{K}^+ + [\text{Fe(CN)}_6]^{4-}$ ($i = 5$) Therefore, $\text{K}_4[\text{Fe(CN)}_6]$ has the same van't Hoff factor ($i=5$) as $\text{Al}_2(\text{SO}_4)_3$.