Chemistry - Liquid Solution Question with Solution | TestHub

The value of van't Hoff factor of the given molecule is as follows : $\begin{array}{rl}{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3& \rightleftharpoons 2{\mathrm{Al}}^{3+}+3{\mathrm{SO}}_4^{2-},\mathrm{i}=5\\ {\mathrm{K}}_2{\mathrm{SO}}_4& \rightleftharpoons 2{\mathrm{K}}^{+}+{\mathrm{SO}}_4^{2-},\mathrm{i}=3\\ {\mathrm{K}}_3\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]& \rightleftharpoons 3{\mathrm{K}}^{+}+{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3+},\mathrm{i}=4\\ \mathrm{Al}{\left({\mathrm{NO}}_3\right)}_3& \rightleftharpoons {\mathrm{Al}}^{3+}+3{\mathrm{NO}}_3^{-},\mathrm{i}=4\end{array}$ The van't Hoff factor for ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ is ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]\rightleftharpoons 4{\mathrm{K}}^{+}+{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-},\mathrm{i}=5$ $\therefore {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ has the same value of van't Hoff factor as that of ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$.