Number of moles of  ions produced in 0.4 moles of Potash alum are

$$\begin{gathered} {\mathrm{K}}_2{\mathrm{SO}}_4.{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3.24{\mathrm{H}}_2\mathrm{O}....\mathrm{Potash} \mathrm{Alum} ; \\ {\mathrm{K}}_2{\mathrm{SO}}_4.{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3.24{\mathrm{H}}_2\mathrm{O}\left(1\mathrm{mole}\right)\rightleftharpoons 2{\mathrm{K}}^{+}\left(2 \mathrm{moles}\right)+2{\mathrm{Al}}^{3+}\left(2 \mathrm{moles}\right)+4{\mathrm{SO}}_4^{2-}\left(4 \mathrm{moles}\right)+24{\mathrm{H}}_2\mathrm{O} ; \\ 1 \mathrm{mole} \mathrm{ion} \left(\mathrm{or}\right) 1 \mathrm{gram} \mathrm{ion}={\mathrm{N}}_{\mathrm{A}} \mathrm{ions}=6.023\times {10}^{23} \mathrm{ions} ; \\ \mathrm{Total} \mathrm{ions}/\mathrm{mole}=8{\mathrm{N}}_{\mathrm{A}} ; \\ {\mathrm{K}}^{+}\mathrm{ions}/\mathrm{mole}=2{\mathrm{N}}_{\mathrm{A}}; \\ {\mathrm{Al}}^{3+}\mathrm{ions}/\mathrm{mole}=2{\mathrm{N}}_{\mathrm{A}}; \\ {\mathrm{SO}}_4^{2-}\mathrm{ions}/\mathrm{mole}=4{\mathrm{N}}_{\mathrm{A}}; \end{gathered}$$