If the ionic equivalent conductivities of $K^{+}A{l}^{3+},\text{and}S{O}_4^{2-}$ ions are 4, 4, and 1 $Sc{m}^{2}E{q}^{-1}$ , respectively. Calculate the value of ${\wedge }_{eq}^0$ for Potash alum $(K_{2}S{O}_4.A{l}_2{\left(S{O}_4\right)}_3.24H_{2}O)$

As per the expression:
$\frac{{\lambda }_M^o}{{\lambda }_{eq}^o}=n-factor$
${\lambda }_{M_{K^{+}}}^o={\lambda }_{e{q}_{K^{+}}}^o\times 1=4Sc{m}^{2}mo{l}^{-1}$
Similarly,
${\lambda }_{M_{A{l}^{3+}}}^o={\lambda }_{e{q}_{A{l}^{3+}}}^o\times 3=12Sc{m}^{2}mo{l}^{-1}$
and
${\lambda }_{M_{S{O}_4^{2-}}}^o={\lambda }_{e{q}_{S{O}_4^{2-}}}^o\times 2=2Sc{m}^{2}mo{l}^{-1}$
$\therefore$
${\Lambda }_{M_{K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O}}^o$ = $2\times {\lambda }_{M_{K^{+}}}^o+{\lambda }_{M_{S{O}_4^{2-}}}^o+2\times {\lambda }_{M_{A{l}^{3+}}}^o+3\times {\lambda }_{M_{S{O}_4^{2-}}}^o$
$\Rightarrow$ $2\times 4+1\times 2+2\times 12+3\times 2$
$\Rightarrow$ $40Sc{m}^{2}mo{l}^{-1}$
*NOTE: Since, $H_{2}O$ is neutral therefore will not account for conductivity.
$\therefore$
${\Lambda }_{e{q}_{K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O}}^o$ = $\frac{{\Lambda }_{M_{K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O}}^o}{n-factor}$
n-factor of Potash Alum is = Total charge on cations
$\Rightarrow$ n-factor = $(+1)\times no.of{K}^{+1}+(+3)\times no.ofA{l}^{3+}$
$\Rightarrow$ $1\times 2+3\times 2$
$\Rightarrow$ 8
$\therefore$ Putting values:
${\Lambda }_{e{q}_{K_{2}S{O}_4.A{l}_2(S{O}_4{)}_3.24H_{2}O}}^o$ = $\frac{40}{8}$
= $5Sc{m}^{2}E{q}^{-1}$
Hence, the correct answer is $5Sc{m}^{2}E{q}^{-1}$