Ionic conductivity at infinite dilution of $A{l}^{3+}$ and $S{O_4}^{2-}$ ions are $189Sc{m}^{2}mo{l}^{-1}$ and $160Sc{m}^{2}mo{l}^{-1}$ respectively the equivalent and molar conductivity of $A{l}_2{\left(S{O}_4\right)}_3$ at infinite dilution is $x$ and $y.$ What is $x+y$

Equalent conductance ${\lambda }^0=\frac{1}{n^{+}}.{{\lambda }_{+}}^0+\frac{1}{n^{-}}.{{\lambda }_{-}}^0$

Where $n^{+}$ and $n^{-}$ are charge on each ion

$\begin{array}{l}{{\lambda }^0}_{A{l}_2{\left(s{o}_4\right)}_3}=\frac{1}{n^{+}}{{\lambda }^0}_{A{l}^{3+}}+\frac{1}{n^{-}}{{\lambda }^0}_{s{o_4}^{2-}}\\ =\frac{1}{3}{{\lambda }^0}_{A{l}^{3+}}+\frac{1}{2}{{\lambda }^0}_{s{o_4}^{2-}}\\ =\frac{1}{3}\times 189+\frac{1}{2}\times 160\\ =143Sc{m}^{2}e{q}^{-1}=x\end{array}$

Molar conductance at infinite dilution ${\wedge }^0={\gamma }_{+}.{{\lambda }_{+}}^0+{\gamma }_{-}.{{\lambda }_{-}}^0$

Where ${\gamma }_{+}$ and ${\gamma }_{-}$ are the number of $+ve$ and $-ve$ ions

$\begin{array}{l}{{\wedge }^0}_{A{l}_2{\left(s{o}_4\right)}_3}={\gamma }_{+}.{{\lambda }^0}_{A{l}^{3+}}+{\gamma }_{-}.{{\lambda }^0}_{s{o_4}^{2-}}\\ =2\times 189+3\times 160\\ =858Sc{m}^{2}mo{l}^{-1}=y\\ x+y=143+858=1001\end{array}$