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Chemistry Question on Electrochemistry

What will be the molar conductivity of AL3+AL^{3+} ions at infinite dilution if molar conductivity of Al2(SO4)3Al_2(SO_4)_3 is 858 S cm2 mol−1858\, S \,cm^2 \,mol^{-1} and ionic conductance of SO42−SO^{2-}_4 is 160 S cm2 mol−1160\, S\, cm^2\, mol^{-1} at infinite dilution?

A

189 S cm2 mol−1189\,S\,cm^{2}\,mol^{-1}

B

698 S cm2 mol−1698\,S\,cm^{2}\,mol^{-1}

C

1018 S cm2 mol−11018\,S\,cm^{2}\,mol^{-1}

D

429 S cm2 mol−1429\,S\,cm^{2}\,mol^{-1}

Answer

189 S cm2 mol−1189\,S\,cm^{2}\,mol^{-1}

Explanation

Solution

ΛAl2(SO4)3∘=2ΛAl3+∘+3ΛSo42−∘\Lambda^{\circ}_{Al_2\left(SO_4\right)_3}=2\Lambda^{\circ}_{Al^{3+}}+3\Lambda^{\circ}_{So^{2-}_{4}} ΛAl3+∘=ΛAl2∘(SO4)3−3ΛSO42−∘2\Lambda^{\circ}_{Al^{3+}}=\frac{\Lambda^{\circ}_{Al_2}\left(SO_{4}\right)_{3}-3\Lambda^{\circ}_{SO^{2-}_{4}}}{2} =858−(3×160)2=\frac{858-\left(3\times160\right)}{2} =189 S cm2 mol−1=189\,S\,cm^{2}\,mol^{-1}