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Chemistry Question on Electrochemistry

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3{Al2(SO4)3} ? Given that ∧Al3+∘{\wedge^{\circ}_{Al^{3+}}} and −∧SO42−∘{- \wedge^{\circ}_{SO^{2-}_4}} are the equivalent conductances at infinite dilution of the respective ions

A

∧Al3+∘+∧SO42−∘{ \wedge^{\circ}_{Al^{3+}} + \wedge^{\circ}_{SO_4^{2-}}}

B

(∧Al3+∘+∧SO42−∘)×6 \left({\wedge^{\circ}_{Al^{3+}} + \wedge^{\circ}_{SO_4^{2-}}} \right) \times 6

C

13∧Al3+∘+12∧SO42−∘\frac{1}{3}{\wedge^{\circ}_{Al^{3+}} + \frac{1}{2} \wedge^{\circ}_{SO_4^{2-}}}

D

2∧Al3+∘+3∧SO42−∘{2 \wedge^{\circ}_{Al^{3+}} + 3 \wedge^{\circ}_{SO_4^{2-}}}

Answer

∧Al3+∘+∧SO42−∘{ \wedge^{\circ}_{Al^{3+}} + \wedge^{\circ}_{SO_4^{2-}}}

Explanation

Solution

Al2(SO4)32Al3++3SO42−A{{l}_{2}}{{(S{{O}_{4}})}_{3}}2A{{l}^{3+}}+3SO_{4}^{2-}.
Since equivalent conductance's are given only for ions, the equivalent conductance at infinite dilution,
Λeq∞=ΛAl 3+o+Λso42−o\Lambda _{eq}^{\infty }=\Lambda _{Al\,3+}^{o}+\Lambda _{so_{4}^{2-}}^{o}