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Question: What will be the molar conductivity of \(Al^{3 +}\) ions at infinite dilution if molar conductivity ...

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 cm2mol−1cm^{2}mol^{- 1} and ionic conductance of SO42−SO_{4}^{2 -} is 160 S cm2mol−1cm^{2}mol^{- 1} at infinite dilution?

A

189Scm2mol−1189Scm^{2}mol^{- 1}

B

698Scm2mol−1698Scm^{2}mol^{- 1}

C

1018Scm2mol−11018Scm^{2}mol^{- 1}

D

429Scm2mol−1429Scm^{2}mol^{- 1}

Answer

189Scm2mol−1189Scm^{2}mol^{- 1}

Explanation

Solution

ΔºAl2(SO4)3=2ΔºAl3++3ΔºSO42−\Delta º_{Al_{2}(SO_{4})_{3}} = 2\Delta º_{Al^{3 +}} + 3\Delta º_{SO_{4}^{2 -}}

ΔºAl3+=ΔºAl2(SO4)3−3ΔºSO42−2\Delta º_{Al^{3 +}} = \frac{\Delta º_{Al_{2}(SO_{4})_{3}} - 3\Delta º_{SO_{4}^{2 -}}}{2}

=858−(3×160)2=189Scm2mol−1= \frac{858 - (3 \times 160)}{2} = 189Scm^{2}mol^{- 1}