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Equivalent conductivity of $Fe_2(SO_4)_3$ is given by

$\begin{array}{1 1}\Lambda^{\infty}_{eq}=\Lambda^{\infty}_{eq}(Fe^{3+})+\Lambda^{\infty}_{eq}(SO_4^{2-})\\\Lambda_{eq}^{\infty}=2\Lambda_m^{\infty}(Fe^{3+})+\Lambda_{eq}^{\infty}(SO_4^{2-})\\\Lambda^{\infty}_{eq}=\Lambda^{\infty}_{eq}(Fe^{3+})+3\Lambda^{\infty}_{eq}(SO_4^{2-})\\\Lambda^{\infty}_{eq}=2\Lambda^{\infty}_{eq}(Fe^{3+})+3\Lambda^{\infty}_{eq}(SO_4^{2-})\end{array} $

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Answer : $\Lambda^{\infty}_{eq}=\Lambda^{\infty}_{eq}(Fe^{3+})+\Lambda^{\infty}_{eq}(SO_4^{2-})$
Equivalent conductivity of $Fe_2(SO_4)_3$ is given by $\Lambda^{\infty}_{eq}=\Lambda^{\infty}_{eq}(Fe^{3+})+\Lambda^{\infty}_{eq}(SO_4^{2-})$
answered Jul 18, 2014 by sreemathi.v
 

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