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# Molar conductivity of $Fe_2(SO_4)_3$ is given by

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

Can you answer this question?

Answer : $\Lambda_m^{\infty}=2\lambda_{m}^{\infty}(Fe^{3+})+3\lambda_{m}^{\infty}(SO_4^{2-})$
Molar conductivity of $Fe_2(SO_4)_3$ is given by $\Lambda_m^{\infty}=2\lambda_{m}^{\infty}(Fe^{3+})+3\lambda_{m}^{\infty}(SO_4^{2-})$
answered Jul 18, 2014