Equivalent conductivity of Cr_2(SO_4)_3 is re | vedclass

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Question

(D) The relationship between molar conductivity $(\Lambda_m)$ and equivalent conductivity $(\Lambda_{eq})$ is given by the formula: $\Lambda_{eq} = \f

Vedclass · Accepted Answer

$\Lambda_{eq} = \frac{\Lambda_m}{6}$

Vedclass · Answer

(D) The relationship between molar conductivity $(\Lambda_m)$ and equivalent conductivity $(\Lambda_{eq})$ is given by the formula: $\Lambda_{eq} = \frac{\Lambda_m}{n-factor}$.For $Cr_2(SO_4)_3$,the dissociation is $Cr_2(SO_4)_3 ightarrow 2Cr^{3+} + 3SO_4^{2-}$.The total positive charge is $2 	imes (+3) = +6$ and the total negative charge is $3 	imes (-2) = -6$.Thus,the $n-factor$ (valency factor) is $6$.Therefore,$\Lambda_{eq} = \frac{\Lambda_m}{6}$.

Equivalent conductivity of $Cr_2(SO_4)_3$ is related to molar conductivity by the expression

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