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

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(D) The relationship between molar conductivity $(\Lambda _m)$ and equivalent conductivity $(\Lambda _{eq})$ is given by the formula: $\Lambda _{eq} =

Vedclass · Accepted Answer

$\Lambda _{eq} = \frac{\Lambda _m}{6}$

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(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 $n-factor$ 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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