The conductivity of a saturated solution of B | vedclass

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(D) The relationship between equivalent conductance $(\lambda_{eq})$,conductivity $(K)$,and solubility $(S)$ is given by: $\lambda_{eq} = \frac{1000 \

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$4 	imes 10^{-6}$

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(D) The relationship between equivalent conductance $(\lambda_{eq})$,conductivity $(K)$,and solubility $(S)$ is given by: $\lambda_{eq} = \frac{1000 	imes K}{S}$.Given: $K = 3.06 	imes 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and $\lambda_{eq} = 1.53 \ \Omega^{-1} \ cm^{2} \ equivalent^{-1}$.Substituting the values: $1.53 = \frac{1000 	imes 3.06 	imes 10^{-6}}{S}$.$S = \frac{3.06 	imes 10^{-3}}{1.53} = 2 	imes 10^{-3} \ mol \ L^{-1}$.For $BaSO_4$,the solubility product $K_{sp} = S^2$.$K_{sp} = (2 	imes 10^{-3})^2 = 4 	imes 10^{-6}$.

The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^{2} \ equivalent^{-1}$. The $K_{sp}$ of the $BaSO_4$ will be

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