For a saturated solution of Ag_{2}CrO_{4} at | vedclass

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(D) The molar conductivity at infinite dilution for $Ag_{2}CrO_{4}$ is given by Kohlrausch's law: $\lambda_{m}^{\infty}(Ag_{2}CrO_{4}) = 2\lambda_{m}^

Vedclass · Accepted Answer

$2.56 	imes 10^{-4}$

Vedclass · Answer

(D) The molar conductivity at infinite dilution for $Ag_{2}CrO_{4}$ is given by Kohlrausch's law: $\lambda_{m}^{\infty}(Ag_{2}CrO_{4}) = 2\lambda_{m}^{\infty}(Ag^{+}) + \lambda_{m}^{\infty}(CrO_{4}^{2-})$.Substituting the values: $\lambda_{m}^{\infty} = 2 	imes 127 + 246 = 500 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.The solubility $(S)$ in $mol \ L^{-1}$ is calculated using the formula: $S = \frac{\kappa 	imes 1000}{\lambda_{m}^{\infty}}$,where $\kappa = 2 	imes 10^{-2} \ \Omega^{-1} \ cm^{-1}$.$S = \frac{2 	imes 10^{-2} 	imes 1000}{500} = \frac{20}{500} = 0.04 \ mol \ L^{-1}$.The solubility product $K_{sp}$ for $Ag_{2}CrO_{4}$ is given by $K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2-}] = (2S)^{2}(S) = 4S^{3}$.$K_{sp} = 4 	imes (0.04)^{3} = 4 	imes 6.4 	imes 10^{-5} = 2.56 	imes 10^{-4}$.

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