The conductivity of a saturated solution of A | vedclass

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Question

(C) The molar conductivity at infinite dilution for $AgCl$ is given by: $\lambda_m^\infty (AgCl) = \lambda_{Ag^+}^\infty + \lambda_{Cl^-}^\infty = 61.

Vedclass · Accepted Answer

$1.435 	imes 10^{-3}$

Vedclass · Answer

(C) The molar conductivity at infinite dilution for $AgCl$ is given by: $\lambda_m^\infty (AgCl) = \lambda_{Ag^+}^\infty + \lambda_{Cl^-}^\infty = 61.9 + 76.3 = 138.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.The solubility $(S)$ in $mol \ L^{-1}$ is calculated using the formula: $S = \frac{\kappa 	imes 1000}{\lambda_m^\infty} = \frac{1.382 	imes 10^{-6} 	imes 1000}{138.2} = 10^{-5} \ mol \ L^{-1}$.To convert solubility to $g \ L^{-1}$,multiply by the molar mass of $AgCl$ $(143.5 \ g \ mol^{-1})$: $S = 10^{-5} \ mol \ L^{-1} 	imes 143.5 \ g \ mol^{-1} = 1.435 	imes 10^{-3} \ g \ L^{-1}$.

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