At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:

The conductivity of a saturated aqueous solution of AgCl is $$1.40\times 10^{-6}$$ S cm$$^{-1}$$ at 300 K. If the solubility of AgCl in water at 300 K is $$X$$ mol L$$^{-1}$$, then $$\log_{10}(X^{-1})$$ is

(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

In the context of the data table provided, $$A$$ (sometimes written as $$b$$) represents the Kohlrausch constant in the Debye-Hückel-Onsager relation for strong electrolytes:

$$\Lambda_m=\Lambda_m^\circ-A\sqrt{c}$$

Where:

$$\Lambda_m$$ = Molar conductivity at a given concentration

$$\Lambda_m^\circ$$ = Limiting molar conductivity (at infinite dilution)

$$c$$ = Molar concentration of the solution

1. Find $$\Lambda_m^\circ$$ for $$\text{AgCl}$$ using Kohlrausch's Law:

$$\Lambda_m^\circ(\text{AgCl})=\Lambda_m^\circ(\text{AgNO}_3)+\Lambda_m^\circ(\text{NaCl})-\Lambda_m^\circ(\text{NaNO}_3)$$

$$\Lambda_m^\circ(\text{AgCl})=134+127-121=140\text{ S cm}^2\text{ mol}^{-1}$$

2. Calculate Molar Solubility ($$X$$):

Using the formula discussed earlier:

$$X=\frac{\kappa\times1000}{\Lambda_m^\circ}=\frac{(1.40\times10^{-6}\text{ S cm}^{-1})\times1000}{140\text{ S cm}^2\text{ mol}^{-1}}=10^{-5}\text{ mol L}^{-1}$$

3. Compute the final log value:

$$X^{-1}=\frac{1}{10^{-5}}=10^5$$

$$\log_{10}(X^{-1})=\log_{10}(10^5)=5$$