At 300 K, the molar conductivities of the aqueous solutions of...

Using the Debye-Hückel-Onsager equation, $\Lambda_m = \Lambda^\circ_m - b\sqrt{C}$, we can find the limiting molar conductivity ($\Lambda^\circ_m$) for each salt.

For $\text{NaNO}_3$:
$\Lambda^\circ_m - b\sqrt{0.01} = 111 \Rightarrow \Lambda^\circ_m - 0.1b = 111$
$\Lambda^\circ_m - b\sqrt{0.04} = 101 \Rightarrow \Lambda^\circ_m - 0.2b = 101$
Subtracting the two equations gives $0.1b = 10 \Rightarrow b = 100$.
$\Lambda^\circ_m(\text{NaNO}_3) = 111 + 100(0.1) = 121 \text{ S cm}^2\text{ mol}^{-1}$.

For $\text{NaCl}$:
$\Lambda^\circ_m - 0.1b = 117$
$\Lambda^\circ_m - 0.2b = 107$
Subtracting gives $0.1b = 10 \Rightarrow b = 100$.
$\Lambda^\circ_m(\text{NaCl}) = 117 + 100(0.1) = 127 \text{ S cm}^2\text{ mol}^{-1}$.

For $\text{AgNO}_3$:
$\Lambda^\circ_m - 0.1b = 125$
$\Lambda^\circ_m - 0.2b = 116$
Subtracting gives $0.1b = 9 \Rightarrow b = 90$.
$\Lambda^\circ_m(\text{AgNO}_3) = 125 + 90(0.1) = 134 \text{ S cm}^2\text{ mol}^{-1}$.

According to Kohlrausch's law of independent migration of ions:
$\Lambda^\circ_m(\text{AgCl}) = \Lambda^\circ_m(\text{AgNO}_3) + \Lambda^\circ_m(\text{NaCl}) - \Lambda^\circ_m(\text{NaNO}_3)$
$\Lambda^\circ_m(\text{AgCl}) = 134 + 127 - 121 = 140 \text{ S cm}^2\text{ mol}^{-1}$.

The molar conductivity of the saturated $\text{AgCl}$ solution is equal to its limiting molar conductivity, so $\Lambda_m(\text{AgCl}) = 140 \text{ S cm}^2\text{ mol}^{-1}$.

The relationship between molar conductivity, conductivity ($\kappa$), and solubility ($X$ in $\text{mol L}^{-1}$) is:
$\Lambda_m = \dfrac{\kappa \times 1000}{X}$

Substituting the given values:
$140 = \dfrac{1.40 \times 10^{-6} \times 1000}{X}$
$140 = \dfrac{1.40 \times 10^{-3}}{X}$
$X = \dfrac{1.40 \times 10^{-3}}{140} = 10^{-5} \text{ mol L}^{-1}$.

We need to find $\log_{10}(X^{-1})$:
$X^{-1} = \dfrac{1}{10^{-5}} = 10^5$
$\log_{10}(X^{-1}) = \log_{10}(10^5) = 5$.

Answer: $5$