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

Salt	Concentration (M)	Molar conductivity (S cm2 mol−1)
NaNO3	0.01	111
	0.04	101
NaCl	0.01	117
	0.04	107
AgNO3	0.01	125
	0.04	116

The conductivity of a saturated aqueous solution of AgCl is 1.40 × 10⁻⁶ S cm⁻¹ at 300 K. If the solubility of AgCl in water at 300 K is X mol L⁻¹, then log₁₀(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.)

The correct answer is 5.

Step 1: Understand the limiting molar conductivity relationship (Kohlrausch's Law of Independent Migration of Ions)
According to Kohlrausch's law, the limiting molar conductivity of silver chloride (${\mathrm{AgCl}}_{}$) can be determined by combining the limiting molar conductivities of appropriate strong electrolytes:

${\Lambda }_m^{\circ }\left(\mathrm{AgCl}\right)={\Lambda }_m^{\circ }\left({\mathrm{AgNO}}_3\right)+{\Lambda }_m^{\circ }\left(\mathrm{NaCl}\right)-{\Lambda }_m^{\circ }\left({\mathrm{NaNO}}_3\right)$

Step 2: Calculate the limiting molar conductivity (${\Lambda }_m^{\circ }$) using Debye-Hückel-Onsager equation
For strong electrolytes, molar conductivity varies with concentration according to the Debye-Hückel-Onsager equation:

${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{C}$

For each salt, let's set up the equation using the given concentrations $C=0.01\text{ M}$ and $C=0.04\text{ M}$:

Notice that $\sqrt{0.01}=0.1$ and $\sqrt{0.04}=0.2$.

Difference in concentration square roots: $\sqrt{C}=0.2-0.1=0.1$.

1. For ${\mathrm{NaNO}}_3$:
At $C=0.01$, ${\Lambda }_m=111={\Lambda }_m^{\circ }-A\left(0.1\right)$
At $C=0.04$, ${\Lambda }_m=101={\Lambda }_m^{\circ }-A\left(0.2\right)$
Subtracting the two equations gives: $A\left(0.1\right)=111-101=10$
Therefore, ${\Lambda }_m^{\circ }\left({\mathrm{NaNO}}_3\right)=111+10=121\text{ S cm}2\text{ mol}-1$.

2. For $\mathrm{NaCl}$:
At $C=0.01$, ${\Lambda }_m=117={\Lambda }_m^{\circ }-A\left(0.1\right)$
At $C=0.04$, ${\Lambda }_m=107={\Lambda }_m^{\circ }-A\left(0.2\right)$
Subtracting the two equations gives: $A\left(0.1\right)=117-107=10$
Therefore, ${\Lambda }_m^{\circ }\left(\mathrm{NaCl}\right)=117+10=127\text{ S cm}2\text{ mol}-1$.

3. For ${\mathrm{AgNO}}_3$:
At $C=0.01$, ${\Lambda }_m=125={\Lambda }_m^{\circ }-A\left(0.1\right)$
At $C=0.04$, ${\Lambda }_m=116={\Lambda }_m^{\circ }-A\left(0.2\right)$
Subtracting the two equations gives: $A\left(0.1\right)=125-116=9$
Therefore, ${\Lambda }_m^{\circ }\left({\mathrm{AgNO}}_3\right)=125+9=134\text{ S cm}2\text{ mol}-1$.

Step 3: Calculate limiting molar conductivity of AgCl
Substituting the calculated values into Kohlrausch's formula:

${\Lambda }_m^{\circ }\left(\mathrm{AgCl}\right)=134+127-121=140\text{ S cm}{2}^{\mathrm{}}\text{ mol}{\mathrm{-1}}^{\mathrm{}}$

Step 4: Find the solubility X
The molar conductivity of a saturated solution of a sparingly soluble salt is related to its specific conductivity ($\kappa$) and solubility ($X$ in mol L⁻¹) by:

${\Lambda }_m=\frac{\kappa \times 1000}{X}$

Given:
$\kappa =1.40\times {10}^{-6}\text{ S cm}-1$
${\Lambda }_m={\Lambda }_m^{\circ }=140\text{ S cm}2\text{ mol}-1$

Substitute these values to find $X$:

$140=\frac{1.40\times {10}^{-6}\times {10}^3}{X}$

$140=\frac{1.40\times {10}^{-3}}{X}$

$X=\frac{1.40\times {10}^{-3}}{140}={10}^{-5}\text{ mol L}{\mathrm{-1}}^{\mathrm{}}$

Step 5: Calculate ${\log }_{10}\left(X^{-1}\right)$
Now, find the value of $X^{-1}$:

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

Finally, taking the logarithm base 10:

${\log }_{10}\left(X^{-1}\right)={\log }_{10}\left({10}^5\right)=5$

Thus, the final required value is 5.