For a saturated solution of `AgCl` at `25^(@)C,k=3.4xx10^(-6)ohm^(-1)cm^(-1)` and that of `H_(2)O(l)` used is `2.02xx10^(-6)ohm^(-1)cm^(-1). lambda_(m)^(@)` for `AgCl` is `138ohm^(-1)cm^(2)mol^(-1)` then the solubility of AgCl in moles per liter will be

Options

A. `10^(-5)`
B. `10^(-10)`
C. `10^(-14)`
D. `10^(-16)`

Explanation

To find the solubility of AgCl in moles per liter, we will follow these steps:

### Step 1: Understand the Given Data
We have the following information:
- Conductivity of saturated AgCl solution, \( \kappa = 3.4 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \)
- Conductivity of water, \( \kappa_{\text{water}} = 2.02 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \)
- Molar conductivity of AgCl at infinite dilution, \( \lambda_m^0 = 138 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \)

### Step 2: Calculate the Conductivity of AgCl Solution
The conductivity of the AgCl solution can be calculated by subtracting the conductivity of water from the conductivity of the saturated AgCl solution:
\[ \kappa_{\text{AgCl}} = \kappa - \kappa_{\text{water}} = 3.4 \times 10^{-6} - 2.02 \times 10^{-6} = 1.38 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Use the Formula for Molar Conductivity
The relationship between molar conductivity, conductivity, and solubility is given by:
\[ \lambda_m^0 = \frac{\kappa \times 1000}{S} \]
Where:
- \( \lambda_m^0 \) is the molar conductivity at infinite dilution,
- \( \kappa \) is the conductivity of the solution,
- \( S \) is the solubility in moles per liter.

### Step 4: Rearrange the Formula to Find Solubility
Rearranging the formula to solve for solubility \( S \):
\[ S = \frac{\kappa \times 1000}{\lambda_m^0} \]

### Step 5: Substitute the Values
Substituting the values we have:
\[ S = \frac{1.38 \times 10^{-6} \times 1000}{138} \]

### Step 6: Calculate the Solubility
Calculating the above expression:
\[ S = \frac{1.38 \times 10^{-3}}{138} = 1.0 \times 10^{-5} \, \text{mol/L} \]

### Conclusion
The solubility of AgCl in moles per liter is:
\[ \boxed{1.0 \times 10^{-5} \, \text{mol/L}} \]