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 138 `ohm^(-1) cm^(2) "mol"^(-1)` then the solubility of AgCl in "mole"s per litre 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 can follow these steps:

### Step 1: Understand the relationship between conductivity and solubility
For a saturated solution of a sparingly soluble salt like AgCl, the molar conductivity (λ_M) can be expressed in terms of the specific conductance (k) and the solubility (S) of the salt:
\[ \lambda_M = \frac{k \times 1000}{S} \]
where:
- \( \lambda_M \) is the molar conductivity of AgCl,
- \( k \) is the specific conductance of the saturated solution,
- \( S \) is the solubility in moles per liter.

### Step 2: Calculate the specific conductance of pure AgCl
The specific conductance of the saturated solution of AgCl is given as \( k = 3.4 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \). The specific conductance of pure water is given as \( k_{H_2O} = 2.02 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \).

To find the specific conductance of pure AgCl, we subtract the conductance of water from that of the saturated solution:
\[ k_{AgCl} = k - k_{H_2O} = (3.4 \times 10^{-6}) - (2.02 \times 10^{-6}) = 1.38 \times 10^{-6} \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 3: Set up the equation for molar conductivity
Now we can use the molar conductivity of AgCl at infinite dilution, which is given as \( \lambda_M = 138 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \). We can substitute the values into the equation:
\[ 138 = \frac{1.38 \times 10^{-6} \times 1000}{S} \]

### Step 4: Solve for solubility (S)
Rearranging the equation to solve for S:
\[ S = \frac{1.38 \times 10^{-6} \times 1000}{138} \]
\[ S = \frac{1.38 \times 10^{-3}}{138} \]
\[ S = 1.0 \times 10^{-5} \, \text{moles per liter} \]

### Final Answer
The solubility of AgCl in moles per liter is \( 1.0 \times 10^{-5} \, \text{mol/L} \).

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