The specific conductivity of a standard solution of AgCl is `1.40 xx 10^(-6) "ohm"^(-1) cm^(-1)` at `25^(@)C`. If `lambda_(Ag^(+))^(@) = 62.3 "ohm"^(-1) cm^(2) mol^(-1)` & `lambda_(Cl^(-1)) = 67.7 "ohm"^(-1) cm^(2) "mol"^(-1)`, the solubility of AgCl at `25^(@) C` is:

Options

A. `2.6 xx 10^(-5) M`
B. `4.5 xx 10^(-3) M`
C. `3.6 xx 10^(-5) M`
D. `3.6 xx 10^(-1) M`

Explanation

To find the solubility of AgCl at 25°C given the specific conductivity and the molar conductivities of its ions, we can follow these steps:

### Step 1: Understand the given data
- Specific conductivity of AgCl (\( \kappa \)) = \( 1.40 \times 10^{-6} \, \Omega^{-1} \, \text{cm}^{-1} \)
- Molar conductivity of \( Ag^+ \) (\( \lambda_{Ag^+} \)) = \( 62.3 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)
- Molar conductivity of \( Cl^- \) (\( \lambda_{Cl^-} \)) = \( 67.7 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)

### Step 2: Calculate the molar conductivity of AgCl
The molar conductivity of a salt at infinite dilution (\( \Lambda_m \)) is the sum of the molar conductivities of its ions:
\[ \Lambda_m = \lambda_{Ag^+} + \lambda_{Cl^-} \]
Substituting the values:
\[ \Lambda_m = 62.3 + 67.7 = 130.0 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

### Step 3: Use the formula for solubility
The solubility (\( S \)) of the salt can be calculated using the formula:
\[ S = \frac{\kappa \times 1000}{\Lambda_m} \]
Where:
- \( \kappa \) is the specific conductivity in \( \Omega^{-1} \, \text{cm}^{-1} \)
- \( \Lambda_m \) is the molar conductivity in \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)

### Step 4: Substitute the values into the formula
Substituting the known values:
\[ S = \frac{1.40 \times 10^{-6} \times 1000}{130.0} \]

### Step 5: Calculate the solubility
Calculating the numerator:
\[ 1.40 \times 10^{-6} \times 1000 = 1.40 \times 10^{-3} \]
Now, divide by \( 130.0 \):
\[ S = \frac{1.40 \times 10^{-3}}{130.0} \approx 1.0769 \times 10^{-5} \, \text{mol/cm}^3 \]

### Step 6: Convert to molarity
Since \( 1 \, \text{cm}^3 = 1 \, \text{mL} \), we can express the solubility in molarity (M):
\[ S \approx 1.08 \times 10^{-5} \, \text{M} \]

### Final Answer
The solubility of AgCl at 25°C is approximately \( 1.08 \times 10^{-5} \, \text{mol/L} \).

---