The conductivity of a solution of AgCl at 298 K is found to be `1.382xx10^(-6)Omega^(-1)cm^(-1)` the ionic conductance of `Ag^(+)` and `Cl^(-)` at infinite dilution are `61.9Omega^(-1)cm^(2)col^(-1)` ad `76.3Omega^(-1)cm^(2)mol^(-1)` respectively the solubility of AgCl is

Options

A. `1.4xx10^(-5)molL^(-1)`
B. `1xx10^(-2)molL^(-1)`
C. `1xx10^(-5)molL^(-1)`
D. `1.9xx10^(-5)molL^(-1)`

Explanation

To find the solubility of AgCl based on the given conductivity and ionic conductance values, we can follow these steps:

### Step 1: Understand the given values
- Conductivity of AgCl solution, \( k = 1.382 \times 10^{-6} \, \Omega^{-1} \text{cm}^{-1} \)
- Ionic conductance of \( Ag^+ \) at infinite dilution, \( \lambda_{Ag^+} = 61.9 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- Ionic conductance of \( Cl^- \) at infinite dilution, \( \lambda_{Cl^-} = 76.3 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)

### Step 2: Calculate the molar conductivity of AgCl at infinite dilution
The molar conductivity \( \Lambda_m \) at infinite dilution is the sum of the ionic conductances of the ions:
\[ \Lambda_m = \lambda_{Ag^+} + \lambda_{Cl^-} \]
Substituting the values:
\[ \Lambda_m = 61.9 + 76.3 = 138.2 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Step 3: Relate conductivity to molar conductivity and solubility
The relationship between conductivity \( k \), molar conductivity \( \Lambda_m \), and solubility \( S \) is given by:
\[ k = \Lambda_m \times S \]
Rearranging this equation to solve for solubility \( S \):
\[ S = \frac{k}{\Lambda_m} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ S = \frac{1.382 \times 10^{-6} \, \Omega^{-1} \text{cm}^{-1}}{138.2 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1}} \]

### Step 5: Calculate the solubility
Calculating \( S \):
\[ S = \frac{1.382 \times 10^{-6}}{138.2} = 1.000 \times 10^{-8} \, \text{mol/cm}^3 \]
To convert this to mol/L:
\[ S = 1.000 \times 10^{-8} \, \text{mol/cm}^3 \times 1000 \, \text{cm}^3/\text{L} = 1.000 \times 10^{-5} \, \text{mol/L} \]

### Final Answer
The solubility of AgCl is:
\[ S = 1.000 \times 10^{-5} \, \text{mol/L} \]

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