The specific conductance of a saturated solution of AgCl at `25^@C` is `1.821xx10^(-5)` mho `cm^(-1)`. What is the solubility of AgCl in water (in g `L^(-1)` ) , if limiting molar conductivity of AgCl is 130.26 mho `cm^(2) mol^(-1)` ?

Options

A. `1.89xx10^(-3) g L^(-1)`
B. `2.78xx10^(-2) g L^(-1)`
C. `2.004xx10^(-2) g L^(-1)`
D. `1.43xx10^(-3) g L^(-1)`

Explanation

To find the solubility of AgCl in water (in g L^(-1)), we can follow these steps:

### Step 1: Understand the relationship between specific conductance, solubility, and molar conductivity.
The specific conductance (κ) of a solution is related to its solubility (S) and the limiting molar conductivity (Λm) by the formula:
\[ S = \frac{\kappa \times 1000}{\Lambda_m} \]
where:
- \(S\) is the solubility in mol L^(-1),
- \(\kappa\) is the specific conductance in mho cm^(-1),
- \(\Lambda_m\) is the limiting molar conductivity in mho cm^2 mol^(-1).

### Step 2: Substitute the given values into the formula.
Given:
- \(\kappa = 1.821 \times 10^{-5} \, \text{mho cm}^{-1}\)
- \(\Lambda_m = 130.26 \, \text{mho cm}^2 \text{mol}^{-1}\)

Now substituting these values into the formula:
\[ S = \frac{1.821 \times 10^{-5} \times 1000}{130.26} \]

### Step 3: Calculate the solubility in mol L^(-1).
Perform the calculation:
\[ S = \frac{1.821 \times 10^{-2}}{130.26} \approx 1.396 \times 10^{-4} \, \text{mol L}^{-1} \]

### Step 4: Convert the solubility from mol L^(-1) to g L^(-1).
To convert mol L^(-1) to g L^(-1), we need the molar mass of AgCl. The molar mass of AgCl is:
- Molar mass of Ag = 107.87 g/mol
- Molar mass of Cl = 35.45 g/mol
- Therefore, molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol

Now, we can calculate the solubility in g L^(-1):
\[ \text{Solubility in g L}^{-1} = S \times \text{Molar mass of AgCl} \]
\[ \text{Solubility in g L}^{-1} = 1.396 \times 10^{-4} \times 143.32 \approx 0.0200 \, \text{g L}^{-1} \]

### Final Answer:
The solubility of AgCl in water is approximately **0.0200 g L^(-1)**.

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