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, we can follow these steps:

### Step 1: Understand the given values
- Specific conductance (k) of saturated AgCl solution = \(1.821 \times 10^{-5} \, \text{mho cm}^{-1}\)
- Limiting molar conductivity (\(\lambda_0\)) of AgCl = \(130.26 \, \text{mho cm}^2 \text{mol}^{-1}\)

### Step 2: Convert specific conductance to molarity
The relationship between specific conductance (k), molarity (S), and limiting molar conductivity (\(\lambda_0\)) is given by the formula:
\[ S = \frac{k \times 1000}{\lambda_0} \]
where \(S\) is the solubility in mol/L (molarity) and \(k\) is in mho/cm.

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ S = \frac{1.821 \times 10^{-5} \, \text{mho cm}^{-1} \times 1000}{130.26 \, \text{mho cm}^2 \text{mol}^{-1}} \]

### Step 4: Calculate the solubility in mol/L
Calculating the above expression:
\[ S = \frac{1.821 \times 10^{-2}}{130.26} \approx 1.397 \times 10^{-4} \, \text{mol/L} \]

### Step 5: Convert molarity to grams per liter
To convert molarity to grams per liter, we need the molar mass of AgCl. The molar mass of AgCl is calculated as:
\[ \text{Molar mass of AgCl} = 108 \, \text{g/mol (Ag)} + 35.5 \, \text{g/mol (Cl)} = 143.5 \, \text{g/mol} \]

Now, we can calculate the solubility in grams per liter:
\[ \text{Solubility (g/L)} = S \times \text{Molar mass} = 1.397 \times 10^{-4} \, \text{mol/L} \times 143.5 \, \text{g/mol} \]

### Step 6: Final calculation
Calculating the above expression:
\[ \text{Solubility (g/L)} \approx 1.997 \times 10^{-2} \, \text{g/L} \]

### Conclusion
Thus, the solubility of AgCl in water is approximately \(2.004 \times 10^{-2} \, \text{g/L}\).

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