The specific conductivity of a saturated solution of AgCl is `3.40xx10^(-6) ohm^(-1) cm^(-1)` at `25^(@)C`. If `lambda_(Ag^(+)=62.3 ohm^(-1) cm^(2) "mol"^(-1)` and `lambda_(Cl^(-))=67.7 ohm^(-1) cm^(2) "mol"^(-1)`, the solubility of AgC at `25^(@)C` is:

Options

A. `2.6xx10^(-5) "mol" L^(-1)`
B. `3.731xx10^(-3) "mol" L^(-1)`
C. `3.731xx10^(-5) "mol" L^(-1)`
D. `2.6xx10^(-3)g L^(-1)`

Explanation

To find the solubility of AgCl in a saturated solution at 25°C, we will follow these steps:

### Step 1: Calculate the Molar Conductivity of AgCl
The molar conductivity (λ) of AgCl can be calculated using the formula:
\[ \lambda_{AgCl} = \lambda_{Ag^+} + \lambda_{Cl^-} \]
where:
- \(\lambda_{Ag^+} = 62.3 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)
- \(\lambda_{Cl^-} = 67.7 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)

Substituting the values:
\[ \lambda_{AgCl} = 62.3 + 67.7 = 130 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

### Step 2: Relate Specific Conductivity to Solubility
The specific conductivity (k) of the solution is related to the molar conductivity and the solubility (s) by the formula:
\[ \lambda = \frac{k \times 1000}{s} \]
Rearranging this gives us:
\[ s = \frac{k \times 1000}{\lambda} \]

### Step 3: Substitute the Values
Given that the specific conductivity \(k = 3.40 \times 10^{-6} \, \text{ohm}^{-1} \, \text{cm}^{-1}\), we can substitute the values into the equation:
\[ s = \frac{3.40 \times 10^{-6} \times 1000}{130} \]

### Step 4: Calculate the Solubility
Calculating the numerator:
\[ 3.40 \times 10^{-6} \times 1000 = 3.40 \times 10^{-3} \]
Now, substituting this into the equation for s:
\[ s = \frac{3.40 \times 10^{-3}}{130} \]
Calculating this gives:
\[ s = 2.615384615 \times 10^{-5} \, \text{mol L}^{-1} \]
Rounding this to two significant figures:
\[ s \approx 2.6 \times 10^{-5} \, \text{mol L}^{-1} \]

### Final Answer
The solubility of AgCl at 25°C is:
\[ \boxed{2.6 \times 10^{-5} \, \text{mol L}^{-1}} \]

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