The specific conductivity of a saturated solution of silver chloride is `2.30 xx 10^(-6) m ho cm^(-1) " at " 25^(@)C`. Calculate the solubility of silver chloride at `25^(@)C " if " lamda_(Ag^(+)) = 61.9 m ho cm^(2) mol^(-1) and lamda_(Cl^(-)) = 76.3 m ho cm^(2) mol^(-1)`

Explanation

To solve the problem of calculating the solubility of silver chloride (AgCl) at 25°C given the specific conductivity and molar conductivities of the ions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Specific conductivity of AgCl solution, \( \kappa = 2.30 \times 10^{-6} \, \text{mho cm}^{-1} \)
- Molar conductivity of \( \text{Ag}^+ \), \( \lambda_{\text{Ag}^+} = 61.9 \, \text{mho cm}^2 \text{mol}^{-1} \)
- Molar conductivity of \( \text{Cl}^- \), \( \lambda_{\text{Cl}^-} = 76.3 \, \text{mho cm}^2 \text{mol}^{-1} \)
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