The specific conductivity of a saturated `AgCl` solution if found to be `2.12 xx 10^(-6)S cm^(-1)` and that for water is `6 xx 10^(-8)S cm^(-1)`. The solubility of `AgCl` is: `(lambda_(eq)^(prop) = 103 S "equiv"^(-1) cm^(2))`

Explanation

To find the solubility of AgCl from the given specific conductivities, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Specific conductivity of saturated AgCl solution, \( K_{AgCl} = 2.12 \times 10^{-6} \, S \, cm^{-1} \)
- Specific conductivity of water, \( K_{H_2O} = 6 \times 10^{-8} \, S \, cm^{-1} \)
- Equivalent conductivity at infinite dilution, \( \lambda_{eq} = 103 \, S \, equiv^{-1} \, cm^{2} \)
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