The specific conductivity of a saturated solution of silver chloride at `18^(@)C` is `1.24 xx 10^(-6)` mho after subtracing that of water. Ionic conductances at infinite dilute of `Ag^(+)` and `Cl^(-)` ions at this temperature are 53.8 and 65.3 respectively. Calculate the solubility of silver chloride in gram per litre.

Answer

To solve the problem of calculating the solubility of silver chloride (AgCl) in grams per liter from the given specific conductivity and ionic conductances, we can follow these steps:

### Step 1: Understand the given data
- Specific conductivity of saturated AgCl solution at 18°C: \( \kappa = 1.24 \times 10^{-6} \, \text{mho/m} \)
- Ionic conductance at infinite dilution:
- \( \lambda_{Ag^+} = 53.8 \, \text{mho cm}^2/\text{mol} \)
- \( \lambda_{Cl^-} = 65.3 \, \text{mho cm}^2/\text{mol} \)

### Step 2: Calculate the total ionic conductance of AgCl
The total ionic conductance at infinite dilution for AgCl can be calculated by adding the individual ionic conductances of the ions:
\[ \lambda_{AgCl} = \lambda_{Ag^+} + \lambda_{Cl^-} = 53.8 + 65.3 = 119.1 \, \text{mho cm}^2/\text{mol} \]

### Step 3: Use the formula for solubility
The solubility (S) of AgCl can be calculated using the formula:
\[ S = \frac{\kappa \times 1000}{\lambda_{AgCl}} \]
Where:
- \( \kappa \) is the specific conductivity in mho/m,
- \( \lambda_{AgCl} \) is the total ionic conductance in mho cm²/mol,
- The factor of 1000 is used to convert the units appropriately.

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ S = \frac{1.24 \times 10^{-6} \, \text{mho/m} \times 1000}{119.1 \, \text{mho cm}^2/\text{mol}} \]

### Step 5: Calculate the solubility
First, convert \( \text{mho/m} \) to \( \text{mho cm} \) by noting that \( 1 \, \text{mho/m} = 100 \, \text{mho cm} \):
\[ S = \frac{1.24 \times 10^{-6} \times 1000}{119.1} = \frac{1.24 \times 10^{-3}}{119.1} \]
Now, performing the division:
\[ S \approx 1.041 \times 10^{-5} \, \text{mol/L} \]

### Step 6: Convert moles to grams
To find the solubility in grams per liter, we need the molar mass of AgCl. The molar mass of AgCl is approximately \( 143.32 \, \text{g/mol} \):
\[ \text{Solubility in grams/L} = S \times \text{molar mass of AgCl} = 1.041 \times 10^{-5} \, \text{mol/L} \times 143.32 \, \text{g/mol} \]
Calculating this gives:
\[ \text{Solubility} \approx 1.494 \times 10^{-3} \, \text{g/L} \]

### Final Answer
The solubility of silver chloride in a saturated solution at 18°C is approximately \( 1.494 \times 10^{-3} \, \text{g/L} \).

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