Solubility product of AgCl is `1 xx 10^(-6)` at 298 K. Its solubility in mole `"litre"^(-1)` would be

Options

A. `1 xx 10^(-6)` mol/ litre
B. `1 xx 10^(-3)` mol/litre
C. `1 xx 10^(-12)` mol/litre
D. None of these

Explanation

To find the solubility of AgCl in moles per liter given its solubility product (Ksp), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Dissociation of AgCl**:
AgCl dissociates in water according to the following equation:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
When 1 mole of AgCl dissolves, it produces 1 mole of Ag⁺ ions and 1 mole of Cl⁻ ions.

2. **Define Solubility**:
Let the solubility of AgCl be \( S \) moles per liter. This means that at equilibrium:
- The concentration of Ag⁺ ions = \( S \)
- The concentration of Cl⁻ ions = \( S \)

3. **Write the Expression for Ksp**:
The solubility product (Ksp) is defined as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations from the solubility:
\[ K_{sp} = S \times S = S^2 \]

4. **Substitute the Given Ksp Value**:
We are given that \( K_{sp} \) of AgCl is \( 1 \times 10^{-6} \):
\[ S^2 = 1 \times 10^{-6} \]

5. **Solve for S**:
To find the solubility \( S \), we take the square root of both sides:
\[ S = \sqrt{1 \times 10^{-6}} = 10^{-3} \]

6. **Express the Final Answer with Units**:
The solubility of AgCl in moles per liter is:
\[ S = 10^{-3} \, \text{mol/L} \]

### Final Answer:
The solubility of AgCl at 298 K is \( 1 \times 10^{-3} \, \text{mol/L} \).

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