The solubility product of AgCl is `1.5625xx10^(-10)` at `25^(@)C` . Its solubility in g per litre will be :-

Options

A. 143.5
B. 108
C. `1.57xx10^(-8)`
D. `1.79xx10^(-3)`

Answer

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

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of silver chloride (AgCl) in water can be represented as:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

2. **Define Solubility**:
Let the solubility of AgCl be \( S \) moles per liter. When AgCl dissolves, it produces \( S \) moles of Ag\(^+\) ions and \( S \) moles of Cl\(^-\) ions.

3. **Write the Expression for Ksp**:
The solubility product (Ksp) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \times S = S^2 \]
Given \( K_{sp} = 1.5625 \times 10^{-10} \).

4. **Calculate Solubility (S)**:
To find \( S \), take the square root of the Ksp:
\[ S = \sqrt{K_{sp}} = \sqrt{1.5625 \times 10^{-10}} \]
Calculating this gives:
\[ S = 1.25 \times 10^{-5} \text{ moles per liter} \]

5. **Convert Moles to Grams**:
To convert the solubility from moles per liter to grams per liter, we need the molar mass of AgCl. The molar mass of AgCl is calculated as follows:
- Molar mass of Ag = 107.87 g/mol
- Molar mass of Cl = 35.45 g/mol
- Therefore, molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol (rounded to 143.5 g/mol for simplicity).

Now, multiply the solubility in moles by the molar mass:
\[ \text{Solubility in g/L} = S \times \text{Molar mass} = (1.25 \times 10^{-5} \text{ mol/L}) \times (143.5 \text{ g/mol}) \]
\[ \text{Solubility in g/L} = 1.796875 \times 10^{-3} \text{ g/L} \approx 1.80 \times 10^{-3} \text{ g/L} \]

6. **Final Answer**:
The solubility of AgCl in grams per liter is approximately \( 1.80 \times 10^{-3} \text{ g/L} \).