The solubility product of `AgCl` is `1.56xx10^(-10)`find solubility in g/ltr

Options

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

Explanation

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

### Step 1: Write the dissociation equation for AgCl
AgCl dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Define the solubility
Let the solubility of AgCl be \( S \) mol/L. At equilibrium, the concentrations of the ions will be:
- \([ \text{Ag}^+ ] = S\)
- \([ \text{Cl}^- ] = S\)

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product expression for AgCl is:
\[ K_{sp} = [ \text{Ag}^+ ][ \text{Cl}^- ] = S \cdot S = S^2 \]

### Step 4: Substitute the given Ksp value
Given that \( K_{sp} = 1.56 \times 10^{-10} \):
\[ S^2 = 1.56 \times 10^{-10} \]

### Step 5: Solve for S
To find \( S \), take the square root of both sides:
\[ S = \sqrt{1.56 \times 10^{-10}} \]
\[ S = 1.25 \times 10^{-5} \text{ mol/L} \]

### Step 6: Convert solubility from 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 is calculated as:
- Molar mass of Ag = 108 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Now, we can convert the solubility:
\[ \text{Solubility in g/L} = S \times \text{Molar mass of AgCl} \]
\[ \text{Solubility in g/L} = 1.25 \times 10^{-5} \text{ mol/L} \times 143.5 \text{ g/mol} \]

### Step 7: Calculate the final solubility in g/L
\[ \text{Solubility in g/L} = 1.25 \times 10^{-5} \times 143.5 \]
\[ \text{Solubility in g/L} = 1.79 \times 10^{-3} \text{ g/L} \]

### Final Answer
The solubility of AgCl in grams per liter is:
\[ \text{Solubility} = 1.79 \times 10^{-3} \text{ g/L} \]

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