The `K_(sp)` of `AgC1` at `25^(@)C` is `1.6 xx 10^(-9)`, find the solubility of salt in `gL^(-1)` in water.

Explanation

To find the solubility of AgCl in grams per liter (g/L) given its solubility product constant (Ksp), we will follow these steps:

### Step 1: Write the dissociation equation and Ksp expression
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

The Ksp expression for this equilibrium is:
\[ K_{sp} = [Ag^+][Cl^-] \]

### Step 2: Define the solubility (s)
Let the solubility of AgCl in water be \( s \) mol/L. At equilibrium, the concentrations of Ag⁺ and Cl⁻ ions will both be \( s \) mol/L.

### Step 3: Substitute into the Ksp expression
Substituting the concentrations into the Ksp expression gives:
\[ K_{sp} = s \cdot s = s^2 \]

### Step 4: Set up the equation with the given Ksp value
Given that \( K_{sp} = 1.6 \times 10^{-9} \), we can write:
\[ s^2 = 1.6 \times 10^{-9} \]

### Step 5: Solve for s
Taking the square root of both sides:
\[ s = \sqrt{1.6 \times 10^{-9}} \]
\[ s = 4.0 \times 10^{-5} \, \text{mol/L} \]

### Step 6: Convert to grams per liter
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 approximately 143.32 g/mol.

Now, we can calculate the solubility in g/L:
\[ \text{Solubility (g/L)} = s \times \text{Molar mass of AgCl} \]
\[ \text{Solubility (g/L)} = 4.0 \times 10^{-5} \, \text{mol/L} \times 143.32 \, \text{g/mol} \]
\[ \text{Solubility (g/L)} = 0.00572 \, \text{g/L} \]

### Final Answer
The solubility of AgCl in water at 25°C is approximately **0.00572 g/L**.

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