The `K_(sp)` of `AgCl` at `25^(@)C` is `2.56 xx 10^(-10)`. Then how much
volume of `H_(2)O` is required to dissolve `0.01` mole of salt ?

Options

A. `800 L`
B. `400 L`
C. `625 L`
D. `50 L`

Explanation

To solve the problem of how much volume of water is required to dissolve 0.01 moles of AgCl 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: Write the expression for Ksp
The solubility product (Ksp) for AgCl can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 3: Define solubility (S)
Let the solubility of AgCl in water be \( S \) moles per liter. At equilibrium, the concentration of Ag\(^+\) and Cl\(^-\) ions will both be \( S \):
\[ K_{sp} = S \times S = S^2 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} = 2.56 \times 10^{-10} \):
\[ S^2 = 2.56 \times 10^{-10} \]

### Step 5: Solve for S
Taking the square root of both sides to find \( S \):
\[ S = \sqrt{2.56 \times 10^{-10}} \]
\[ S \approx 1.6 \times 10^{-5} \text{ moles per liter} \]

### Step 6: Relate moles to volume
We are given that we need to dissolve 0.01 moles of AgCl. The relationship between moles, volume, and solubility can be expressed as:
\[ \text{Solubility} = \frac{\text{moles}}{\text{volume}} \]
Thus,
\[ S = \frac{0.01 \text{ moles}}{V} \]

### Step 7: Substitute S into the equation
Substituting \( S \) into the equation:
\[ 1.6 \times 10^{-5} = \frac{0.01}{V} \]

### Step 8: Solve for V
Rearranging the equation to solve for volume \( V \):
\[ V = \frac{0.01}{1.6 \times 10^{-5}} \]
\[ V \approx 625 \text{ liters} \]

### Conclusion
The volume of water required to dissolve 0.01 moles of AgCl is approximately **625 liters**.

---