The solubility product of `AgCl` is `10^(-10)M^(2)`. The minimum volume ( in `m^(3))` of water required to dissolve `14.35 mg` of `AgCl` is approximately `:`

Options

A. `0.01`
B. `0.1`
C. 100
D. 10

Answer

To solve the problem, we need to determine the minimum volume of water required to dissolve 14.35 mg of AgCl, given that the solubility product (Ksp) of AgCl is \(10^{-10} \, \text{M}^2\).

### Step-by-Step Solution:

1. **Determine the solubility (S) of AgCl:**
The solubility product expression for AgCl can be written as:
\[ K_{sp} = [Ag^+][Cl^-] \]
Since AgCl dissociates into one Ag\(^+\) ion and one Cl\(^-\) ion, we can let the solubility \(S\) be the concentration of Ag\(^+\) and Cl\(^-\) ions:
\[ K_{sp} = S \cdot S = S^2 \]
Given \(K_{sp} = 10^{-10}\):
\[ S^2 = 10^{-10} \implies S = \sqrt{10^{-10}} = 10^{-5} \, \text{M} \]

2. **Convert the mass of AgCl to moles:**
The molar mass of AgCl is approximately 143.5 g/mol. First, convert the mass from mg to grams:
\[ 14.35 \, \text{mg} = 14.35 \times 10^{-3} \, \text{g} \]
Now, calculate the number of moles of AgCl:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{14.35 \times 10^{-3}}{143.5} \approx 1.000 \times 10^{-4} \, \text{mol} \]

3. **Calculate the volume of water needed:**
We know the concentration of AgCl in solution is equal to the solubility \(S\):
\[ [Ag^+] = [Cl^-] = S = 10^{-5} \, \text{M} \]
Using the formula for concentration:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{Number of moles}}{S} = \frac{1.000 \times 10^{-4}}{10^{-5}} = 10 \, \text{L} \]

4. **Convert volume from liters to cubic meters:**
Since \(1 \, \text{m}^3 = 1000 \, \text{L}\):
\[ \text{Volume (m}^3\text{)} = \frac{10 \, \text{L}}{1000} = 0.01 \, \text{m}^3 \]

### Final Answer:
The minimum volume of water required to dissolve 14.35 mg of AgCl is approximately \(0.01 \, \text{m}^3\).