If excess of `AgNO_(3)` solution is added to 100mL of a 0.024 M solution of dichlorobis (ethylenediamine) cobalt (III) chloride . How many mole of AgCl be precipitated ?

Options

A. `0.0012`
B. `0.0016`
C. `0.0024`
D. `0.0048`

Explanation

To solve the problem, we need to determine how many moles of AgCl will be precipitated when excess AgNO₃ is added to a solution of dichlorobis(ethylenediamine)cobalt(III) chloride.

### Step-by-Step Solution:

1. **Identify the given data**:
- Volume of the dichlorobis(ethylenediamine)cobalt(III) chloride solution = 100 mL = 0.1 L
- Molarity of the solution = 0.024 M

2. **Calculate the number of moles of dichlorobis(ethylenediamine)cobalt(III) chloride**:
- Use the formula:
\[ n = \text{Molarity} \times \text{Volume} \]
- Substitute the values:
\[ n = 0.024 \, \text{mol/L} \times 0.1 \, \text{L} = 0.0024 \, \text{moles} \]

3. **Determine the number of chloride ions**:
- The formula for dichlorobis(ethylenediamine)cobalt(III) chloride indicates that there are 2 chloride ions per molecule. Therefore, the number of moles of chloride ions is:
\[ \text{Moles of Cl}^- = 2 \times 0.0024 = 0.0048 \, \text{moles} \]

4. **Reaction with AgNO₃**:
- When excess AgNO₃ is added, it will react with the chloride ions to form AgCl. The reaction is:
\[ \text{Cl}^- + \text{Ag}^+ \rightarrow \text{AgCl} \]
- Since we have 0.0048 moles of chloride ions, this means that 0.0048 moles of AgCl will be precipitated.

5. **Conclusion**:
- The total moles of AgCl precipitated will be **0.0048 moles**.

### Final Answer:
0.0048 moles of AgCl will be precipitated.