How many mloes of of `AgCI` would be obtained when `100mL` of `0.1 M Co(NH_(3))_(5)Cl_(3)` is treated with excess of `AgNO_(3)` ?
(a) 0.01
(b) `0.02`
(c ) `0.03`
(d )None of these .

Explanation

To determine how many moles of AgCl would be obtained when 100 mL of 0.1 M Co(NH3)5Cl3 is treated with excess AgNO3, we can follow these steps:

### Step 1: Calculate the number of moles of Co(NH3)5Cl3
The formula to calculate the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
Given:
- Molarity = 0.1 M
- Volume = 100 mL = 0.1 L

Now, substituting the values:
\[ \text{Number of moles} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]

### Step 2: Determine the number of free chloride ions
The complex Co(NH3)5Cl3 can dissociate to release chloride ions. The dissociation can be represented as:
\[ \text{Co(NH3)5Cl3} \rightarrow \text{Co(NH3)5}^{3+} + 3 \text{Cl}^- \]
From this dissociation, we see that 1 mole of Co(NH3)5Cl3 produces 3 moles of Cl⁻ ions.

### Step 3: Calculate the total moles of chloride ions
Since we have 0.01 moles of Co(NH3)5Cl3, the number of moles of chloride ions produced will be:
\[ \text{Moles of Cl}^- = 0.01 \, \text{moles of Co(NH3)5Cl3} \times 3 = 0.03 \, \text{moles of Cl}^- \]

### Step 4: Determine the moles of AgCl formed
When AgNO3 is added, each Cl⁻ ion will react with Ag⁺ to form AgCl:
\[ \text{Cl}^- + \text{Ag}^+ \rightarrow \text{AgCl} \]
Thus, the number of moles of AgCl formed will be equal to the number of moles of Cl⁻ ions present:
\[ \text{Moles of AgCl} = 0.03 \, \text{moles of Cl}^- = 0.03 \, \text{moles of AgCl} \]

### Conclusion
Therefore, the number of moles of AgCl that would be obtained is **0.03 moles**.

### Answer
(c) 0.03

---