The volume (in `mL`) of `0.1M Ag NO_(3)` required for complete precipitation of chloride ions present in `30 mL` of `0.01M` solution of `[Cr(H_(2)O)_(5)Cl]Cl_(2)`, as silver chloride is close to:

Explanation

To solve the problem of determining the volume of 0.1 M AgNO₃ required for the complete precipitation of chloride ions from a 30 mL solution of 0.01 M [Cr(H₂O)₅Cl]Cl₂, we can follow these steps:

### Step 1: Calculate the moles of chloride ions (Cl⁻) in the solution
Given:
- Volume of the solution = 30 mL = 30/1000 L = 0.030 L
- Molarity of the solution = 0.01 M

Using the formula for moles:
\[ \text{Moles of Cl}^- = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of Cl}^- = 0.01 \, \text{mol/L} \times 0.030 \, \text{L} = 0.0003 \, \text{mol} \]

### Step 2: Determine the number of moles of chloride ions
Since the complex [Cr(H₂O)₅Cl]Cl₂ contains 3 chloride ions (1 from the complex and 2 from the counter ions), we multiply the moles of Cl⁻ by 3:
\[ \text{Total moles of Cl}^- = 3 \times 0.0003 \, \text{mol} = 0.0009 \, \text{mol} \]

### Step 3: Calculate the volume of AgNO₃ required
We know that 1 mole of Cl⁻ reacts with 1 mole of AgNO₃ to form AgCl. Therefore, the moles of AgNO₃ required will be equal to the moles of Cl⁻:
\[ \text{Moles of AgNO}_3 = 0.0009 \, \text{mol} \]

Now we can use the molarity of AgNO₃ to find the volume required:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]
\[ \text{Volume (L)} = \frac{0.0009 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.009 \, \text{L} \]

### Step 4: Convert the volume from liters to milliliters
To convert liters to milliliters, we multiply by 1000:
\[ \text{Volume (mL)} = 0.009 \, \text{L} \times 1000 = 9 \, \text{mL} \]

### Final Answer
The volume of 0.1 M AgNO₃ required for complete precipitation of chloride ions is **9 mL**.

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