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, we need to determine the volume of `0.1 M AgNO3` required to completely precipitate the chloride ions from the given complex `[Cr(H2O)5Cl]Cl2`. Here’s the step-by-step solution:

### Step 1: Determine the number of moles of chloride ions in the complex
The complex `[Cr(H2O)5Cl]Cl2` contains 2 chloride ions. Given that the concentration of the complex solution is `0.01 M` and the volume is `30 mL`, we can calculate the number of moles of the complex first.

\[ \text{Number of moles of complex} = \text{Molarity} \times \text{Volume (in L)} \]

Convert the volume from mL to L:

\[ 30 \, \text{mL} = 0.030 \, \text{L} \]

Now calculate the moles:

\[ \text{Number of moles of complex} = 0.01 \, \text{mol/L} \times 0.030 \, \text{L} = 0.0003 \, \text{moles} \]

Since each complex releases 2 chloride ions, the total number of moles of chloride ions is:

\[ \text{Number of moles of Cl} = 2 \times 0.0003 = 0.0006 \, \text{moles} \]

### Step 2: Determine the moles of AgNO3 required
The precipitation reaction between AgNO3 and Cl^- ions is:

\[ Ag^+ + Cl^- \rightarrow AgCl \, (s) \]

From the reaction, we see that 1 mole of AgNO3 is required for 1 mole of Cl^-. Therefore, the number of moles of AgNO3 required is equal to the number of moles of chloride ions:

\[ \text{Moles of AgNO3 required} = 0.0006 \, \text{moles} \]

### Step 3: Calculate the volume of AgNO3 solution required
We know the molarity of the AgNO3 solution is `0.1 M`. We can use the formula:

\[ \text{Volume (in L)} = \frac{\text{Number of moles}}{\text{Molarity}} \]

Substituting the values:

\[ \text{Volume (in L)} = \frac{0.0006 \, \text{moles}}{0.1 \, \text{mol/L}} = 0.006 \, \text{L} \]

Convert the volume from liters to milliliters:

\[ 0.006 \, \text{L} = 6 \, \text{mL} \]

### Final Answer
The volume of `0.1 M AgNO3` required for complete precipitation of chloride ions is **6 mL**.

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