The volume (in mL) of 1.0 M `AgNO_3` required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of `[Cr(H_2O)_6]Cl_3`, as silver chloride is close to

Explanation

To solve the problem, we need to calculate the volume of 1.0 M AgNO3 required to completely precipitate the chloride ions from a given solution of [Cr(H2O)6]Cl3.

### Step-by-Step Solution:

1. **Identify the number of chloride ions in the compound:**
The compound [Cr(H2O)6]Cl3 contains 3 chloride ions (Cl-) because the subscript indicates that there are 3 Cl ions associated with the complex.

2. **Calculate the total moles of chloride ions:**
We are given the molarity (0.01 M) and volume (30 mL) of the [Cr(H2O)6]Cl3 solution. To find the total moles of chloride ions:
\[ \text{Moles of Cl-} = \text{Molarity} \times \text{Volume (in L)} \]
Convert volume from mL to L:
\[ 30 \text{ mL} = 0.030 \text{ L} \]
Now calculate:
\[ \text{Moles of Cl-} = 0.01 \text{ M} \times 0.030 \text{ L} = 0.0003 \text{ moles} \]
Since each [Cr(H2O)6]Cl3 provides 3 Cl- ions:
\[ \text{Total moles of Cl-} = 3 \times 0.0003 = 0.0009 \text{ moles} \]

3. **Convert moles of Cl- to millimoles:**
\[ \text{Millimoles of Cl-} = 0.0009 \text{ moles} \times 1000 = 0.9 \text{ mmol} \]

4. **Determine the moles of Ag+ required:**
For complete precipitation, the moles of Ag+ required will equal the moles of Cl-:
\[ \text{Moles of Ag+} = 0.0009 \text{ moles} \]

5. **Calculate the volume of 1.0 M AgNO3 needed:**
We know that molarity (M) is defined as moles of solute per liter of solution. Therefore, we can rearrange the formula to find the volume:
\[ \text{Volume (L)} = \frac{\text{Moles of Ag+}}{\text{Molarity of AgNO3}} = \frac{0.0009 \text{ moles}}{1.0 \text{ M}} = 0.0009 \text{ L} \]
Convert this volume to mL:
\[ 0.0009 \text{ L} = 0.9 \text{ mL} \]

### Final Answer:
The volume of 1.0 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)6]Cl3 is **0.9 mL**.