Volume of Silver Nitrate for Chloride Precipitation in Coordination Complex

To find the volume of $\text{AgNO}_3$ required for the complete precipitation of chloride ions, we first consider the ionization of the coordination complex in an aqueous solution.

The given complex is pentaaquachloridochromium(III) chloride, with the chemical formula:
$[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2$

Only the chloride ions present outside the coordination sphere (the counter ions) ionize in solution:
$[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\text{ (aq)} \longrightarrow [\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]^{2+}\text{ (aq)} + 2\text{Cl}^-\text{ (aq)}$

The chloride ion inside the coordination sphere is directly bonded to the chromium metal center and does not ionize to form a precipitate with silver nitrate under normal conditions. Thus, each mole of the complex yields $2\text{ moles}$ of precipitable chloride ions:
$\text{Moles of } \text{Cl}^- = 2 \times \text{Moles of } [\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2$

Next, we calculate the millimoles of the coordination complex present in the solution:
$\text{Millimoles of complex} = M \times V = 0.01\text{ M} \times 30\text{ mL} = 0.3\text{ mmol}$

The millimoles of precipitable $\text{Cl}^-$ ions in the solution are:
$\text{Millimoles of } \text{Cl}^- = 2 \times 0.3\text{ mmol} = 0.6\text{ mmol}$

The precipitation reaction with $\text{AgNO}_3$ proceeds as:
$\text{Ag}^+\text{ (aq)} + \text{Cl}^-\text{ (aq)} \longrightarrow \text{AgCl}\text{ (s)}\downarrow$

From the stoichiometry of the reaction, $1\text{ mole}$ of $\text{Ag}^+$ reacts with $1\text{ mole}$ of $\text{Cl}^-$:
$\text{Millimoles of } \text{AgNO}_3 \text{ required} = \text{Millimoles of } \text{Cl}^- = 0.6\text{ mmol}$

Let $V$ be the volume (in $\text{mL}$) of $0.1\text{ M } \text{AgNO}_3$ solution required. Then:
$\text{Molarity of } \text{AgNO}_3 \times V = 0.6\text{ mmol}$
$0.1\text{ M} \times V = 0.6\text{ mmol}$
$V = \frac{0.6}{0.1} = 6\text{ mL}$

Therefore, the volume of $0.1\text{ M } \text{AgNO}_3$ required is 6.