The volume (in $\mathrm{mL}$) of $0.1 \mathrm{M} ~\mathrm{AgNO}_{3}$ required for complete precipitation of chloride ions present in $20 \mathrm{~mL}$ of $0.01 \mathrm{M}$ solution of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ as silver chloride is __________.

The given solution contains a complex ion $\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2$ which contains one chloride ion per complex. Therefore, when this complex is treated with silver nitrate, each mole of the complex will consume 2 moles of silver nitrate to form two moles of silver chloride. <br/><br/> The first step is to write the balanced chemical equation for the reaction between the complex and silver nitrate, as follows: <br/><br/> $$\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2 + 2 \mathrm{AgNO}_3 \longrightarrow \left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) + 2 \mathrm{AgCl} + 2 \mathrm{NO}_3^-$$ <br/><br/> From the equation, we can see that 1 mole of the complex consumes 2 moles of silver nitrate to form 2 moles of silver chloride. Therefore, the number of millimoles of chloride ions present in the given solution can be calculated as: <br/><br/> $$\text{Millimoles of } \mathrm{Cl}^- \text{ ions} = \text{concentration} \times \text{volume} = 0.01\ \mathrm{M} \times 2 \times 10\ \mathrm{mL} = 0.2\ \mathrm{mmol}$$ <br/><br/> To calculate the volume of 0.1 M silver nitrate required, we can use the formula: <br/><br/> $$\text{Millimoles of silver nitrate required} = \text{Millimoles of chloride ions} \times 2 = 0.4\ \mathrm{mmol}$$ <br/><br/> We can then use the formula: <br/><br/> $$\text{Volume of silver nitrate solution} = \frac{\text{Millimoles of silver nitrate required}}{\text{Molarity of silver nitrate solution}}$$ <br/><br/> Substituting the values, we get: <br/><br/> $$\text{Volume of silver nitrate solution} = \frac{0.4\ \mathrm{mmol}}{0.1\ \mathrm{M}} = 4\ \mathrm{mL}$$ <br/><br/> Therefore, the volume of 0.1 M silver nitrate required for complete precipitation of chloride ions present in 20 mL of 0.01 M $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_{2}$ solution as silver chloride is $\boxed{4\ \mathrm{mL}}$.