1. Identify ionisable chloride ions

The complex is: $\left[\mathrm{Cr}(\mathrm{H_2O})_5\mathrm{Cl}\right]\mathrm{Cl}_2$

Here, one $\mathrm{Cl^-}$ is inside the coordination sphere, and two $\mathrm{Cl^-}$ ions are outside the coordination sphere.

Only the chloride ions outside the coordination sphere ionise in solution and get precipitated immediately by $\mathrm{AgNO_3}$ as $\mathrm{AgCl}$.

So, each mole of $\left[\mathrm{Cr}(\mathrm{H_2O})_5\mathrm{Cl}\right]\mathrm{Cl}_2$ gives 2 moles of precipitable $\mathrm{Cl^-}$.

2. Moles of the complex in 20 mL of 0.01 M solution

Given:

• Volume $= 20\,\mathrm{mL} = 0.020\,\mathrm{L}$
• Molarity $= 0.01\,\mathrm{M}$

Moles of complex: $n = M \times V = 0.01 \times 0.020 = 2.0 \times 10^{-4}\,\mathrm{mol}$

3. Moles of chloride ions precipitated

Since 1 mole of complex gives 2 moles of free $\mathrm{Cl^-}$: $n(\mathrm{Cl^-}) = 2 \times 2.0 \times 10^{-4} = 4.0 \times 10^{-4}\,\mathrm{mol}$

4. Reaction with silver nitrate

The precipitation reaction is: $\mathrm{Ag^+ + Cl^- \rightarrow AgCl(s)}$

Thus, moles of $\mathrm{AgNO_3}$ required $=$ moles of $\mathrm{Cl^-}$ required to be precipitated: $n(\mathrm{AgNO_3}) = 4.0 \times 10^{-4}\,\mathrm{mol}$

5. Volume of 0.1 M $\mathrm{AgNO_3}$ required

Using: $V = \frac{n}{M} = \frac{4.0 \times 10^{-4}}{0.1} = 4.0 \times 10^{-3}\,\mathrm{L}$

Convert to mL: $4.0 \times 10^{-3}\,\mathrm{L} = 4\,\mathrm{mL}$

6. Final answer

The required volume of $0.1\,\mathrm{M}$ $\mathrm{AgNO_3}$ is: $\boxed{4\,\mathrm{mL}}$