Identify the complex and the chloride ions present

The compound is: $[Cr(H_2O)_5Cl]Cl_2$

Here:

• One $Cl^-$ is inside the coordination sphere: $[Cr(H_2O)_5Cl]^{2+}$
• Two $Cl^-$ are outside the coordination sphere as counter ions

Only the ionisable chloride ions outside the coordination sphere precipitate immediately with $AgNO_3$.

So, each mole of $[Cr(H_2O)_5Cl]Cl_2$ gives: $2 ext{ moles of } Cl^-$

Calculate moles of the complex in 30 mL of 0.01 M solution

Given:

• Volume $= 30 \, \text{mL} = 0.030 \, \text{L}$
• Molarity $= 0.01 \, \text{M}$

Moles of complex: $n = M \times V = 0.01 \times 0.030 = 3.0 \times 10^{-4} \text{ mol}$

Calculate moles of precipitable chloride ions

Since 1 mole of complex gives 2 moles of free $Cl^-$, $n(Cl^-) = 2 \times 3.0 \times 10^{-4} = 6.0 \times 10^{-4} \text{ mol}$

Use reaction with silver nitrate

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

Thus, moles of $AgNO_3$ required = moles of $Cl^-$ required to be precipitated: $n(AgNO_3) = 6.0 \times 10^{-4} \text{ mol}$

Find the volume of 0.1 M $AgNO_3$

$V = \frac{n}{M} = \frac{6.0 \times 10^{-4}}{0.1} = 6.0 \times 10^{-3} \text{ L}$

Convert to mL: $6.0 \times 10^{-3} \text{ L} = 6 \text{ mL}$

Final Answer

The required volume of $0.1\,M$ $AgNO_3$ is: $\boxed{6}$