This question involves calculating the volume of AgNO₃ solution needed to precipitate chloride ions from a coordination compound. Let's break it down step by step.

Step 1: Identify the Source of Chloride Ions

The compound is $\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]{\mathrm{Cl}}_2$.

In coordination compounds, the ions inside the square bracket $[...]$ are coordinated to the central metal atom and are not free ions. The ions outside the bracket are free ions and can be precipitated.

Therefore, in $\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]{\mathrm{Cl}}^{+}$, the Cl inside the bracket is part of the complex ion. The two Cl^- ions outside the bracket are free and available for precipitation with Ag⁺ ions from AgNO₃.

So, 1 mole of $\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]{\mathrm{Cl}}_2$ provides 2 moles of free Cl^- ions.

Step 2: Calculate Moles of Complex Compound

We are given 30 mL of a 0.01 M solution.

Molarity (M) = moles of solute / volume of solution in liters.

Moles of complex compound = Molarity × Volume (in L)

Volume = 30 mL = 30/1000 = 0.03 L

Moles of complex = $0.01\mathrm{mol}{L}^{-1}$ × $0.03L$ = $3\times {10}^{-4}\mathrm{mol}$

Step 3: Calculate Moles of Chloride Ions

Since 1 mole of complex gives 2 moles of Cl^-, the moles of Cl^- are:

Moles of Cl^- = 2 × Moles of complex = 2 × $3\times {10}^{-4}$ = $6\times {10}^{-4}\mathrm{mol}$

Step 4: Apply the Concept of Moles for Precipitation

The precipitation reaction is:

${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \mathrm{AgCl}\left(s\right)$

This shows a 1:1 molar ratio between Ag⁺ and Cl^-.

Therefore, moles of Ag⁺ required = moles of Cl^- present = $6\times {10}^{-4}\mathrm{mol}$

Step 5: Calculate the Volume of AgNO₃ Solution

The AgNO₃ solution is 0.1 M, which means it contains 0.1 moles of Ag⁺ per liter.

We can use the formula: $M_1{V}_1=n_1$, where M₁ is molarity, V₁ is volume in liters, and n₁ is moles.

Rearranging for volume: $V_1=\frac{n_1}{M_1}$

Substituting the values: $V_1=\frac{6\times {10}^{-4}\mathrm{mol}}{0.1\mathrm{mol}{L}^{-1}}=6\times {10}^{-3}L$

Convert liters to milliliters: $6\times {10}^{-3}L\times 1000\mathrm{mL}{L}^{-1}=6\mathrm{mL}$

Final Answer

The volume of 0.1 M AgNO₃ required is 6 mL.

Related Concepts & Formulae

Molarity (M): The number of moles of solute per liter of solution.
Formula: $M=\frac{n}{V}$ (n = moles, V = volume in liters)

Stoichiometry: The calculation of reactants and products in chemical reactions based on the balanced equation. The 1:1 ratio from ${\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\to \mathrm{AgCl}$ was crucial.

Coordination Compounds: Understanding the difference between ionization spheres (counter ions outside the bracket) and coordination spheres (ligands directly bonded to the metal inside the bracket) is essential for determining which ions are free to react.