An excess of AgNO$$_3$$ is added to 100 mL of a 0.05 M solution of tetraaquadichloridochromium (III) chloride. The number of moles of AgCl precipitated will be __________ $$\times 10^{-3}$$. (Nearest integer)

The coordination compound is named tetraaquadichloridochromium(III) chloride.

From the nomenclature:

• The central metal ion is (Cr^{3+}).
• The ligands present inside the coordination sphere are four water molecules and two chloride ions.
• One chloride ion remains outside the coordination sphere as the counter ion.

Hence, the formula of the complex is

$$[Cr(H_2O)_4Cl_2]Cl.$$

When the complex is dissolved in water, it dissociates as

$$[Cr(H_2O)_4Cl_2]Cl \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^-.$$

Only the chloride ion present outside the coordination sphere is free to react with silver nitrate. The reaction with (AgNO_3) is

$$Cl^- + AgNO_3 \rightarrow AgCl(s) + NO_3^-.$$

Thus, one mole of the complex produces one mole of (AgCl).

The number of moles of the complex present is

$$\text{Moles} = M \times V = 0.05 \times 0.1 = 0.005\ \text{mol}.$$

Since the stoichiometric ratio is (1:1),

$$\text{Moles of }AgCl = 0.005\ \text{mol}.$$

Expressing this in the required form,

$$0.005 = 5 \times 10^{-3}\ \text{mol}.$$

Hence, the value of (x) is 5.