An excess of AgNO _3 is added to 100 mL of a 0.05 M solution of...

The formula of tetraaquadichloridochromium (III) chloride is $[Cr(H_2O)_4Cl_2]Cl$.

In this complex, there is one ionizable chloride ion present outside the coordination sphere.

The dissociation reaction is:

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

Number of moles of the complex $= M \times V = 0.05 \times \dfrac{100}{1000} = 0.005$ moles.

Since one mole of the complex yields one mole of chloride ions, the number of moles of $Cl^-$ ions produced is $0.005$ moles.

When excess $AgNO_3$ is added, the chloride ions precipitate as $AgCl$:

$Ag^+ + Cl^- \rightarrow AgCl \downarrow$

Number of moles of $AgCl$ precipitated $= 0.005 = 5 \times 10^{-3}$ moles.

The value is $5$.

Answer: $5$