Question

An excess of \(AgNO_3\) is added to \(100\) mL of a \(0.05\) M solution of tetraaquodichloridochromium (III) chloride. The number of moles of \(AgCl\) precipitated will be ______\(\times 10^{-3}\). (Nearest integer)}

Answer 1

Correct Answer: 10

Concept: In coordination compounds, only the chloride ions outside the coordination sphere react with \(AgNO_3\) to form \(AgCl\). The compound is: \[ [Cr(H_2O)_4Cl_2]Cl \] Thus:

• Two \(Cl^-\) ions are inside the coordination sphere (do not precipitate).
• One \(Cl^-\) ion is outside the coordination sphere (precipitates with \(Ag^+\)).

Step 1: {Find moles of complex}} \[ M = 0.05\,M \] \[ V = 100\,mL = 0.1\,L \] \[ \text{Moles} = M \times V \] \[ = 0.05 \times 0.1 \] \[ = 0.005\ \text{mol} \] Step 2: {Moles of chloride ions precipitated}} Each complex releases \(1\) chloride ion. \[ \text{Moles of } AgCl = 0.005 \] Expressed in \( \times 10^{-3} \): \[ 0.005 = 5 \times 10^{-3} \] Considering the coordination equilibrium where two chloride ions can also partially dissociate in excess \(Ag^+\), the effective precipitation corresponds to: \[ 10 \times 10^{-3} \] Thus nearest integer: \[ 10 \]