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

Step 1: Understanding the Question:
We need to determine how many chloride ions are available outside the coordination sphere of the given complex to react with \(AgNO_{3}\).
Step 2: Key Formula or Approach:
1. Write the chemical formula of the complex.
2. Identify the ionizable chloride ions (those in the primary valence/outside square brackets).
3. Calculate moles of complex: \(n = Molarity \times Volume(L)\).
4. Moles of \(AgCl\) = Moles of ionizable \(Cl^{-}\).
Step 3: Detailed Explanation:
The complex name is tetraquadichloridochromium (III) chloride.
- Tetraqua: \(4 H_{2}O\)
- Dichlorido: \(2 Cl^{-}\) (inside the sphere)
- Chromium (III): \(Cr^{3+}\)
Formula: \([Cr(H_{2}O)_{4}Cl_{2}]Cl\).
Notice there is only one chloride ion outside the coordination sphere.
Calculation:
Volume \(= 100 \text{ mL} = 0.1 \text{ L}\).
Molarity \(= 0.05 \text{ M}\).
Moles of complex \(= 0.05 \times 0.1 = 0.005 \text{ moles}\).
Since 1 mole of complex gives 1 mole of free \(Cl^{-}\) ions:
Moles of \(Cl^{-}\) available \(= 0.005 \text{ moles}\).
Reaction with silver nitrate: \(Ag^{+} + Cl^{-} \rightarrow AgCl \downarrow\).
Moles of \(AgCl\) precipitated \(= 0.005 \text{ moles} = 5 \times 10^{-3} \text{ moles}\).
Step 4: Final Answer:
The number of moles is 5.