Question

The volume (in mL) of 0.1 M AgNO₃ required for complete precipitation of chloride ions present in 20 mL of 0.01 M solution of [Cr(H₂O)₅CI]Cl₂ as silver chloride is___

Hint:

Determine the total number of reactive ions and use stoichiometric relationships to find reagent volume.

Answer 1

Correct Answer: 4

Reaction Involved:

\[ [Cr(H_2O)_5Cl]Cl_2 + 2AgNO_3 \rightarrow [Cr(H_2O)_5Cl](NO_3)_2 + 2AgCl \downarrow \]

Data Given: 

• Molarity of \( [Cr(H_2O)_5Cl]Cl_2 \): \( 0.01 \, \text{M} \)
• Volume of \( [Cr(H_2O)_5Cl]Cl_2 \): \( 20 \, \text{mL} \)
• Molarity of \( \text{AgNO}_3 \): \( 0.1 \, \text{M} \)

Step-by-Step Calculation:

1. Calculate moles of the complex:
   • \( \text{Moles of complex} = \text{Molarity} \times \text{Volume (in L)} \)
   • \( \text{Moles of complex} = 0.01 \times 0.020 = 0.0002 \, \text{moles} \, (= 0.2 \, \text{millimoles}) \)
2. Determine moles of \( \text{AgNO}_3 \) required:
   • Each complex molecule releases 2 moles of \( \text{Cl}^- \), requiring 2 moles of \( \text{AgNO}_3 \).
   • \( \text{Moles of AgNO}_3 = 2 \times \text{Moles of complex} \)
   • \( \text{Moles of AgNO}_3 = 2 \times 0.0002 = 0.0004 \, \text{moles} \, (= 0.4 \, \text{millimoles}) \)
3. Calculate the volume of \( \text{AgNO}_3 \) solution required:
   • \( \text{Volume (in L)} = \frac{\text{Moles}}{\text{Molarity}} \)
   • \( \text{Volume (in L)} = \frac{0.0004}{0.1} = 0.004 \, \text{L} \)
   • \( \text{Volume (in mL)} = 0.004 \times 1000 = 4 \, \text{mL} \)

Final Answer:

The volume of \( 0.1 \, \text{M AgNO}_3 \) required is 4 mL.