Question

An excess of AgNO₃ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be

• A. 0.001
• B. 0.002
• C. 0.003
• D. 0.01

Answer 1

The Correct Option is A

To determine the number of moles of AgCl precipitated, let's first understand the chemical reaction involved.

The compound dichlorotetraaquachromium(III) chloride has the formula [Cr(H₂O)₄Cl₂]Cl. When dissolved in water, it dissociates as follows:

[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+\text{Cl}^− \rightarrow [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ + \text{Cl}^−

Each dichlorotetraaquachromium(III) complex molecule releases 1 Cl^- ion. These Cl^- ions react with AgNO₃ to form AgCl:

\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \downarrow

Now, calculate using the provided data:

1. Volume of solution = 100 mL = 0.1 L
2. Concentration of dichlorotetraaquachromium(III) chloride = 0.01 M
3. Moles of dichlorotetraaquachromium(III) chloride = Volume × Molarity = 0.1 \times 0.01 = 0.001 \, \text{mol}
4. Since each molecule provides one Cl^- ion, moles of Cl^- = 0.001 mol.

Given an excess of AgNO₃, all Cl^- ions will react to form AgCl precipitate.

Thus, the number of moles of AgCl precipitated = number of moles of Cl^- ions = 0.001 mol.

Therefore, the correct answer is 0.001.