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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:

1. 0.002
2. 0.003
3. 0.01
4. 0.001

Correct answer: 0.003

Solution

Dichlorotetraaquachromium (III) chloride has the formula [Cr(H2O)4Cl2]Cl. It dissociates to give 1 Cl⁻ ion from the outer sphere and 2 Cl⁻ ions from the coordination sphere, totaling 3 Cl⁻ ions per molecule. For 0.01 M solution in 100 mL, moles of the compound = 0.01 × 0.1 = 0.001. Total moles of Cl⁻ = 0.001 × 3 = 0.003. Each Cl⁻ reacts with Ag⁺ to form AgCl, so 0.003 moles of AgCl precipitate.

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