Question

An excess of $AgNO_3$ 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. 0.01
• B. 0.001
• C. 0.002
• D. 0.003

Answer 1

Answer: (B)

0.001

Answer 2

${AgNO_3 + [Cr(H_2O)_4 Cl_2 ]Cl -> AgCl}$
Excess mole = $MV_{(ilt)}$
$= 0.01 \times \frac{100}{1000}$
$= 0.001$