Question

An excess of $AgN{O}_3$ 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. 0.002
• B. 0.003
• C. 0.01
• D. 0.001

Answer 1

The correct option is D 0.001

$AgN{O}_3+\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\to AgCl+[Cr\left(H_{2}O{)}_{4}C{l}_2\right]N{O}_3$
millimole
$100\times 0.01=1--excess011$
$\therefore$ mole of AgCl = $1\times {10}^{-3}=0.001$