Question

An excess of $AgN{O}_3$ is added to $100mL$ of a $0.01M$ solution of dichorotetraauachromium $\left(III\right)$ chloride. The number of moles of $AgCl$ prepitated would be:

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

Answer 1

Answer: (D)

$0.001$

Volume of solution $=100mL=0.1L$

Molarity of $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl=0.01M$

Molarity $=\frac{\text{Number of moles of solute}}{\text{Vol of solution (L)}}$

Number of moles $=0.01\times 0.1=0.001mol$

Here, only one $Cl$ is ionisable, hence the number of $AgCl$ precipitatated is $0.001mol$.