Question

The fraction of chlorine precipitated by $\text{AgN}{\text{O}}_{\text{3}}$ solution from $\left[\text{Cu}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{3}}\text{Cl}\right]\text{C}{\text{l}}_{\text{2}}$ is:

• A. 2/3
• B. 1/3
• C. 1/4
• D. 1/2

Answer 1

Answer: (A)

2/3

$\left[Cu\right(N{H}_3{)}_{3}Cl]C{l}_2+\underset{Excess}{\underset{}{AgN{O}_3}}\to [Cu\left(N{H}_3{)}_{3}Cl\right](N{O}_3{)}_2+2AgCl$

There are three chlorine atoms initially.

Finally, there is only 1 chlorine atom.

So, $\frac{2}{3}rd$ fraction of chlorine is removed.

Hence, the correct option is $A$