Question

How many moles of $AgCl$ would be obtained, when 100 ml of 0.1 m $CoC{l}_3(N{H}_3{)}_5$ is treated with excess of $AgN{O}_3$?

• A. 0.01
• B. 0.02
• C. 0.03
• D. None of these

Answer 1

The correct option is B 0.02

$\left[\underset{0.1M}{Co\left(N{H}_3\right)Cl}\right]C{l}_2+2AgN{O}_3⟶\underset{2\times 0.1=0.2M}{2AgCl}$

Molarity= $\frac{n}{V}\times 1000$

$\Rightarrow 0.2=\frac{n}{V}\times 1000$

$\Rightarrow n=\frac{0.2\times 100}{1000}=0.02$ moles