Question

The relative decrease in the vapour pressure of an aqueous solution containing 2 mol $\left[Cu{\left(N{H}_3\right)}_{3}Cl\right]Cl$ in 3 mol $H_{2}O$ is 0.50.On reaction with $AgN{O}_3$, this solution will form:

• A. 1 mol $AgCl$
• B. 0.25 mol $AgCl$
• C. 2 mol $AgCl$
• D. 0.40 mol $AgCl$

Answer 1

The correct option is A 1 mol $AgCl$

$\frac{\Delta P}{P^o}=0.50=\frac{n_{1}i}{n_{1}i+n_2}$
$0.50=\frac{2i}{2i+3}$
$i=1.5=1+x$
$x=0.5$
Hence, 2 mol $NaCl$ will exists as 1 mol ${Cl}^{\ominus }$ due to 50% ionization.