Question

If $PC{l}_5$ is 80% dissociated at 523 $K$, what is the vapour density of the equilibrium mixture at 523 $K$ ?

• A. 57.92
• B. 67.92
• C. 77.93
• D. 45.45

Answer 1

The correct option is A 57.92

$PC{l}_5\left(s\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$n=2$
$\alpha =80\%=\frac{80}{100}=0.8$
theoretical vapour/Normal vapour density,
$D=\frac{Normalmolarmass}{2}=\frac{208.5}{2}=104.25$

vapour density of the equilibrium mixture, $d=?$
we have the relation
$\alpha =\frac{D-d}{(n-1)d}$

$0.8=\frac{104.25-d}{(2-1)d}$
$0.8d=104.25-d$
$0.8d+d=104.25$
$1.8d=104.25$
$d=\frac{104.25}{1.8}=57.92$