Question

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

Answer 1

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$\alpha \text{(Degree of dissociation)}=80\%=\frac{80}{100}=0.8$
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,
$\alpha =\frac{D-d}{(n-1)d}$

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