Question

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

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

Answer 1

The correct option is A $57.92$

The Equation,
$PC{l}_5\left(s\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$n=2$
$\alpha =80\%=\frac{80}{100}=0.8$
theoretical vapour/Normal vapour density,
$D=\frac{\text{Normal molar mass}}{2}=\frac{208.5}{2}=104.25$
vapour density of the equilibrium mixture, $d=?$
we have the relation
${\alpha }_{\left(\text{dissociation constant}\right)}=\frac{D-d}{(n-1)d}$
$0.8=\frac{104.25-d}{(2-1)d}$
$0.8d=104.25-d$
$0.8d+d=104.25$
$1.8d=104.25$
$d=\frac{104.25}{1.8}=57.92$