Question

The vapour density of $PC{l}_5$ is $104.25$ but when heated to ${230}^{o}C,$ its vapour density is reduced to $62$.
The decomposition reaction of $PC{l}_5$ on heating is
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
Calculate the degree of dissociation of $PC{l}_5$ at this temperature :

• A. $0.086$
• B. $0.68$
• C. $0.068$
• D. $0.86$

Answer 1

The correct option is B $0.68$

The reaction is:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$n=2$
$\alpha =\frac{D-d}{(n-1)d}=\frac{104.25-62}{(2-1)\times 62}$
$=0.68$