Question

Vapour density of $PC{l}_5$ is 104.16 but when heated at 230${}^{o}C$ its vapour density is reduced to 62. The degree of dissociation of $PC{l}_5$ at this temperature will be:

• A. $6.8$%
• B. $68$%
• C. $46$%
• D. $60$%

Answer 1

Answer: (B)

$68$%

$PC{l}_5\leftrightarrow PC{l}_3+C{l}_2$

Initial $C$ $0$ $0$

At eq.$C(1-\alpha )$ $C\alpha$ $C\alpha$

Total moles at equilibrium $=C(1-\alpha )+C\alpha +C\alpha =C(1+\alpha )$

For a reaction at equilibrium, Vapour density is inversely proportional to the number of moles.

Therefore,

$\frac{Totalmolesofeq.}{initialtotalmoles}=\frac{vapourdensityinitial}{vapourdensityateq.}$

$\frac{C(1+\alpha )}{C}=\frac{D}{d}$

$1+\alpha =\frac{D}{d}$

$\alpha =\frac{D}{d}-1$

$D=104.16$

$d=62$

$\therefore \alpha =\frac{104.16}{16}-1=0.68$

$\therefore ,\alpha =68$ %