Vapour densit...

Question

Vapour density of $P C l_{5}$ is 104.16 but when heated at 230 $^{o} C$ its vapour density is reduced to 62. The degree of dissociation of $P C l_{5}$ at this temperature will be:

6.8

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68

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46

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60

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Solution

The correct option is D $68$ %
$P C l_{5} \leftrightarrow P C l_{3} + C l_{2}$

Initial $C$ $0$ $0$

At eq. $C (1 - α)$ $C α$ $C α$

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

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

Therefore,
$\frac{T o t a l m o l e s o f e q .}{i n i t i a l t o t a l m o l e s} = \frac{v a p o u r d e n s i t y i n i t i a l}{v a p o u r d e n s i t y a t e q .}$

$\frac{C (1 + α)}{C} = \frac{D}{d}$

$1 + α = \frac{D}{d}$

$α = \frac{D}{d} - 1$

$D = 104.16$

$d = 62$

$∴ α = \frac{104.16}{16} - 1 = 0.68$

$∴, α = 68$ %

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