IF $P C l_{5}$ dissociate to an extent of 80% at 523 K.Calculate the vapour density of the equilibrium mixture at 523 K.

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Solution

$P C l_{5} (g) ⟷ P C l_{3} (g) + C l_{2} (g)$

$80 %$ $20 %$ $20 %$

Assuming the pressure is 1 atm.

The density of the mixture will have an $80 %$ contribution from $P C l_{5}$

$20 %$ contribution from $P C l_{3}$ and

$20 %$ contribution from $C l_{2}$

Density of $P C l_{5}$ at $25^{o} C = \frac{208 g / m o l}{22.4 L} \times \frac{273 K}{523 K} \times 0.80 = 3.87 g / L$

Density of $P C l_{3} a t 25^{o} C = \frac{137 g / m o l}{22.4 L} \times \frac{273 K}{523 K} \times 0.20 = 0.63 g / L$

Density of $C l_{2} a t 25^{o} C = \frac{70.9 g / m o l}{22.4 L} \times \frac{273 K}{523 K} \times 0.20 = 0.33 g / L$

Density of mixture $= 3.87 + 0.63 + 0.33 = 4.83 g / L$

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