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Raoult's Law

At 353 K, t...

Question

At $353 K$ , the vapour pressure of pure ethylene bromide and propylene bromide are $22.93$ and $16.93 k N M^{- 2}$ , respectively, and these compounds form a nearly ideal solution. $3 m o l$ of ethylene bromide and $2 m o l$ of propylene bromide are equilibrated at $553 K$ and a total pressure of $20.4 k N M^{- 2}$ .
(a)what is the composition of the liquid phase?
(b)what amount of each compound is present in the vapour phase?

(a)

0.578, 0.422

(b)

0.9967 m o l, 0.5374 m o l

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(a)

0.345, 0.422

(b)

0.9967 m o l, 0.434 m o l

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(a)

0.578, 0.212

(b)

0.9967 m o l, 0.5374 m o l

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(a)

0.578, 0.212

(b)

0.497 m o l, 0.5374 m o l

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Solution

The correct option is A (a) $0.578, 0.422$ (b) $0.9967 m o l, 0.5374 m o l$
(a)Total number of moles $= 3 + 2 = 5$
Mole fraction of ethylene bromide $\frac{3}{5} = 0.6$
Mole fraction of propylene bromide $= 1 - 0.6 = 0.4$
(b) Partial pressure of ethylene bromide is the product of its mole
fraction and vapour pressure of pure ethylene bromide. It is $0.6 \times 22.93 k N m^{- 2} = 13.758 k N m^{- 2}$
The partial pressure of propylene bromide is $0.4 \times 16.93 k N m^{- 2} = 6.772 k N m^{- 2}$
Total pressure is $13.758 + 6.772 = 20.53 k N m^{- 2}$
The mole fraction of ethylene bromide in the vapour phase is the
ratio of its partial pressure to
total pressure. It is $\frac{13.758}{20.53} = 0.67$
The mole fraction of propyelene bromide in the vapour phase is $1 - 0.67 = 0.33$

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