Question

At 353 K, the vapour pressure of pure ethylene bromide and propylene bromide is 22.93 and 16.93 $N{m}^{-2}$ respectively and these compounds form nearly ideal solutions. 3 moles of ethylene bromide and 2 moles of propylene bromide are equilibrated at 353 K and at a total pressure of 20.4 $N{m}^{-2}$, mark the correct option(s):

• A. $X_A=0.578$
• B. $X_B=0.578$
• C. Number of moles of A in vapour phase is 0.99
• D. Number of moles of B in the vapour phase is 0.99

Answer 1

Answer: (A), (C)

The correct options are
A $X_A=0.578$
C Number of moles of A in vapour phase is 0.99

Let
Ethylene bromide = A; $P_A^o=22.93N{m}^{-2};n_A=3$
Propylene bromide = B; $P_B^o=16.93N{m}^{-2};n_B=2$
Total pressure = $P_T=20.4N{m}^{-2}$
$\therefore$ Acc. to Raoult's law
$P_T=P_A^o{X}_A+P_B^o{X}_B$

$P_T=P_A^o{X}_A+P_B^o(1-X_A)$

$P_T=(P_A^o-P_B^o)X_A+P_B^o$

$X_A=\frac{P_T-P_B^o}{P_A^o-P_B^o}$

$X_A=\frac{20.4-16.93}{22.93-16.93}=0.578$

$X_B=1-X_A=0.422$

Let mole fraction in vapour phase = $Y_A$
$Y_A=\frac{P_A^o}{P_T}$

$Y_A=\frac{22.93\times 0.578}{20.4}=0.649$

Assuming that number of moles of A and B that are vapourised are a and b then
$Y_A=\frac{a}{a+b}=0.649...\left(1\right)$
But composition of A in liquid phase,
$X_A=\frac{3-a}{(3-a)+(2-b)}=0.578$
$X_A=\frac{3-a}{5-(a+b)}=0.578...\left(2\right)$
solving
a = 0.99
b = 0.53