Question

Q. Calculate the mole fraction of
component B in the vapour phase in
equilibrium with an ideal liquid mixture
of A and B, with a mole fraction of 0.5
for the component A in the liquid phase.
p° of pure A = 119 torr and p0 of pure B
= 37 torr (at same temperature T)
(a) 0.237 (b) 0.327
(c) 0.372 (d) 0.273

Answer 1

Answer:

Given:

 

T = 30 °C

 

PB0 = 119 torr

PT0 = 37 torr

PT  = XT PT0  

Mole fraction of benzene = 0.4

 

Mole fraction of ttoluene = 1 - 0.4  

 

= 0.6

 

According to Raoult's law,

 

PB = XBP0  

PT  = XT PT0  

 

Total pressue = PB + PT  ---(1)

 

By putting values in eq (1)

 

PTot = XBPB0 + XT PT0  

 

      = (0.4 × 119) +  (0.6×37)

 

      = 69.8  torr

 

Mole fraction of toluene in vapour phase, then see attachment, please mark as brainliest