Question

The vapour pressure of pure liquids a and b are 450 and 6 700 respectively at 3:50 kelvin

Answer 1

Answer:

Vapour pressure of pure liquid A, PoA= 450 mm of Hg

Vapour pressure of pure liquid A, PoA= 700 mm of Hg

Total vapour pressure, ptotal= 600 mm of Hg

Use the formula of Raoult’s law

600 = (450 – 700) XA+ 700

250 XA= 100

XA= 100/250 = 0.4

Use formula

XB= 1 - XA

Plug the values we get

XB= 1 − 0.4 = 0.6

use formula

PA= PoA× XA= 450 × 0.4= 180 mm of Hg

PB= PoB× XB= 700 × 0.6 = 420 mm of Hg

Now, in the vapour phase:

Mole fraction of liquid A

= 180 /(180+ 420))= 0.30

Mole fraction of liquid B, =1 – 0 .30 = 0.70

Explanation: