1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The vapour pressure of pure liquids 1 and 2 are 550 mm Hg and 800 mm Hg respectively at 350 K. What will be the mole fraction of component 2 in liquid phase when total vapour pressure of the solution is 600 mm Hg?

A
0.1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 0.2By Raoult's Law : PT=p∘1χ1+p∘2χ2 where PT is the total pressure of the solution. p∘1 is the vapour pressure of pure component 1 p∘2 is the vapour pressure of pure component 2 χ1 is the mole fraction of component 1 in liquid phase. χ2 is the mole fraction of component 2 in liquid phase. Putting the values: 600=550χ1+800χ2600=550χ1+800(1−χ1)600=800−250χ1χ1=0.8 χ2=1−0.8=0.2

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program