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

# In a binary mixture of two ideal volatile liquids A and B, calculate the mole fraction of component A in vapour phase if: Vapour pressure of A in pure state is 500 mm Hg Vapour pressure of B in pure state is700 mm Hg Total pressure (PT) of the solution is 600 mm Hg

A
0.25
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.42
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.75
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.58
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 0.42χA=yA×PTp0A χA is the mole fraction of component A in liquid phase χB=yB×PTp0B χB is the mole fraction of component B in liquid phase Also χA+χB=1 So, 1PT=yAp0A+yBp0B1PT=yAp0A+1−yAp0B Solving, PT=p0A×p0Bp0A+(p0B−p0A)×yA Here , PT=300 mm Hgp0A=400 mm Hgp0B=700 mm Hg 600=500×700500+(700−500)×yA6=355+2yA30+12yA=35yA=512=0.416≈0.42

Suggest Corrections
2
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program