CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
187
You visited us 187 times! Enjoying our articles? Unlock Full Access!
Question

Volatile liquids A and B forming an azeotrope (at a particular temperature range) have vapour pressures of 280 mm and 200 mm respectively. If the mole fraction of A in the solution mixture is 0.3, and the mixture has a total vapour pressure of 300 mm.

Choose the correct statement(s) for the following mixture.

A
It forms a minimum boiling azeotrope.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
It forms a maximum boiling azeotrope.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
First minimum and then maximum boiling azeotrope is formed.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
First maximum and then minimum boiling azeotrope is formed.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A It forms a minimum boiling azeotrope.
Given,
Pure vapour pressure of A , pA=280 mmPure vapour pressure of B , pA=200 mm

Mole fraction of A , XA=0.3Mole fraction of B, XB=10.3=0.7

By Raoult's law
Total vapour pressure of solution,
PT=XA.pA+XB.pB
PT=280×0.3+200×0.7
=224 mm.

PT>XA.pA+XB.pB
300 mm>224 mm
Thus, the given mixture shows positive deviation from the Raoult's law and forms minimum boiling azeotrope

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon