Vapour pressure of C 6 H 6 and C 7 H 8 mixture at 50ºC is given P (mm Hg) = 180X B + 90, where X B is the mole fraction of C 6 H 6 . A solution is prepared by mixing 12 mol benzene and 8 mol toluene and if vapours over this solution are removed and condensed into liquid and again brought to the temperature 50ºC, what would be mole fraction of C 6 H 6 in the vapour state. (At. wt. of C = 12, H = 1)
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Y’ B = 0.932.
P(mm Hg) = 180X B + 90
(Benzene) = 180 + 90 (X B = 1) = 270 mm Hg
(Toluene) = 90 mm Hg (X B = 0)
moles of C 6 H 6 = 12, X B = 0.6
moles of C 6 H 5 CH 3 = 8 X T = 0.4
Vapour Pressure of solution P S = X B 
+ X T 
P S = 198 mm Hg
mole fraction of Benzene in vapour state Y B = 
Y B = 
= 
= 0.82
Mole fraction of Toluene in vapour state
Y T = 0.18
Now this vapour when condensed
P S = Y T 
+ Y B 
= 0.18 × 90 + 0.82 × 270 = 237.6 mm Hg.
Now mole fraction of benzene in vapour state is
= 
= 
Ans. Y’ B = 0.932
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems