wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For a solution if p0A=600mmHg, p0B=840mmHg under atmospheric conditions and vapour pressure of solution is 1 atm then find
i) Composition of solution
ii) Composition of vapour in equilibrium with solution.

Open in App
Solution

(i) According to Raoult's law
PAαxA
also xA+xB=1
P0AxA+P0BxB=P
from question : P=1 atm
P0A=600mm Hg
P0B=840mm Hg
760mm Hg=1 atm
600mmHg=1760×600atm=6076 atm
840mmHg=8476 atm
So, 1=6076(1xB)+8476(xB)
xB=1624
xB=2/3;xA=1/3
composition of vapour phase in equilibrium with solution is-
yA=PAPtotal=P0AxA1=P0AxA1=2076=0.26
yB=PBPtotal=P0BxB1=2838=0.74
(ii) Relative lowerins of vapour pressure =ΔPP0=P0APAP0A=xB=WBMAMBWA
vapour pressure of water =360mm Hg
Volume of water =200 ml
mass =200g
ΔPP0=0.5=WB×18200×60
WB=200×60×0.518=333.33 grams.

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermochemistry
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon