the vapour pressure of a solvent decreasesby 20mm hg when a non volatile solute was added.the molefraction of solute at this condition was 0.4.when the mole fraction wasX,then then decrease in pressure is 40mm hg.find the value of pessure exerted by vapour of pure solventi.e X
kalyan , 5 Years ago
Grade 12th pass
2 Answers
Arun
Last Activity: 5 Years ago
Dear student
As we learned
For very dilute solutions -
- wherein
Hope it helps
In case of any difficulty, please feel free to ask
Regards
Arun
Vikas TU
Last Activity: 5 Years ago
Lowering in vapour pressure is directly proportional to mole fraction of solute 20 mm of Hg / 40 mm of Hg = 0.4 / x x = 0.8 = mole fraction of solute , Mole fraction of solvent = 1-0.8 = 0.20 Preesure exerted by vapour of pure solvent
Po - Ps / Po = 0.40 20 / Po = 0.4 Po = 50 mm
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