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Grade 12th passPhysical Chemistry

the vapour pressure of a solvent decreases by 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

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Profile image of kalyan
6 Years agoGrade 12th pass
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2 Answers

Profile image of Arun
6 Years ago
Dear student
 

As we learned 

 

For very dilute solutions -

\frac{\Delta P}{P^{0}}= \frac{n_{solute}}{n_{solvent}}

 

- wherein

since \: \: n_{solute}+n_{solvent}\simeq n_{solvent}

 

 

 \therefore P^{0}-P_{s}=P^{0}\times mole\; fraction \; solute

10=P^{0}\times 0.2;\; 20=P^{0}\times n\Rightarrow n=0.4\therefore N=0.6.


 
Hope it helps
 
In case of any difficulty, please feel free to ask 
 
Regards
Arun
Profile image of Vikas TU
6 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