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Grade: 12th pass
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
one month ago

Answers : (2)

22607 Points
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 
one month ago
Vikas TU
8845 Points
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
one month ago
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