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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
anser 2 Answers
Arun

Last Activity: 5 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

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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