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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

11 months ago

Arun
24739 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

Regards
Arun
11 months ago
Vikas TU
11766 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
11 months ago
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