How to Calculate the Mole Fraction in a Pentane-Hexane Solution?

AI Thread Summary
The discussion revolves around a chemistry problem involving vapor pressure and Raoult's law, specifically regarding a pentane-hexane solution at 25°C. The problem states that the mole fraction of pentane in the vapor phase is 0.15, and the vapor pressures of pure pentane and hexane are given as 511 torr and 150 torr, respectively. The user is attempting to determine the mole fraction of pentane in the liquid solution using Raoult's law but is struggling to find a solution. They set up equations based on the relationship between the total pressure and the partial pressures of the components, but their attempts have not yielded a conclusive answer. The key focus is on applying Raoult's law correctly to solve for the mole fraction in the solution.
filou
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Hi
I have difficulties with a problem that my teacher gave in assignement. Teh teacher said it was a bit higher than our level but I think it far above what I am able to do! so here it is :

The vapor pressure in equilibrium with pentane-hexane solution at 25°C has a mole fraction of pentane equal to 0.15 at that temperature. What is the mole fraction in the solution?

Up to now I found that the vapor pressure of pentane is 511 torr and the one of hexane is 150 torr. I think that I should use Raoult's law but my attempts were not concluant... I hope someone will be able to answer my question.
 
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P^o (P) = 511~,~~P^o (H) = 150

P(P) = \chi (P.solution) * P^o (P)= 0.15P(tot)

=>~0.15P(tot) = \chi (P)* 511

And~0.85P(tot) = (1 - \chi(P))*150
 
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