[NathanielZhu]

What is the mole fraction of pentane, CH3(CH2)3CH3, in the vapor phase from a solution consisting of a mixture of pentane and hexane, CH3(CH2)4CH3, at 25.00 °C, if the total pressure above the solution is 264.3 torr? The vapor pressures of pentane and hexane at 25.00 °C are 511.0 torr and 150.0 torr, respectively. Hint; assuming these compounds form nearly ideal solutions when mixed together, determine the mole fraction of pentane in the liquid phase first; note that the mole fraction of pentane in the vapor phase will be its vapor pressure divided by the total pressure.
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Unlike the others I posted, I have absolutely no clue what this question is talking about.
I did what the hint said to get the mole fraction of pentane in liquid phase which is (511/(511+150)) but I don't know what to do after that.

[ramboacid]

For this problem, you have to remember that the partial pressures of pentane and hexane must add up to the total vapor pressure, and that the partial pressures of the pentane and hexane are proportional to their composition in vapor.

For example, if liquid X and liquid Y have vapor pressures of 100 mmHg and 300 mmHg, and the observed vapor pressure is 150 mmHG for a mixture of the two solutions, then we would set up the following equations:

x+y = 1               
 - This says that the two substances compose 100% of the vapor, and x and y are the percents
100x+300y = 150
 - This calculates the actual vapor pressure when x and y are percent compositions of the vapor

Solving gives that x = 0.75 and y = 0.25, so the mole fraction of liquid X in the vapor is 3/4 and the mole fraction of liquid Y in the vapor is 1/4.