What is the mole fraction of benzene C6H6 in the VAPOR PHASE from a solution consisting of a mixture of benzene and n-hexame ch3(ch2)4ch3 at 27celsius, if the total pressure above the solution is 135.2 torr? The vapor pressures of benzene and n-hexame at 27 celsius are 103 and 161.7 torr respectively. Hint: Assuming these compounds form nearly ideal solutions when mixed together, determine the mole fraction of benzene in the liquid phase first, note that the mole fraction of benzene in the vapor phase will be its vapor pressure divided by total pressure.
I get x for benzene to be 0.451 but its incorrect. What should i be doing differently? Thanks
To calculate the mole fraction of benzene in the vapor phase, we need to follow a series of steps.
Step 1: Calculate the mole fraction of benzene in the liquid phase.
To do this, we can use Raoult's Law, which states that the vapor pressure of each component of an ideal solution is directly proportional to its mole fraction in the liquid phase.
The mole fraction of benzene in the liquid phase can be calculated using the formula:
Mole fraction of benzene in liquid phase = Vapor pressure of benzene / Total pressure
Given:
Vapor pressure of benzene = 103 torr
Vapor pressure of n-hexane = 161.7 torr
Total pressure = 135.2 torr
Mole fraction of benzene in liquid phase = 103 torr / 135.2 torr = 0.761
Step 2: Calculate the mole fraction of benzene in the vapor phase.
The mole fraction of benzene in the vapor phase is equal to its vapor pressure divided by the total pressure.
Let's denote the mole fraction of benzene in the vapor phase as x.
x = Vapor pressure of benzene / Total pressure
Given:
Vapor pressure of benzene = 103 torr (from previous information)
Total pressure = 135.2 torr (given in the question)
x = 103 torr / 135.2 torr ≈ 0.762
Therefore, the correct mole fraction of benzene in the vapor phase is approximately 0.762, not 0.451.
Please recheck your calculations or ensure you applied Raoult's Law correctly.
benzene................hexane
|________________|_________|
103............x=135.2....161.7
<.....161.7-103=58.7.......>
<.....32.2......>|<..26.5...>
26.5/58.7 = 0.451 = Xbenzene
32.2/58.7 = 0.548 = Xhexane
We can check that out.
0.451*103 = 46.45 mm benzene
0.548*161.7 = 88.61 mm hexane
total = 135.1 which is close.
Xbenzene in vapor phase = Pbenzene/total P = 46.45/135.2 = ??