The vapor pressure of ethanol at 40 degrees C is about 35 mm Hg, and the vapor pressure of water at 40 degrees C is 55 mm Hg. A sample of wine at this temperature has a mole fraction of ethanol of 0.10, and a mole fraction of water of 0.90. What is the vapor pressure of ethanol and water of this wine, and what is the mole fraction of these components in the vapor?
According to Raoult’s law, the vapor pressure of a solution is just the vapor pressure of its parts added up and multiplied by their mole fraction.
For example, since ethanol is 35 mm Hg and its mole fraction is 0.1, ethanol’s contribution is 3.5 mm Hg.
Since water is 55 mm Hg and its mole fraction is 0.9, water’s contribution is 49.5 mm Hg.
Adding them up, the total vapor pressure is 3.5 + 49.5=53 mm Hg
Note that this makes the assumption that the solution is ideal. You would need to know the heat of solution in order to predict changes from this value. Since that’s not given, this is the best you can do.
The mole fraction of the components in the vapor is just the vapor pressure of each component divided by the total vapor pressure:
Ethanol: 3.5/53=.066
Water: 49.5/53=.933
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According to Raoult’s law, the vapor pressure of a solution is just the vapor pressure of its parts added up and multiplied by their mole fraction.
For example, since ethanol is 35 mm Hg and its mole fraction is 0.1, ethanol’s contribution is 3.5 mm Hg.
Since water is 55 mm Hg and its mole fraction is 0.9, water’s contribution is 49.5 mm Hg.
Adding them up, the total vapor pressure is 3.5 + 49.5=53 mm Hg
Note that this makes the assumption that the solution is ideal. You would need to know the heat of solution in order to predict changes from this value. Since that’s not given, this is the best you can do.
The mole fraction of the components in the vapor is just the vapor pressure of each component divided by the total vapor pressure:
Ethanol: 3.5/53=.066
Water: 49.5/53=.933