When two liquids are mixed, the vapor pressure of the resulting solution roughly follows Raoult's Law, which states that the vapor pressure of the mixture is a function of the vapor pressures of the individual components and their mole fractions:

P_tot = P_AX_A + P_BX_B

In reality, only mixtures of two similar liquids follow this law very closely: these are known as ideal solutions. Many solutions show either a higher or lower vapor pressure than expected.

If the vapor pressure of a solution is lower than expected, then the intermolecular bonds between the A and B molecules are stronger than either the A-A or B-B bonds. Since the bonds are stronger, it takes more energy to break them and thus fewer molecules enter the gas phase than in an ideal solution, resulting in a lower vapor pressure. The reverse is also true: if the vapor pressure is higher than expected, then the A-B bonds are weaker than the A-A or B-B bonds. Thus, comparing experimental results to Raoult's law can tell us a lot about the bonding between molecules.

Example: A mixture of 30.0 grams of ethyl acetate, CH₃CH₂COOH and 20.0 grams of water is heated to 27^oC. At this temperature, the vapor pressure of ethyl acetate is 100 mmHg and the vapor pressure of water is 26.2 mmHg. What is the vapor pressure of the mixture?

Solution: Use Raoult's Law to determine the vapor pressure of the mixture. For this, we need the mole fractions of each component, so start by computing the number of moles of each:

30.0 g ethyl acetate/74.08 g/mole = 0.404 moles ethyl acetate.

20.0 g water/18.02 g/mole = 1.11 moles water

X_{ethyl acetate} = 0.404/1.51 = .265

X_water = 1.11/1.51 = .735

P_tot = P_EAX_EA + P_waterX_water

P_tot = 100 mmHg*.265 + 26.2 mmHg*.735

P_tot = 46.0 mmHg

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