Vapor Pressure Lowering
The vapor pressure of a liquid is the equilibrium pressure of gas molecules from that liquid (i.e., the results of evaporation) above the liquid itself. A glass of water placed in an open room will evaporate completely (and thus never reach equilibrium); however, if a cover is placed on the glass, the space above the liquid will eventually contain a constant amount of water vapor. How much water vapor is present depends on the temperature, but not on the amount of liquid that is present at equilibrium (provided some liquid is present at equilibrium). (At room temperature, the vapor pressure of pure water is about 20 Torr, which is about one-fortieth of the total atmospheric pressure on a "normal" day at sea level.)
If, instead of pure water, an aqueous solution is placed in the glass, the equilibrium pressure will be lower than it would be for pure water. Raoult's law states that the vapor pressure of the solvent over the solution is proportional to the fraction of solvent molecules in the solution; that is, if twothirds of the molecules are solvent molecules, the vapor pressure due to the solvent is approximately two-thirds of what it would be for pure solvent at that temperature. If the solute has a vapor pressure of its own, then the total vapor pressure over the solution would be:
P vap = ⅔ (pure solvent vapor pressure) + ⅓ (pure solute vapor pressure)
Generally, one expects that solutes which are liquids in their pure form (such as ethyl alcohol) will have some vapor pressure of their own, whereas ionic compounds (such as sodium chloride) will not contribute to the total vapor pressure over the solution.
One consequence of this lowering in vapor pressure may be observed in a spilled can of soda. As the water evaporates, the soda becomes more sugar and less water, until the vapor pressure of the water is so low that it barely evaporates. As a result, the spilled soda remains sticky for a long time. Contrast this behavior with that of a water spill.