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Randy Holmes-Farley
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The reason it evaporates faster is, as folks have noted, an effect of Raoult's law which relates to the vapor pressure (and hence evaporation rate) of mixtures. Seawater deviates from Raoult's law since it is not an ideal mixture, but not enough to alter the direction of the effect.
Here's another rundown of Raoult's law, but in general, it says that in a mixture, the vapor pressure is directly determined by the vapor pressure that component (here, water) would have if pure, times the mole fraction of the component in the mixture (i.e., the fraction of water in the mixture)). The mole fraction of water in seawater is less than in freshwater (where it is 1.0000), so the vapor pressure is less and evaporation is slower.
https://en.wikipedia.org/wiki/Raoult's_law
Here's another rundown of Raoult's law, but in general, it says that in a mixture, the vapor pressure is directly determined by the vapor pressure that component (here, water) would have if pure, times the mole fraction of the component in the mixture (i.e., the fraction of water in the mixture)). The mole fraction of water in seawater is less than in freshwater (where it is 1.0000), so the vapor pressure is less and evaporation is slower.
https://en.wikipedia.org/wiki/Raoult's_law
). It was a fun and good question that you asked.
