The question asks: If you have 2 beakers, one with a completely volatile solvent (Beaker A), and other beaker has a non-volatile solvent mixed with a volatile solvent (Beaker B).. what would happen if you left those 2 beakers in equilibrium with each other?
The answer they gave was that all the solvent will go to Beaker B... because beaker B has a lower vapor pressure. How can solvents just "go" to another beaker?
For what it's worth, this question would be written better on the MCAT, even if they were trying to make it ambiguous. Don't get too concerned with this particular question, but absorb the concept.
If you have a beaker with a volatile solution, it will evaporate and escape to the environment. This implies that it has a high vapor pressure. If you have a non-volatile solution, it will not escape. This implies it has a low vapor pressure.
This question looks like it's based on a chemistry demo where they put two half-filled beakers under a domed lid, making it a closed system. The beaker with the volatile solution evaporates away and condenses where it can, mostly in the other beaker, where once condensed it will be difficult to reevaporate.
In the end, the first beaker will be empty, with some of its contants in the atmosphere and some condensed in the other beaker. It's not true that "all of the solvent will go to Beaker B" unless the system was saturated with the solvent to begin with. I doubt they specified that, so the best answer would be that it would completely leave the first beaker, leaving it empty over time.
On that note, this should be in the Q and A section.
