There are two open beakers placed in a sealed box at constant temperature. Beaker 1 has 25 ml of .1 M NaCl in water. Beaker 2 has 45 ml of .05 M NaCl in water. What happens as the system goes towards equilibrium and what is the final contents in each beaker?

Answer 1

First is what happens.
1L H2O = 1000g/18.015 = 55.5 mol
So in 0.05M XH2O = 55.5-0.05/55.5 = 0.9991 so pH2O from the beaker will be X*Ptotal and if we take a convenient number like 1 for Ptotal then pH2O = 0.9991

What about the 0.1M.
XH2O = (55.5-0.1)/55.5 = 0.0082
The vapor pressure of the 0.05M solution is greater so the two solution will exchange vapor until they are the same which means the concn of each is the same.
If we let x = volume H2O lost from 0.05M solution and gained by the 0.1M solution and set them equal we have the final volume for 0.05M will be 45-x and the final volume for the 0.1M solution will be 25+x.
Then concn 0.05 will be 0.05 x [45/(45-x)] and the 0.1M will be 0.1 x [25/(25+x)]. Set those equal to each other and solve for x. I get about 11.8 mL but you should do it a little more accurately. So the final contents of each beaker will be 25+x and 45-x.
Interesting question, eh?

Answer 2

So does that mean that at the very end, all of the solution will be in Beaker 1 and none of the solution will be in Beaker 2?

Answer 3

Sorry, ignore that last question, I was confusing myself. But I'm still a bit confused on the process. For the XH20 of the .1 M, wouldn't the answer to that be .998 and not .0082? Even if it changes, it still doesn't change the rest of the solution, right, since the vapor pressure of .05 M solution is still greater?