The vapor pressure of pure water at 70°C is 231 mm Hg. What is the vapor pressure depression of a solution of 115 g of the antifreeze ethylene glycol, C2H6O2, a nonvolatile compound, in 205 g of water? Use molar masses with at least as many significant figures as the data given.
I got 200. by using P(solution)= X(mole fraction of solvent) * P(solvent). I still got it wrong.
This is what I did: P(sln)= (11.4/13.2) * 231
8 answers
You have calculated the Psolution, and that looks ok; however, the problem asks for the "depression" of the vapor pressure.
What's the difference between Psol and Pvap depression? How would I find the Pvap depression?
You can do it two ways. One is to use what you've already done, then lowering is
Posolvent -P soln = delta P.
OR the second way is
delta P = Xsolute*Posolvent
You should get the same answer either way.
Posolvent -P soln = delta P.
OR the second way is
delta P = Xsolute*Posolvent
You should get the same answer either way.
Okay so I used the equation dP= Xsolute * Psol. I got 32.3 and it was correct.
I used the exact same equation for this one, and I got 21.7. For some reason it was incorrect. What am I doing wrong?
The vapor pressure of pure water at 70°C is 239 mm Hg. What is the vapor pressure depression of a solution of 135 g of CaCl2 in 220. g of water? Use molar masses with at least as many significant figures as the data given.
dP= (1.22/13.42) * 239
dP = 21.7 mm Hg
I used the exact same equation for this one, and I got 21.7. For some reason it was incorrect. What am I doing wrong?
The vapor pressure of pure water at 70°C is 239 mm Hg. What is the vapor pressure depression of a solution of 135 g of CaCl2 in 220. g of water? Use molar masses with at least as many significant figures as the data given.
dP= (1.22/13.42) * 239
dP = 21.7 mm Hg
Just a quick glance but at first glance I would say that you didn't use the van't Hoff factor for CaCl2 which is 3.
In addition I note that the first problem has v.p. at 70 C = 231 and the second one shows 239 mm at the same T.
When using the van't Hoff factor, I multiply the number of moles of CaCl2 by the factor right? So when finding the mole fraction of CaCl2 in the solution, it would be 3(1.22 mol CaCl2)/[3(1.22) + 12.2] which equals about .231. So then it would be dP = .231 * 239 = 55.2 mm Hg. This was still an incorrect answer.
The method look ok to me and I obtained 55 when I worked it. Since the method seems ok I would look to make sure the number of significant digits are ok. I've seen a lot of databases reject values because too many or too few s.f. were reported. I also wonder about the two values of 231 vs 239 vapor pressure for H2O at 70 C.
2 answers