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
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.
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
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.
To calculate the vapor pressure depression, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
First, let's find the mole fraction of the solvent, water, which is represented by X(solvent).
Mole fraction (X) is calculated by dividing the moles of the component by the total moles of the solution.
To find the moles of water, we need to convert the given mass of water (205 g) into moles. We can use the molar mass of water, which is approximately 18.015 g/mol.
moles of water = mass of water / molar mass of water
moles of water = 205 g / 18.015 g/mol
moles of water = 11.387 mol (rounded to 3 decimal places)
To find the moles of ethylene glycol, which is the solute, we can use the given mass of ethylene glycol (115 g) and its molar mass.
The molar mass of ethylene glycol (C2H6O2) is approximately 62.068 g/mol.
moles of ethylene glycol = mass of ethylene glycol / molar mass of ethylene glycol
moles of ethylene glycol = 115 g / 62.068 g/mol
moles of ethylene glycol = 1.853 mol (rounded to 3 decimal places)
Now, we can calculate the mole fraction of water (X(solvent)).
X(solvent) = moles of water / total moles of solution
X(solvent) = 11.387 mol / (11.387 mol + 1.853 mol)
X(solvent) = 0.860 (rounded to 3 decimal places)
Finally, we can use Raoult's law to calculate the vapor pressure depression (ΔP).
ΔP = X(solvent) * P(solvent)
Given: P(solvent) = 231 mm Hg
ΔP = 0.860 * 231 mm Hg
ΔP = 198.46 mm Hg (rounded to 3 decimal places)
Therefore, the vapor pressure depression of the solution is approximately 198.46 mm Hg.