[WhoCares357]

The partial water vapor pressure over the solution of glucose (C₂H₁₂O₆) in H₂O is 745 torr at 100 degrees Celsius. Find the freezing temperature of this solution.

I know I have to use these three equations:
Raoult's law -> Vapor pressure of solvent = (Vapor pressure of pure solvent) * (Mole fraction of solvent)
boiling point elevation -> (new boiling point) - (boiling point of pure solvent) = (molality of solute) * (constant [degrees celsius/molality])
freezing point depression -> (freezing point of pure solvent) - (new freezing point) = (molality of solute) * (constant [degrees celsius/molality])

I started by setting up the freezing point depression formulat as such:
(0 degrees Celsius) - x degrees celsius = (molality of glucose) * (1.86 celsius/molality)

The only thing I'm missing in the equation is molality of glucose. This is where I am right now. I don't really know how to find the molality of glucose. I understand that it has something to do with the 745 torr at 100 celsius, but I don't know what.

I cannot use raoult's law because I don't know the mole fraction of the solvent. I also can't use the boiling point elevation formula because I do not know the "new boiling point" with the given pressure.

Can anyone please point me in the right direction?

Thank you.

[MrTeo]

What happens when you heat water up to 100ºC?
What's the vapor pressure of water at that temperature?

[WhoCares357]

Okay so I followed your hint and set up the equation: 745 torr = (760 torr) * (moles of H₂O/moles of solution)
I solved the equation and found that the molar fraction of H₂O is ~0.980. This means that there are ~0.020 moles of glucose. To find molality I divided 0.020 moles by the molar weight of water (18 g/mol * .980 mol = .018kg). Molarity is .020 moles/.018kg ~1.134m.

I plugged the molality into the equation I presented in the first post and got a final freezing temperature of ~-2.109 Celsius.

Can anyone verify that the answer is correct?

[MrTeo]

This means that there are ~0.020 moles of glucose.

Nope, this only means that:



as you can see 0.020 is an adimensional value.
Consider that the solution is probably diluted (otherwise Raoult's law wouldn't work) so the amount of glucose is very small if compared with that of water...

[WhoCares357]

So I can't assume a constant for the moles of solution?

I just assumed that the whole solution was one mole. Since water was 98% moles of the solution it was .98 moles; glucose was .02. Is the problem that .02 is too small? Why does that matter? If I assumed the whole solution was 100 moles wouldn't I still get the same molality?

[MrTeo]

So I can't assume a constant for the moles of solution?

I just assumed that the whole solution was one mole. Since water was 98% moles of the solution it was .98 moles; glucose was .02. Is the problem that .02 is too small? Why does that matter? If I assumed the whole solution was 100 moles wouldn't I still get the same molality?

As I said the moles of solution can be considered equal to the moles of water due to dilution... but you can't say that there are 0.020 mol of glucose as you don't know.
Anyway your idea is quite right, but how many moles are in a liter of water?

[WhoCares357]

There are 55.556 moles of water in 1 liter of water. So there are ~1.134 moles of glucose / 1 kg of water.

I used a proportion to find the number of moles (if 55.556 is 98% how much is 2%) and set up the equation for molality.

So my answer was correct, but I took the incorrect route to it.

[MrTeo]

OK with me