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#### Borek

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##### Re: Not a problem of the week, but similar level of difficulty
« Reply #15 on: October 01, 2017, 08:59:56 PM »

I would like to try and solve this, but I was curious to know if all the information needed to solve the problem is already present?

Not exactly, but I assume the freezing point of water and fact that it is almost identical with the triple point of water are a common knowledge. Perhaps it should be spelled out in the problem.

Quote
Also, I was wondering what the latter three pieces of information referred to. Are they for ice/water, or the glucose solution?

Ice/water.
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#### Borek

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##### Re: Not a problem of the week, but similar level of difficulty
« Reply #16 on: October 01, 2017, 09:08:21 PM »

For (1), we additionally need to know the molar weight and normal freezing point for water, as well as the universal gas constant (8.315).

These are between things that I assume to be a general knowledge, easily available. Again, you are right they could be spelled out in the problem.

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I finally tried to find the normal freezing point of water using the saturated vapour pressure piece of information. I tried to use the Clausius-Clapeyron relation, but it didn't seem to be valid...

That's actually the path I used to solve the problem originally - using C-C relation twice to find saturated pressures above water and ice, comparing them and solving for T.

I admit I completely forgot about the Kf formula you listed, the question arose from a long discussion on the vapor pressures above solids and liquids and conclusions we can get from them, so I was a bit narrow minded.

Still, chances are Kf formula can be derived this way. I have long lost my notes related to the problem and sadly I have no time to solve it again at the moment. Perhaps later today.
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#### XeLa.

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##### Re: Not a problem of the week, but similar level of difficulty
« Reply #17 on: November 18, 2017, 05:15:40 PM »

Okay, I managed to solve this question using the Clausius-Clapeyron equation...

First, consider the equation...

ln(P1/P2)=(ΔHfus/R) × (1/T2 - 1/T1)

ΔHfus is found to be 6010 J/mol using the provided data.

Let P1 = vapour pressure of saturated water at T1, and P2 = vapour pressure of 1 molal glucose solution at T2.

P1 = 611 Pa
T1 = 273 K
P2 = xsolvent × P°solvent = (1 - 0.0177) × 611 = 600.185 Pa (round for final answer)
T2 = ? K

Find T2...

ln(600.185/611) = (6010/8.314) × (1/273 - 1/T2)

T2 = 271.17 K

T2 would be the effective temperature of the glucose solution at its freezing threshold... Given this, we can use the equation:

ΔT = Kf · m

ΔT = 273 K - 271.17 K = 1.83 K
m = 1 mol/kg

Kf = 1.83/1 = 1.83 kg · K/mol

As such, the cryoscopic constant for water is 1.83 kg · K/mol.

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#### Borek

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##### Re: Not a problem of the week, but similar level of difficulty
« Reply #18 on: November 19, 2017, 12:34:32 AM »

Yes, that is close to what I had on mind, just my approach took using Claussius-Clapeyron equation for sublimation and evaporation and an additional assumption (quite easy to defend). Had to solve it again to close the case

From the information given we know at the melting point pressure of the vapor is 611 Pa. What we also know (not from the question itself, but I assume it to be a common knowledge), is that the melting point (0 °C) is very close to the triple point (0.01 °C apart) - that means we can safely assume 611 Pa to be also a saturated vapor over the solid.

Next things were already said earlier, so I am just listing them here for completeness. At the melting point saturated vapor pressures over the solid and the liquid are identical:

$$p_{liquid} = p_{solid}$$

we also know that the

$$p_{liquid} = (1-x)p_{liquid}^0$$

and finally we have two Claussius-Clapeyron equations ($t$ stands for the melting point temperature):

$$\log\left(\frac{p_{solid}}{611 Pa}\right) = \frac {\Delta H_{sub}}R\left(\frac 1 {273.16} - \frac 1 t\right)$$

and

$$\log\left(\frac{p_{liquid}^0}{611 Pa}\right) = \frac {\Delta H_{vap}}R\left(\frac 1 {273.16} - \frac 1 t\right)$$

plugging $p_{liquid}^0$ into the second equation

$$\log\left(\frac{\frac{p_{liquid}}{1-x}}{611 Pa}\right) = \frac {\Delta H_{vap}}R\left(\frac 1 {273.16} - \frac 1 t\right)$$

We have two equations, easy to solve for $t$ and $p_{liquid}$. I was too lazy to do that manually, so I just plugged them into Wolfram Alpha, and got t=271.33.

Again, solution is 1 molal, so ΔT=273.16-171.33=1.83 K×kg/mol.
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