# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on June 03, 2017, 04:57:11 PM

Title: Not a problem of the week, but similar level of difficulty
Post by: Borek on June 03, 2017, 04:57:11 PM
Estimate cryoscopic constant for water, knowing that:

In 1 molal glucose solution molar fraction of the glucose is 0.0177.

Saturated vapor pressure over pure water at 0°C is 611 Pa.

ΔHvap is 45.05 kJ/mol (evaporation)

ΔHsub is 51.06 kJ/mol (sublimation)
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on July 01, 2017, 07:09:34 AM
Almost a month has passed and nobody even tried?

Hint: what is the saturated vapor pressure over the ice at 0°C?
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on July 12, 2017, 05:18:44 PM
Another hint: what can you tell about vapor pressures over the ice and the solution at the freezing point?
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Vidya on July 27, 2017, 09:14:26 AM
Another hint: what can you tell about vapor pressures over the ice and the solution at the freezing point?
Vaporpressure of ice and solution at the freezing point are equal.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on July 28, 2017, 05:17:51 PM
Vaporpressure of ice and solution at the freezing point are equal.

Yep.

Can you apply this information to the problem?
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Vidya on July 29, 2017, 12:12:55 AM
Yes
First we will calculate new vapor pressure of the Soln
Using equation P solution = Xsolvent. P( pure  water)
Xsolvent = moles of solvent (water)/ ( total moles of Solute and solvent)
Solute here is glucose . Once you have the new vapor pressure of the Solution then we can use phase diagram of water to know new freezing point of water. Now
Δt = m • kf.
Here m is the molality of glucose and  Δt is depression in freezing point.
Kf is the cryoscopic constant.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on July 29, 2017, 03:27:55 AM
While what you wrote is true it still doesn't say how to calculate Kf from the information given (phase diagram is not between them).
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Vidya on July 29, 2017, 03:42:50 AM
If phase diagram is not an option then we can use one more equation
In this equation
R is the universal gas constant
M is the molar mass of water
Tf is the freezing point of pure water
ΔHfusion  is the fusion enthalpy for ice and water.
kf = RMTf2/(1000 x ΔHfusion)
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on July 29, 2017, 04:47:20 AM
ΔHfusion is not given.

Have you read the problem statement?
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Vidya on July 29, 2017, 05:00:35 AM
Do you know the relation between ΔH Sub , ΔH Vap and ΔH fusion ?...use that relationship to work out ΔH fusion
Of course I have read the problem ...how could I even start the problem without reading it?
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on July 29, 2017, 06:16:03 AM
OK, I see what you mean.

Still, as it was typical with the problem of the week I would prefer to see a solution in real numbers, not just an outline.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Vidya on July 29, 2017, 06:28:51 AM
As  a rule in this forum we can discuss and guide but can not give calculated answers.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Arkcon on July 29, 2017, 06:39:29 AM
As  a rule in this forum we can discuss and guide but can not give calculated answers.

A bit confused, aren't you?

Borek: deliberately writes the problem of the week.  He's well past his undergraduate studies.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: XeLa. on October 02, 2017, 12:35:04 AM
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?

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

Thank you,
XeLa
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: XeLa. on October 02, 2017, 02:04:16 AM
Okay, well my initial thoughts are how to approach this question. If one were to derive the expression:

Kf = (M·R·Tf2)/ΔHfus   (1), then one could quite easily solve for Kf.

Using the given information, the heat of fusion for water can be derived as such:

ΔHfus = ΔHsub - ΔHvap

Therefore, the heat of fusion for water is 6.01 kJ/mol.

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).

Using the mole fraction of glucose in a 1-molal solution of the solute, we can derive the molar weight for water using

1 = nglucose/(MWwater·nwater), and 0.0177 = nglucose/(nglucose + nwater)...

We find it to be 0.01802 kg/mol.

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...

If the question assumes that we know the normal freezing point of water, then it should be pretty straightforward...

I would be interested to see your solution for how this question should be solved, Borek.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on October 02, 2017, 02:59:56 AM
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.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on October 02, 2017, 03:08:21 AM
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.

Quote
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.
Title: Re: Not a problem of the week, but similar level of difficulty
Post by: XeLa. on November 18, 2017, 11:15:40 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.

Title: Re: Not a problem of the week, but similar level of difficulty
Post by: Borek on November 19, 2017, 06:34:32 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 (http://www.wolframalpha.com/input/?i=ln(p%2F(1-0.0177)%2F611)%3D45050%2F8.31*(1%2F273.16-1%2Ft),+ln(p%2F611)%3D51060%2F8.31*(1%2F273.16-1%2Ft)), and got t=271.33.

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