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Topic: Not a problem of the week, but similar level of difficulty  (Read 22581 times)

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Offline Borek

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Not a problem of the week, but similar level of difficulty
« 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)
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Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #1 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?
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Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #2 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?
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Offline Vidya

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Re: Not a problem of the week, but similar level of difficulty
« Reply #3 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.

Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #4 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?
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Offline Vidya

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Re: Not a problem of the week, but similar level of difficulty
« Reply #5 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.

Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #6 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).
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Offline Vidya

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Re: Not a problem of the week, but similar level of difficulty
« Reply #7 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)

Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #8 on: July 29, 2017, 04:47:20 AM »
ΔHfusion is not given.

Have you read the problem statement?
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Offline Vidya

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Re: Not a problem of the week, but similar level of difficulty
« Reply #9 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?

Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #10 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.
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Offline Vidya

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Re: Not a problem of the week, but similar level of difficulty
« Reply #11 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.

Offline Arkcon

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Re: Not a problem of the week, but similar level of difficulty
« Reply #12 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.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline XeLa.

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Re: Not a problem of the week, but similar level of difficulty
« Reply #13 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

Offline XeLa.

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Re: Not a problem of the week, but similar level of difficulty
« Reply #14 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.

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