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Topic: Instantly Evaporating Polyvinyl Alcohol Solvent?  (Read 8072 times)

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Online Borek

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Re: Evaporating Chloroform?
« Reply #15 on: October 28, 2015, 03:12:32 AM »
Chloroform by itself evaporates easily, why do you want to complicate things using additional solvent?
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Offline billnotgatez

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Re: Evaporating Chloroform?
« Reply #16 on: October 28, 2015, 06:37:41 AM »
@dylanail
Is this thread related to your other thread
http://www.chemicalforums.com/index.php?topic=82450.0

Offline Arkcon

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Re: Evaporating Chloroform?
« Reply #17 on: October 28, 2015, 06:42:22 AM »
@dylanail
Is this thread related to your other thread
http://www.chemicalforums.com/index.php?topic=82450.0

Why not merge the two threads?  I just did.

Briefly, adding a solute to something cause boiling point elevation, according to colligative properties. https://en.wikipedia.org/wiki/Colligative_properties
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Offline dylanail

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Re: Instantly Evaporating Polyvinyl Alcohol Solvent?
« Reply #18 on: October 28, 2015, 08:07:53 PM »
The two threads were unrelated. Thanks though!
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Offline Intanjir

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Re: Instantly Evaporating Polyvinyl Alcohol Solvent?
« Reply #19 on: October 28, 2015, 09:23:56 PM »
Boiling point elevation can be understood as a colligative property and almost independent of the nature of the solute, but there is a major exception.
If the solute has a significant vapor pressure of its own then the boiling point of the solution most definitely depends on the nature of the solute. In the ideal case the total vapor pressure of the solution is just the vapor pressure you would expect from a mole-weighted average of the constituents vapor pressure.
https://en.wikipedia.org/wiki/Raoult%27s_law

BTW many colligative properties must have similar such exceptions. They must in order to account for cases when the solute is identical or near identical to the solvent because, in such cases, there should be no expected deviation. For example, if I increase the molecular weight of oxygen in water from 16 to 18 to make so-called 'heavy-oxygen water' then I expect the freezing point to be essentially the same for both water and it and also for any mixture of the two.

In the case of boiling point the relevant property of the solute we do actually care about it is volatility. In the case of freezing point it is the miscibility of the solute in the solid phase of the solvent. (The difference here is due to the fact that all gases are miscible with one another whereas very few solids are miscible)


Anyways, it is quite possible for the boiling point of a solution to be below that of any of the constituents. At least it is if we assume the components manage to mix despite not being highly attracted to one another, so that it is actually energetically easier for particles to break free of the bulk liquid. It isn't hard to imagine that this is possible thanks to the rather chaotic character of liquids at a molecular scale, but only to a degree. Inevitably at some point the relative lack of mutual attraction will cause immiscibility.

Chloroform will form a positive azeotrope with methanol and ethanol. Unusually the effect might actually be large enough to be useful in the case of methanol:
Chloroform BP=61.1 ˚C
Methanol BP = 64.7 ˚C
87.4% (by weight) Chloroform + 12.6% Methanol (Azeotrope) BP=53.5 ˚C
https://en.wikipedia.org/wiki/Azeotrope_tables

However, just because the boiling point is lower does not mean we have improved on the evaporation rate of of pure chloroform. If methanol evaporates sufficiently slower then chloroform and if there is enough of it in the solution then the solution could potentially evaporate slower as well.
I don't claim to understand evaporation rate stuff well and there is fair amount of methanol in there by mole so as far as I know it really could go either way.
« Last Edit: October 28, 2015, 09:35:55 PM by Intanjir »

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