Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nonfatmintlatte on January 16, 2010, 04:40:07 AM
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I already know that the addition of a solute to a solvent lowers its freezing/melting point. However, if some material that is insoluble is added into the solvent or the solution, how will that affect the change in freezing/melting point of the solvent?
Please help, thanks
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I think the insoluble solute doesn't affect the freezing point of the solvent....{But nothing is 100% insoluble}
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I think the insoluble solute doesn't affect the freezing point of the solvent....{But nothing is 100% insoluble}
Are you sure?
It's easy to think of it like this : If something is insoluble in a compound, I guess you could in a way treat it as an impurity in the original solute. What do you know about impure compounds and their freezing/boiling points?
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I think the insoluble solute doesn't affect the freezing point of the solvent....{But nothing is 100% insoluble}
Are you sure?
It's easy to think of it like this : If something is insoluble in a compound, I guess you could in a way treat it as an impurity in the original solute. What do you know about impure compounds and their freezing/boiling points?
I actually don't know. Of course, an impure substance will not have the exact same freezing/boiling point as its pure substance. What does this mean?
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I think the insoluble solute doesn't affect the freezing point of the solvent....{But nothing is 100% insoluble}
Are you sure?
It's easy to think of it like this : If something is insoluble in a compound, I guess you could in a way treat it as an impurity in the original solute. What do you know about impure compounds and their freezing/boiling points?
NO - If it is insoluble it will not effect the freezing/boiling point.
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DrCMS –
Not to contradict your entry, but to make an observation.
Sometimes questions can be muddy (possible pun).
You have a beaker of water and you put sugar in it until no more will dissolve.
You then boil the water to get a super-saturated solution.
You put in an insoluble string and the sugar solidifies on that string making rock candy.
I would guess dropping in sand might do the same thing except it would be too gritty to be candy.
I can not think of adding an insoluble to a solid and get a liquid.
I assume this observation does not apply to the original question (sort of apples and oranges).
Regards,
Bill
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I think the insoluble solute doesn't affect the freezing point of the solvent....{But nothing is 100% insoluble}
Are you sure?
It's easy to think of it like this : If something is insoluble in a compound, I guess you could in a way treat it as an impurity in the original solute. What do you know about impure compounds and their freezing/boiling points?
Sorry guys, that was a typo... I meant solvent.
NO - If it is insoluble it will not effect the freezing/boiling point.
I'm sorry if I am mistaken, but can you please explain why freezing/boiling point will not be affected?
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Well what we are interested in is freezing point of "solvent". Only soluble materials can affect the freezing point of the "solvent".
Because may be insoluble compounds do not change the vapour pressure of the mixture
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Well, does that mean there is no sort of interaction between an insoluble substance and a solvent whatsoever?
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Yup, but I am not sure...
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If you put 100ml of water in a beaker what do you expect the boiling point to be?
If you put 100ml of water in a beaker with 100g of sand what do you expect the boiling point to be?
If you heated 100g of sand and turned it into glass and added that to the first beaker do you think it would change the boiling point?
If you used 50g of water or 200g of water in the same beaker do you think the boiling point will change?
If an insoluble material like glass changed the boiling point based on the amount of it relative to the solvent all the above examples would have different boiling points.
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@DrCMS
This makes things clear :D
In all the cases you mentioned boiling point is same. Right?
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If you put 100ml of water in a beaker with 100g of sand what do you expect the boiling point to be?
If you put 1000ml of water in a beaker with 10g of thorougly screned flour(which is definitly non-soluble in water) what do you expect the freezing point to be?
All colligative properties are entropic effects(google it or Atkins it). On the other hand colloidal particles are very big, so there not so much of them to raise significantly the degree of disorder of the whole system.
So the answer is YES, but to the very small degree.
IMHO
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can anyone tell me how to phrase the reasons for the answers? like some people say that the answer is yes or the answer is no. i am supposed to answer this for homework and we haven't reached a consensus yet here.
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There is no single and simple answer to the question, unless it is "no". Freezing point depression is proportional to concentration (molality) of the dissolved substances. Molality of insoluble substance is always zero, hence change in freezing point is also zero. But very fine suspensions or colloidal solutions - which are in a way borderline between solution and not-a-solution - may have specific properties different from properties of "normal" solutions. In most cases effects will be probably neglectable, so I would stick with "no" as a first approximation.
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I think the answer is that it does have a very slight change on the freezing/boiling points just because of the intermolecular attractions of the insoluble substance.
BUT, I think it's only changed extremely slightly, because of observation. Let's say you have a pot of water, maybe the pot is made of aluminum. Let's say you have a beaker full of water. Wouldn't they both freeze and boil at the same temperature? However, can't you technically say that the glass and the aluminum are insoluble substances touching the water?
I hope I'm not wrong but this is just my thinking.
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can anyone tell me how to phrase the reasons for the answers? like some people say that the answer is yes or the answer is no. i am supposed to answer this for homework and we haven't reached a consensus yet here.
You can think about the problem in such way.
Entropic factor in other words is the influence of particles other than molecules of solvent, i.e. that have another size, shape, electric and magnetic properties etc. that prevent from molecules of solvent to form regular geometric structures of a solid crystal at low temperatures(by increasing disorder) when there is already no preventing enthalpic factor(i.e. motion of molecules itself does not prevent from solvent to form solid crystal structure).
Now imagine yourself a glass ball of radius 1 cm that you put into a beaker with distilled water. It is obvious that it wouldn't prevent from molecules of water to form crystals in their usual way, excluding a very thin layer adjacent to glass. Now you break the ball into billions of billions of pieces with volume that equals the volume of 10-20 molecules of water and disperse them uniformly in the water(it is not dissolution). These pieces also would prevent from adjacent layers of water to form regular hexagonal patterns at usual freezing point. But now the "adjacent volume of water" appears to be a considerable part of the system so you would have to slow down the motion of molecules a bit more to force your water to freeze.
You can think also about a handful of fullerens or a bundle of nanotubes that were put to a beaker with water.
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do boiling chips change the boiling point
are they soluble