Vapor + atm pressure

[NA19]

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So the answer to this question is apparently D instead of A because: "if you started with pure ethylene glycol, and then added water (move from left to right), the freezing point of ethylene glycol should also drop because water is an impurity."

I'm confused about why the freezing point would start increasing then if water is depressing ethylene glycol's freezing point. How do you know which substance to count as an impurity and which to count as the solvent?

 

[ashtonjam]

Solvent/impurity is all relative. If one is more 'important' or present in higher quantity than the other it could probably be called the solvent. But you could just as easily reverse the roles. Granted, water is usually the solvent so it's just that way by convention.

 

[Trayshawn]

Yeah. General rule of thumb, the one thats in the majority is usually the solvent. That is most often water, but not always.

Easiest way to do this is to consider the two extremes. Water has stronger intermolecular attractions and thus has a higher melting/freezing point. As a result, the right extreme of the graph (100% water) should be above the left extreme (100% glycol).

Then, the graph should decrease as you move away from these two extremes with the added impurities. This gives D. The lowest point on such a graph is usually somewhere in the middle. I believe this is what is known as an azeotrope.

 

[gettheleadout]

Are we looking at the correct figure here? The OP's description doesn't match what I see in the attachment and the thread title doesn't match the post.

 

[NA19]

Sorry - the title doesn't match the post. But the description and figure match each other 🙂

 

[discowisco]

Yeah. General rule of thumb, the one thats in the majority is usually the solvent. That is most often water, but not always.

Easiest way to do this is to consider the two extremes. Water has stronger intermolecular attractions and thus has a higher melting/freezing point. As a result, the right extreme of the graph (100% water) should be above the left extreme (100% glycol).

Then, the graph should decrease as you move away from these two extremes with the added impurities. This gives D. The lowest point on such a graph is usually somewhere in the middle. I believe this is what is known as an azeotrope.

When your moving away from the glycol extreme..it isn't decreasing though? The graph is steady then increases up the curve as shown in the graph D

Also how come in tbr they show graphs of freezing point depression with linear graphs sloping down with a negative slope down with increased impurities? It's not a curved line.

 

[NA19]

I have no idea. The answer key said that D is supposed to look like it's decreasing and then increasing.