Aamc 3 #35

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BPMED

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Can someone please explain the solution to this problem? I thought the answer was A 'the crotonic acid solution, because it contains a lower molar concentration of solute particles'.

This made sense to me because since it had a lower molecular weight, it would have more moles for a given gram of that substance (ex. crotonic acid would have more moles per gram than would oxalic acid).

Help?
 
Can someone please explain the solution to this problem? I thought the answer was A 'the crotonic acid solution, because it contains a lower molar concentration of solute particles'.

This made sense to me because since it had a lower molecular weight, it would have more moles for a given gram of that substance (ex. crotonic acid would have more moles per gram than would oxalic acid).

Help?

Hey bro. I just took AAMC3 and got a 34 (11/12/11)!!! My diagnostic was a 24 and my first full-length practice test was a 23 so I'm pretty freaking excited about that kind of progress.

Anyway, that was a tricky problem. Oxalic acid (without going into too much detail) is a stronger acid. So, more NaOH would need to be added to raise the pH of the solution to 4.7.

Now, remember that when you add a non-volatile solute to a solvent, you a) raise the boiling point, and b) lower the freezing point. Hopefully you understand these phenomena.

Anyway, the point is, more NaOH in the oxalic solution means you end up lowering the freezing point more.

Think of a chef adding salt to a pot of boiling water. This is done to RAISE the boiling point, so that you can cook stuff at a higher temperature.

Think of adding putting salt on an icy road. This is done to LOWER the freezing point, i.e., instead of ice forming at 32F, it will form perhaps at 25F.

Lastly, remember that BPE and FPD depend on the number of molecules, not on the type of molecule.
 
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thanks for the reply and congrats on the progress. After thinking of this problem for awhile, I thought it was more along the lines of oxalic being diprotic and crotonic being monoprotic, which would affect the van't hoff number. this made sense to me since the molarity of both acids were the same, meaning the molalities were pretty much the same as well.

What do you think?
 
thanks for the reply and congrats on the progress. After thinking of this problem for awhile, I thought it was more along the lines of oxalic being diprotic and crotonic being monoprotic, which would affect the van't hoff number. this made sense to me since the molarity of both acids were the same, meaning the molalities were pretty much the same as well.

What do you think?

That would be the opposite of how to approach it. Oxalic acid is the stronger acid, and like you said, it's diprotic.

So, you have two acids of different strengths. We've established that. In the problem stem, it said you added NaOH to make the pH of both solutions some number (can't remember what it was, let's say 6.0). So it takes more NaOH to make the oxalic acid solution have a pH of (6.0), because it's a stronger acid to begin with.

And more NaOH means the freezing points is lowered more.
 
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