# BR question. Confusing~~

Discussion in 'MCAT Study Question Q&A' started by chaser0, 05.05.12.

1. ### chaser0

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Solute X does not dissociate once in solution. If another impurity were chosen instead that can dissociate into two particles in solution, how would the freezing point be affected?
A.) the freezing point will decrease by twice as much as expected, if the impurity does not dissociate.
B.) the FP would decrease by as much as expected, if the impurity does not dissociate.
C.) the freezing point would decrease by half as much as expected, if the impurity does not dissociate.
D.) the FP would be constant.

Anyone want to give an answer and their reason??
Last edited: 05.05.12
2. ### Dasypus

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Last edited: 05.25.12
3. ### chaser0

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Wait what? That doesnt make any sense.

A says it does NOT dissociate, meaning that the new solute has the same effect as the original solute X.
The FP shouldnt be any different at all between X and the new.
4. ### chaser0

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Oh and sorry I wrote the answer choices wrong.

Fixed it
B and C are exclusive
5. ### Dasypus

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Last edited: 05.25.12
6. ### chaser0

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Oh wait i think ur right.

The wording of the answer Choices are funny.
What they mean to say is:
The FP would decrease by twice as much compared to another experiment where the impurities did not dissociate.

So yeah, A haha
7. ### MedPR

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Going with B. FP is a colligative property meaning that only the number of particles matters. If something doesn't dissociate then the vant hoff factor is 1. If something that can dissociate into two things, but doesn't dissociate in the solution given then the van't hoff factor is still 1.
8. ### chaser0

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Yeah answer is A, the problem is just worded extremely poorly~~

I thought it was B too