Geekchick921

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Today's answer confuses me. It seems to me like they wrong answer was marked correct.

Three identical solutions are brought to their cloud point.

Solution 1 is kept at its cloud point and solute is added
Solution 2 is heated above its cloud point
Solution 3 is cooled below its cloud point

If each solution is filtered, which would you expect will have precipitate remain in the filter?

(a) Solution 1 and 2
(b) Solution 2 only
(c) Solution 1 and 3
(d) Solution 3 only

I am pretty confident I understand this concept, as it's pretty simple, and 1 and 3 would be the ones to have a precipitate... certainly not 2, as raising the temperature increases solubility. I submitted C, and it was wrong. I was confused and clicked through the rest of the answers, and B is right, but the explanation also makes me think that there's a mistake today.


Explanation: The cloud point is the temperature at which a solution is on the brink of precipitation. If the temperature is decreased and/or solute is added, the solution will precipitate. If the temperature is increased and/or the solution is diluted, the solution will not precipitate. Hence, Solution 1 and Solution 3 will precipitate.

:confused:
 

BerkReviewTeach

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Today's answer confuses me. It seems to me like they wrong answer was marked correct.

Three identical solutions are brought to their cloud point.

Solution 1 is kept at its cloud point and solute is added
Solution 2 is heated above its cloud point
Solution 3 is cooled below its cloud point

If each solution is filtered, which would you expect will have precipitate remain in the filter?

(a) Solution 1 and 2
(b) Solution 2 only
(c) Solution 1 and 3
(d) Solution 3 only

I am pretty confident I understand this concept, as it's pretty simple, and 1 and 3 would be the ones to have a precipitate... certainly not 2, as raising the temperature increases solubility. I submitted C, and it was wrong. I was confused and clicked through the rest of the answers, and B is right, but the explanation also makes me think that there's a mistake today.


Explanation: The cloud point is the temperature at which a solution is on the brink of precipitation. If the temperature is decreased and/or solute is added, the solution will precipitate. If the temperature is increased and/or the solution is diluted, the solution will not precipitate. Hence, Solution 1 and Solution 3 will precipitate.

:confused:
Your logic and explanation seem correct to me. I assume it's just a typo on their part. It happens.
 

MintJulep

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I just answered it (answered C), and it said it was correct. So perhaps they fixed that.
 

Geekchick921

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I just answered it (answered C), and it said it was correct. So perhaps they fixed that.
Yup! Looks like. You can see the % of user that got it right is pretty low.
 

Flapjacks

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Ya same here, never heard of cloud point before and I've retaken chem 1, 2 twice.
 

AbbyNormal

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I've never heard of a cloud point before in my life.
Me either and I had three semesters of college chemistry.

But now I know.
 

Geekchick921

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I don't think I ever heard the term before either, but judging from the name, and the context of the rest of the question, I figured it would be the point of maximum dissolved solute.
 
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I see where you guys are going... but it seems that this explination is only correct for MOST solutions as most solutions are endothermic (q>o)... thus, lets say nacl(s) + "q" ---> dissolves to na + cl... we know about ksp values and such and i believe the dissociation process is endothermic, then yess more heat shifts equilibrium to the right BUT BUT BUT, if its some other proccess that has q on teh right side say an EXOTHERMIC PROCESS... then heating has the opposite effect (it makes more solid) and cooling allows more percipitate...

IN CONCLUSION, the question basically forgot to tack on a if endothermic or exothermic dissolving processs. ALSO THE CLOUD POINT AND ADDING MORE PERCIPITATE INVOLVES CHANGES DUE TO LE CHATELEAR IN TERMS OF CONCENTRATIONS NOT HEATS SO ENDOTHERMIC OR EXOTHERMIC DOESNT REALLY MATTER BESIDES THE FACT THAT THE SOLUTION CONTAINOR WILL GIVE OFF HEAT OR TAKE IN HEAT WHEN MORE PERCIPTIATE IS ADDED.

PS sorry i just realized i used all caps. lol
 

BerkReviewTeach

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I see where you guys are going... but it seems that this explination is only correct for MOST solutions as most solutions are endothermic (q>o)... thus, lets say nacl(s) + "q" ---> dissolves to na + cl... we know about ksp values and such and i believe the dissociation process is endothermic, then yess more heat shifts equilibrium to the right BUT BUT BUT, if its some other proccess that has q on teh right side say an EXOTHERMIC PROCESS... then heating has the opposite effect (it makes more solid) and cooling allows more percipitate...

IN CONCLUSION, the question basically forgot to tack on a if endothermic or exothermic dissolving processs. ALSO THE CLOUD POINT AND ADDING MORE PERCIPITATE INVOLVES CHANGES DUE TO LE CHATELEAR IN TERMS OF CONCENTRATIONS NOT HEATS SO ENDOTHERMIC OR EXOTHERMIC DOESNT REALLY MATTER BESIDES THE FACT THAT THE SOLUTION CONTAINOR WILL GIVE OFF HEAT OR TAKE IN HEAT WHEN MORE PERCIPTIATE IS ADDED.

PS sorry i just realized i used all caps. lol
Be careful with your reasoning here, because although completely valid at face value, you have forgotten the very significant impact of entropy when going from a crystalline structure to free moving solute particles. The reaction is driven by deltaG, which we get from dG = dH - TdS. dS for solubility is highly positive, so as T goes up, dG goes down, making solubility more favorable. Remember that the endothermic vs. exothermic application of Le Chatelier's Principle is for systems where the entropy change is small compared to the enthalpy change.
 
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Be careful with your reasoning here, because although completely valid at face value, you have forgotten the very significant impact of entropy when going from a crystalline structure to free moving solute particles. The reaction is driven by deltaG, which we get from dG = dH - TdS. dS for solubility is highly positive, so as T goes up, dG goes down, making solubility more favorable. Remember that the endothermic vs. exothermic application of Le Chatelier's Principle is for systems where the entropy change is small compared to the enthalpy change.
Hey, yes although i completely agree with you, i also think that since no numbers are given here, both cases must be assumed. thus no real answer is present... probably not a real mcat question...