Vapor Pressure, Solution Pressure, BP

[royalmarquis]

This question refers to my inability to understand the reasoning behind
EK Chemistry In Class Exam 4 Question 92

When volatile solvents A and B are mixed in equal proportions heat is given off to the surroundings. Pure A has a higher BP than pure B.

1. Will the new mixture have a BP that is higher than both pure A and pure B, regardless of the relative amounts of A and B?

2. What's the relationship of Solution Pressure to Vapor Pressure to BP?

3. Can you explain what the explanation is trying to say? Also, the graph doesn't make sense. (Explanation on page 177)

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[plzNOCarribbean]

Um, I will take a stab;
Ok, so when you mix two pure liquids, typically, if they follow Roault's law, the vapor pressure above the solution should be lower than their individual vapor pressures because the mole fraction of each liquid in the solution is no longer 1 (its a decimal now since they were mixed) However, you can add two solvents and have the attractive forces formed between them be weaker than pure A and pure B, in which case the vapor pressure is increased

Basically, vapor pressure is proportional to the IMF of the liquid; you must consider nonvolatile solutes also, as they will act as anchors to the liquid particles and thus the BP will increase, which will DECREASE the Vapor pressure;

Again, remember, not all mixtures will follow Roault's Law and some will deviate, so refer back to the IMF concept; As far as that question, it says that heat is given off so the new bonds formed between the two liquids must be stronger than pure A or pure B; because the IMF formed are stronger, then the Vapor pressure should be LOWER, and the boiling point should be Higher;

VP INCREASES as BP Increases; VP is inversely prop to IMF;
IMF increase = BP Inc = VP decreases
Hope that helps; anyone else feel free to let me know if any of that was wrong

 

[fingerscrossedd]

About three years later, and I find myself stuck on this question too

So my understanding:
-stronger IMF = lower VP
-higher BP = lower VP
-A has a higher BP than B; A has a lower VP than B
-mixed in equal proportions, so.....

isn't this true: VP(A)<VP(mixture)<VP(B)
?

I'm guessing I'm probably simplifying way too much Maybe I'm not taking mole fractions into account? The correct answer says:
Boiling point Mixture < Boiling Point B

which I understand, but why isn't my point on Vapor Pressure true?

Danke!