Boiling point of mixtures

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m25

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So I always hear how unpure substances have higher boiling point/lower vapor pressure, but is this statement only true when the unpure substance we add have higher boiling points then the main substance?
That is, is it correct to say that by adding unpure substance A, we are actually just getting the average of A and the main substance's boiling point?
So in other words, from the perspective of substance A, it's boiling point actually decreases when it's added to the main substance?
And how does this relate to Raoult's Law for calculating vapor pressure of solutions with two or more substances, if there is any relation at all?
Raoults_Law.png

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The way I understand it is that vapor pressure is the pressure that the vapor above is pushing down on the substance (so it's like the amount of vapor above the substance; hence why you can use mole fractions and Raoult's law to figure out vapor pressure). When you have impurities in the substance it makes it harder for the substance to bounce out and become vapor (the impurities block the substance from bouncing out; think about the top of the substance as a line of beads {no string in between} and you can punch below the beads and they will go flying out, now replace some of the beads with rocks and now when you punch the bottom not as many beads bounce out because some of the rocks go also). Now the boiling point is the temperature at which the liquid becomes vapor (hotter things bounce around more and will be more likely to bounce out of liquid phase and become vapor {adding heat will break up intermolecular forces between the particles also}). So you see that a lower boiling point means a higher vapor pressure right (the lower the temperature it takes for the substance to become vapor means that there is more vapor above pushing down keeping it in equilibrium)? OK, so now onto your questions:

Yes, the boiling point will be higher than the substance with the lower boiling point, but lower than the substance with the higher boiling point (unless of course there are intermolecular forces that are binding the two substances together and you get a negative deviation from Raoult's Law).

Yes, but like I stated earlier there could be intermolecular forces (like hydrogen bonding) that are holding the molecules together and you'd get a higher boiling point with a mixture.

It indirectly correlates, the vapor pressure is simply the pressure that the vapor is exerting on the substance (the amount of vapor above it) so a lower vapor pressure means a higher boiling point (less vapor above substance means that you need a higher temperature to pushing the solution to become vapor) and vice versa. You can even see it in the graph you posted, pure A has a higher vapor pressure (there is more vapor above pure liquid A than there is of vapor B above pure liquid B) so when you mix them you use the mole fraction to get the vapor pressure and thus a relation for the boiling point.
 
Correct me if I'm wrong, but this is how I would look at it
The statement "unpure substances have higher boiling points/lower vapor pressure" is a good statement when you're talking about nonvolatile substances being dissolved in the solution.
The more non-volatile solutes in a solution, the lower the vapor pressure because there are now fewer solvent molecules at the surface of the solution (non volatile substances would be taking up some space at the surface) which minimizes evaporation and re-absorption. Non-volatile solutes do not have vapour pressure. However, when you have a volatile substance, this relates to raoult's law because now you have another "(Pvap)(X)" term added to the equation
 
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