Question

Suppose you had a mixture of 1 moles of methanol and 1 mole of ethanol at a particular temperature. The vapor pressure of pure methanol at this temperature is 81 kPa, and the vapor pressure of pure ethanol is 45 kPa. Using Raoult's Law calculate the total vapor pressure for this solution.

Answer 1

First write down Raoult's law

Answer 2

That's the thing, I do not know what Raoult's Law is. In class we never even talked about Raoult's Law let alone do an example question to see what it was like.

Answer 3

It's \(\sf \large P_{soln}=X_{solvent}*P^o_{solvent}\)

you can read about it here http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Changes_In_Vapor_Pressure,_Raoult's_Law

An example very similar to your question is given at the bottom, example 2

Answer 4

Since methanol and ethanol are already in moles, do I leave them alone?

Answer 5

you dont need to convert them if that's what you're asking.

Answer 6

Okay.

Answer 7

Now for the vapor pressure before I do anything else, do I need to convert them?

Answer 8

Okay, if I understood this right, the total vapor pressure should come out to 126 kPa.

Answer 9

\(\sf \Large P_{total}=\chi_{ethanol}* P^o_{methanol}+\chi_{methanol}* P^o_{ethanol}\)


\(\sf \Large P_{total}=0.5*81 kPa +0.5*45 kPa=63~kPa\)

Answer 10

Oh.. How did you get 0.5 though? I'm confused.

Answer 11

\(\Large \chi_{ethanol}=\dfrac{n_{ethanol}}{n_{ethanol}+n_{methanol}}=\dfrac{1}{1+1}=0.5\)

Answer 12

Ah, okay. Thank you.

Answer 13

no problem, it's just the mole fraction.