Claisen - 22-2-2011 at 07:33
Is lowering of vapour pressure a colligative property?
I tried proving this and found it to be true.
Pa = Poaχa, where a=solvent and b=solute
Substituting χa=1-χb and some rearranging, we get
Poa-Pa=Poaχb
As Poa is constant for a given liquid at a given temperature, lowering of vapour pressure (LHS) is directly proportional to the no. of moles of solute
provided the solution is very dilute.
Is my proof correct?
According to my book, only R.L.V.P. is a colligative property and L.V.P. is not.
smuv - 22-2-2011 at 09:16
Ideally yes, see Raoult's law. I am sure you can find a proof, though I think yours is wrong.
Pa + Pb = Pab (vap pressure of mixture)
Poa*xa = Pa (vap pressure of component A in mixture)
Pob*xb = Pb (vap pressure of component b in mixture)
These are from memory, but I believe them to be correct. Keep in mind lots of mixtures deviate very strongly from Raoult's law.
gsd - 22-2-2011 at 09:16
Yes it is.
http://en.wikipedia.org/wiki/Colligative_properties
gsd
Magpie - 22-2-2011 at 09:16
According to Wiki lowering of vapor pressure is a colligative property.
What are RLVP and LVP, and for that matter LHS?
Define your acronyms! 
smuv - 22-2-2011 at 09:17
lol 3 way tie.
Claisen - 22-2-2011 at 09:28
@smuv
No its not wrong. The component 'b' is a solute (so it does not have any vapour pressure).
@magpie
RLVP= Relative lowering of vapour pressure
LVP= Lowering of vapour pressure 
If you divide Poa-Pa=Poaχb by Poa, you get RLVP on L.H.S.
Anyway my book is wrong. Lowering of vapour pressure is also a colligative property.
Thanks for helping me!!
smuv - 22-2-2011 at 09:51
ok, you never stated that B was assumed to have no vapor pressure.
Claisen - 22-2-2011 at 12:27
Alright, mea culpa.