So what you're saying is that if I want to find the osmotic pressure/bp elevation/fp depression/vapor pressure depression of an aqueous solution of either CH3OH or HCl, I have to use to use i=2??? (so for example, to find vapor pressure depression, ?P = 2*Xsolute * Psolvent)
If yes, I am confused on a problem I have. It asks me to arrange the following aqueous solutions in order of increasing vapor pressure:
0.35 m ethanol (note: molality)
0.50 m sugar
0.20 m KBr
0.20 m Na2SO4
Originally, I had the following answer: Na2SO4 < sugar < KBr < ethanol, which I do not entirely understand if what you say is true. Assuming the kg of solvent is the same for everything (I assumed 1-kg), I will have the following MOLAR quantities:
0.35 mol ethanol --> 0.35 mol CH3CH2O- + 0.35 mol H+ (total: 0.70 mol)
0.50 mol sugar ---> I don't know how much to dissociate since I don't know the sugar (presumable total: 0.5 mol)
0.20 mol KBr ---> 0.20 mol K+ + 0.20 mol Br- (total: 0.40 mol)
0.20 mol Na2SO4 ---> 0.40 mol Na+ + 0.20 mol SO4 2- (total: 0.60 mol)
Now, since the formula for finding the change in vapor pressure is ?P = i*Xsolute * Psolvent, I can assume that the larger the molar concentration of solutes, the larger the vapor pressure lowering will be. With that in mind, the order of deviation of vapor pressure would be as follows: KBr < sugar < Na2SO4 < ethanol.
Since the question asks us to show the INCREASING vapor pressure, and vapor pressure DECREASES with increasing molar concentration, the answer should be the reverse: ethanol <Na2SO4 < sugar < KBr, which does NOT agree with the answer I had written last year (Na2SO4 < sugar < KBr < ethanol). NOTE: The answer I wrote down could have been wrong.....I don't recall if the teacher posted solutions to the problem). Thanks so much.
Ryan