posted by Mima on .
Neat pure ethanol C2H5OH has an enthalpy of vaporization of 39.3 Kj/mol, and a vapor pressure of 0.308 atm at 50 C. What is the boiling point, at atmospheric pressure of 1 atm, of a solution consisting of 35 G of phenol (a non-electrolyte hydrocarbon with a molecular formula C6H6O) in 250 g of ethanol?
Do you have an answer? Is there some reason you can't use the colligative property of solutions (at least I don't see Kb listed nor a normal boiling point for ethanol so they may expect you to go another route.) Here is the best I've been able to come up with but I've not done one like this before so take the information with a grain of salt.
mols phenol = grams/molar mass.
mols ethanol = g/molar mass.
Xethanol = ?
pethanol at normal boiling point = Xethanol*760 and call that p1
Then substitute into the Clausius-Clapeyron equation. I used p2 = 760 and p1 from above.
T1 = 78.4 + 273.15 (The 78.4 is not listed in the problem so this may be off limits.)
Solve for T2. I get very close to 80 C. Interestingly, if I look up the Kb for elevation of boiling point for ethanol and go through that calculation, I also end up with a new boiling point of very close to 80 C.