Posted by Lauren on Saturday, February 18, 2012 at 10:24pm.
You have half the answer. Just finish it. You didn't provide data (densities, vapor pressure) etc so I can't use numbers. I assume the part you have is
pH2O = XH2O*P^{o}_{normal v.p. H2O}
You need to also do
pEtOH = XEtOH*P^{o}_{normal v.p. EtOH}
Then Ptotal = pH2O + pEtOH
Sorry here are the data:
Values at 20 degrees Celcius
ethanol : density (g/mL)= 0.789/ vapor pressure (torr)= 43.9
water: density (g/mL) =0.998/ vapor pressure (torr)= 17.5
I' m still confused on what you are telling me to do :( Can u please clarify...
Sorry here are the data:
Values at 20 degrees Celcius
ethanol : density (g/mL)= 0.789/ vapor pressure (torr)= 43.9
water: density (g/mL) =0.998/ vapor pressure (torr)= 17.5
I' m still confused on what you are telling me to do :( Can u please clarify...
we are the same person LOL
***We are siblings haha
The total vapor pressure of the solution is the partial pressure of the H2O + the partial pressure of the EtOH. I calculated the pH2O and obtained 15.45 torr just as you did. Therefore, you must have used the right mole fraction of H2O. I calculated that to be 0.8828 (I know that's too many places but I can round later). Look in your notes. If that is true than 1.000-0.8828 = 0.11715 for mole fraction EtOH or you can calculate the mole fraction of EtOH from scratch.
Then pEtOH = XEtOH x 43.9 = approximately 5 torr.
So the total pressure will be 15 from the H2O + 5 from the EtOH (of course you know these are not ideal solutions) = about 20 torr.
You need to go back through and come up with better numbers than those I've used. Also remember that you can't have more than 3 s.f. so don't report a 15.45 (or course that will be changed now) since that has 4 s.f.
Thanks DrBob222! You are the best! :]
I had this exact problem for my homework but different numbers. I got it right thanks
Thanks sooo much DrBob222! And I'm glad to help kylie :D