Chemistry
posted by Lauren .
A solution is made by mixing 30.0 mL of ethanol, C2H6O, and 70.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 °C?
I got an answer of 15.45 torr and it marked me incorrect and said "Your answer is the the partial vapor pressure contributed by the water component of the solution. Since ethanol is a volatile liquid, it will also contribute to the vapor pressure."
HELP PLEASE!

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 1.0000.8828 = 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