[ladymonsterface]

Calculate the vapour pressure of a 5% by mass benzoic acid [C7H6O2 (aq)] in ethanol solution at 35˚C. The vapour pressure of pure ethanol at this temperature is 13.40 kPa.

so what I did was

first, I pretend that it was for 1 kg and found that in on kg there would be .409 moles of Benzoic acid and 20.6 moles of ethanol (which seemed like alot)
 
then
Xsolvant=mols of solvant/total moles
Xsolvant=20.6/20.682
Xsolvant=.996



Then
P=Xsolvant*Psolvant
P=.996*13.40
P=13.3


it just doesnt seem right, what did I do wrong?

[Borek]

How many total moles?

[ladymonsterface]

How many total moles?

thats a slightly embarrassing mistake on my part ok, so if total mass is 21.009

then its 20.6/21.009
and that comes to .98

Then
P=Xsolvant*Psolvant
P=.98*13.40
P=13.14


so does it look right now?

[Borek]

Numbers look reasonable - at least they are in the correct ballpark.

[ladymonsterface]

Numbers look reasonable - at least they are in the correct ballpark.

Well check over them again anyway. Thanks alot