posted by Angelica on .
Benzene has a vapor pressure of 100.0 Torr at 26°C. A nonvolatile, nonelectrolyte compound was added to 0.309 mol benzene at 26°C and the vapor pressure of the benzene in the solution decreased to 60.0 Torr. What amount (in moles) of solute molecules were added to the benzene?
100=.309P; P= 323.6Torr
Then plugged into dP=xP --> 40=x (323.6Torr) where x is found to be .1236
I then thought the difference of .309 mol and .1236 mol would be the answer (.1843 mol) but this is incorrect. I've tried different variations of this algebra but am not sure what is wrong
I have tried cross multiplying as someone suggested, but I arrived at the same answers I had tried that were incorrect. If someone could explain how to set up the problem I'd really appreciate it =)
I think I have answered this in detail below where you and a helped tried to figures things out. If you have trouble find it repost and I'll look for it (but add something to the post so I'll know to look for it).