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?
My work:
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).
To solve this question, you need to use Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. The equation representing Raoult's law in this context can be written as:
P = X_solvent * P_solvent,
where P is the vapor pressure of the solution, X_solvent is the mole fraction of the solvent, and P_solvent is the vapor pressure of the pure solvent.
Given that the vapor pressure of pure benzene (P_solvent) is 100.0 Torr and the vapor pressure of the benzene in the solution (P) is 60.0 Torr, we can substitute these values into the Raoult's law equation as follows:
60.0 = X_solvent * 100.0.
Rearranging the equation to solve for the mole fraction of the solvent (X_solvent), we get:
X_solvent = 60.0 / 100.0 = 0.6.
Now, the mole fraction can be defined as the moles of solvent (benzene) divided by the total moles of solute and solvent.
X_solvent = moles of solvent / (moles of solute + moles of solvent).
Given that the moles of benzene are 0.309, we can solve for the moles of solute:
0.6 = 0.309 / (moles of solute + 0.309).
Rearranging the equation to isolate the moles of solute, we have:
0.6 * (moles of solute + 0.309) = 0.309.
0.6 * moles of solute + 0.6 * 0.309 = 0.309.
0.6 * moles of solute = 0.309 - 0.6 * 0.309.
0.6 * moles of solute = 0.309 - 0.1854.
0.6 * moles of solute = 0.1236.
Simplifying further:
moles of solute = 0.1236 / 0.6 ≈ 0.206 mol.
Therefore, approximately 0.206 moles of solute molecules were added to the benzene.