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
just cross multiply. Since the temperature is the same.
x/100torr = .309/60torr
But if I cross multiply wouldnt it be .309/100 Torr since the pressure was 100 Torr when moles were .309. Once I solve for n, that is the amount of solute of Benzen--but I need the amount of solute molecules added. So to find that would I just find the difference in the moles Benzene to be my answer?
I think you're on the wrong track and you have added what you started with to what I gave you earlier and now you have them mixed up, I think.
See if this makes sense?
delta P = Xsolute*Posolvent
delta P = 100-60 = 40
40 = Xsolute*100
Xsolute = 40/100 = 0.40
So Xbenzene = 1.00 - 0.40 = 0.60
The question is not for the mole fraction of the solute but for the moles.
(moles benzene/total moles) = 0.60
(0.309/total moles) = 0.60
(0.309/0.60) = 0.515
So if the total moles = 0.515
and moles benzene = 0.309
then moles solute = 0.515-0.309 = 0.206
Check my method. Check my arithmetic.
You can prove that this is right.
If P normal benzene = 100 and
we add the solute to make P benene solution = 60, then Xbenzene = 0.60 which is what we obtained usint the deota P formula.
To determine the amount of solute molecules added to benzene, you need to use Raoult's Law, which relates the vapor pressure of the solvent in a solution to the mole fraction of the solvent present.
Raoult's Law states that the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by its vapor pressure in the pure state.
Let's denote the moles of solute added to benzene as "n".
According to Raoult's Law, the vapor pressure of the benzene solution containing the solute can be expressed as:
P = (0.309 mol benzene / (0.309 mol benzene + n mol solute)) * 100.0 Torr
And the vapor pressure after the solute is added is given as 60.0 Torr.
So, we can set up the following equation:
60.0 Torr = (0.309 mol benzene / (0.309 mol benzene + n mol solute)) * 100.0 Torr
Now, let's solve for "n".
60.0 Torr = (0.309 / (0.309 + n)) * 100.0 Torr
Dividing both sides of the equation by 100.0 Torr:
0.6 = 0.309 / (0.309 + n)
Cross-multiplying:
0.6 * (0.309 + n) = 0.309
0.1854 + 0.6n = 0.309
0.6n = 0.309 - 0.1854
0.6n = 0.1236
Now, divide both sides by 0.6 to solve for "n":
n = 0.1236 / 0.6
n ≈ 0.206 mol
Therefore, approximately 0.206 moles of solute molecules were added to the benzene.