I am faced with the question:

 

The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol will have a vapor pressure of __________ atm.

 

 

 

If I convert the milimolar to molar I get:

 

0.01 mol Nap.

0.09 mol Eth.

and 0.10 mol total

 

Thus the mole fractions should be:

 

X_Nap. = 0.01 / 0.10 = 0.10

 

X_ Eth. = 0.09 / 0.10 = 0.90

 

 

Which according to Raoult's Law

 

Vapor Pressure of Soln. = Sum of Vapor Partial pressures

 

0.459 atm * X_Nap. -{0.10} or 0.459 * 0.10 = 0.0459 Partial Pressure Nap.

0.459 atm * X_Eth. -{0.90} or 0.459 * 0.90 = 0.4131 Partial Pressure Eth.

 

I sum the two and get: 0.459. Which is consequently the same as the pure vapor pressure. I am assuming that this is related to the fact that naphthalene is supposed to be non-volatile.

 

 

My problem is that this matches none of the answer choices. Any help appreciated. My exam is in 20 mins.

 

Answer Choices:

 

0.498 0.0918 0.367 0.413 0.790

 

 

*Chem software says that 0.413 is the answer. Not sure how.

Naphthalene stinks

(I know it's too late for the exam)

Hi,

they said in the question that: naphthalene is nonvolatile (the nonvolatile compound don't have a vapor pressure..)...so in ur exercise solution u shouldn't take consideration of naphta's pressure.....:
0.459 atm * X_Nap. -{0.10} or 0.459 * 0 = 0 Partial Pressure Nap.
0.459 atm * X_Eth. -{0.90} or 0.459 * 0.90 = 0.4131 Partial Pressure Eth. is the vapor pressure of solution.