Vapor pressure and Raoult's Law?

Question

The vapor pressure of pure ethanol at 60 degrees 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.

Answer 1

Since napthalene is a nonvolatile component hense total vapour pressure of solution is due to vapour pressure of ethanol.

vapour pressure of solution =vapour pressure of ethanol in solution

             =mole fraction of ethanol in solution into vapour pressure of ethanol in pure state

Total no. of milli moles of solution =10 + 90 =100

Mole fraction of ethanol =90\100 = 0.90

VAPOUR PRESSURE OF SOLUTION =0.90 INTO 0.459 = 0.4131 atm.

Answer 2

Total millimoles = 10 + 90 = 100.0 => 0.100 moles

mole fraction ethanol = 0.0900 mol / 0.100 =0.90

p = p° X

p = 0.459 x 0.90 = 0.413 atm