Question

The vapor pressure of pure ethanol at 60 °c is 0.459 atm. raoult's law predicts that asolution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol willhave a vapor pressure of ________ atm.

Answer 1

Raoult's law is stated as the partial vapor pressure of each component in the  mixture of ideal liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. Mathematically, it is expressed as

P = Xi Pi

where P= vapor  pressure of solution
Xi = mole fraction of i component
Pi = vapor pressure of pure i component.

In present case, P1 = 0.459 atm and n1 = number of mole  of ethanol = 0.090
n2 = number of mole of  naphthalene = 0.01

∴ Mole fraction of ethanol = $\frac{n1}{\text{n1 + n2}}$ = $\frac{0.090}{\text{0.090 + 0.010}}$  = 0.9

Thus, vapor pressure of solution = 0.9 X 0.459 = 0.4131 atm. 

Answer: 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 0.4131 atm.