The vapor pressure of methanol is 03 atm at 367 degree C Liq

The vapor pressure of methanol is 0.3 atm at 36.7 degree C. Liquid methanol is kept in a sealed flask @ 36.7 degree C; the gas above the liquid contains air and methanol vapor at a pressure of 1.28 atm. What is the partial pressure of methanol in the gas, the partial pressure of N_2, the mole fraction of acetone in the gas? How would you determine the boiling point of acetone, assuming a constant P_T of 1.28 atm?

Solution

a) Partial pressure of acetone at 36.7 C   = 0.49 atm

Partial presure of methanol = 0.3 atm

Therefore partial pressure of air = 1.28 - 0.3 - 0.49 = 0.49 atm

b) Partial pressure of N2 = 0.79 * partial pressure of air   = 0.3871 atm

c) mole of fraction of acetone = 0.49 / 1.28   = 0.3828

d) Using the Antoine Equation we can obtain the boiling point of acetone at partial vapour pressure of 0.49 atm.