Question

A liquid mixture of 0.400 mole fraction ethanol and 0.600 methanol was placed in an evacuated (i.e., no air) bottle and after many days is now in equilibrium with its vapor. Assuming Raoult's Law applies (actually, both activity coefficients are within 0.02 of unity), what is the mole fraction of each compound in the vapor at 25C? at 40C?

Answer 1

mole fraction methanol = 0.76

mole fraction ethanol = 0.24

Step-by-step explanation:

Raoult´s law gives us the partial vapor pressure of a component in solution as the product of the mole fraction of the component and the value of its pure pressure:

PA = X(A) x Pº(A)

where PA is the partial vapor pressure of component A, X(A) is the mole fraction of A, and Pº(A) its pure vapor pressure.

From reference literature the pure pressures of methanol, and ethanol are at 25 ºC :

PºCH₃OH = 16.96 kPa

PºC₂H₅OH = 7.87 kPa

Given that we already have the mole fractions, we can calculate the partial vapor pressures as follows:

PCH₃OH = 0.600 x 16.96 kPa = 10.18 kPa

PC₂H₅OH = 0.400 x 7.87 kPa = 3.15 kPa

Now the total pressure in the gas phase is:

Ptotal = PCH₃OH + PC₂H₅OH = 10.18 kPa + 3.15 kPa = 13.33 kPa

and the mole fractions in the vapor will be given by:

X CH₃OH = PCH₃OH / Ptotal = 10.18 kPa/ 13.33 kPa = 0.76

X C₂H₅OH = 1 - 0.76 = 0.24