Answer:
mole fraction methanol = 0.76
mole fraction ethanol = 0.24
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
