Solutions Question 11

At 298K, methanol has a vapor pressure of 16.9kPa and ethanol has a vapor pressure of 7.83kPa. If the vapor pressure of methanol above the mixture is 0.35, calculate the mole fraction of ethanol in the mixture.

$mole\; fraction\;methanol=\frac{moles_{methanol}}{total\;moles}=x$

$mole\; fraction\;ethanol=\frac{moles_{(ethanol)}}{total\;moles}=y$

since the total amount of the mole fractions is equal to 1, then:

$y=1-x=mole\;fraction\;ethanol$

This would make the partial pressure equal to:

$pressure\;methanol=x*16.9kPa$

$pressure\;ethanol=(1-x)*7.83kPa$

since the pressure of each component = mole fraction * Vapor Pressure of component.

In the vapor phase, the mole fraction is equivalent to the pressure divided by total pressure.

Therefore the mole fraction of methanol in the vapor phase would be:

$mole\;fractio\;methanol_{(vapor))}=\frac{x*16.9}{(x*16.9)+((1-x)*7.83)}=0.35$

When making x subject of the formula, it would become 0.200.

This would make the mole fraction of ethanol as 1-x, which would be equal to 0.800.