Raoult’s law Question 1

The vapour pressures of ethanol and propanol are 17000Pa and 15450Pa respectively. An ideal solution is formed by mixing 24g of ethanol with 35g of propanol. Calculate the vapour pressure of each component and thus the mole fraction of the two components in the vapour phase.

In order to be able to find the vapour pressure for each component it is important to find the mole fraction of both the ethanol and the propanol.

1 mole ethanol = 46g and 1 mole propanol = 60g

moles\;ethanol = \frac{24}{46}=0.522\;moles
moles\;propanol = \frac{35}{60}=0.583\;moles

This means that the mole fraction of each of the two components can be found:

mf_{(ethanol)}=\frac{moles_{(ethanol)}}{total\;moles}=\frac{0.522}{0.522+0.583}=0.472
mf_{(propanol)}=\frac{moles_{(propanol)}}{total\;moles}=\frac{0.583}{0.522+0.583}=0.528

This would make the vapour pressure of each component as:

VP_{(ethanol)}=mf_{(ethanol)}*VP{(pure\;ethanol)}=0.472*17000=8024Pa
VP_{(propanol)}=mf_{(propanol)}*VP{(pure\;propanol)}=0.528*15450=8150Pa

This means that the mole fraction in the vapour phase can be found as the pressure of ethanol divided by the total pressure:

mf_{(ethanol\;vapour)}=\frac{VP_{(ethanol)}}{total\;pressure}=\frac{8024}{8024+8150}=0.496

and that for propanol vapour can be found either as:

mf_{(propanol\;vapour)}=\frac{VP_{(propanol)}}{total\;pressure}=\frac{8150}{8024+8150}=0.504

or

mf_{(propanol\;vapour)}=1-mf_{(ethanol\;vapour)}=1-0.496=0.504