Raoult’s law Question 3

At a given temperature, the vapour pressure of heptane is 1500 Pa and that of octane is 860 Pa. Assuming ideal behaviour, and that the vapour contains 3 times as many moles of octane as heptane, what is the mole fraction of the heptane in the liquid phase?

Lets assume that the mole fraction of heptane in the solution is equal to x and the mole fraction of octane in the solution is equal to y.

Since the mole fraction is equal to 1, then x + y = 1, which can used to say that y = 1 – x. this can be substituted for the mole fraction of octane.

This would make the vapour pressure of each component as:

$VP_{(heptane)}=mf_{(heptane)}*VP{(pure\;heptane)}=x*1500=1500x$

$VP_{(octane)}=mf_{(octane)}*VP{(pure\;octane)}=(1-x)*860=860(1-x)$

Since the number of moles have been given for the vapour phase, then the mole fractions for the vapour phase can be found:

$mf_{(heptane\;vapour)}=\frac{moles_{(heptane)}}{total\;moles}=\frac{1}{3+1}=0.25$

$mf_{(octane\;vapour)}=\frac{moles_{(octane)}}{total\;moles}=\frac{3}{3+1}=0.75$

This mole fraction in the vapour phase can also be found using the vapour pressure as calculated previously, with the mole fraction of ethanol vapour being equal to:

$mf_{(heptane\;vapour)}=\frac{VP_{(heptane)}}{total\;pressure}=\frac{1500x}{1500x+860(1-x)}$

Since the vapour pressure of heptane is equal to 0.25, then x can be found by using subject of the formula.

$mf_{(heptane\;vapour)}=\frac{1500x}{1500x+860(1-x)}=0.25$

Where x is found to be 0.160.