Raoult’s Law Question 5

Aspirin is very soluble in ether. If 0.372g of aspirin are dissolved in 2.97g of diethyl ether, the vapor pressure of the ether lowers from 0.631 to 0.601 atm at 23^oC. What is the molecular weight of the aspirin?

In order to be able to find the vapour pressure of the solution the mole fraction of both the ether and the aspirin.

1 mole ether = 60g and 1 mole aspirin = ?g

Therefore the moles of each component are:

$moles\;deiethyl\: ether = \frac{2.97}{60}=0.0495\;moles$

$moles\;aspirin = (\frac{0.372}{x})moles$

This means that the mole fraction of ether can be found:

$mf_{(ether)}=\frac{moles_{(ether)}}{total\;moles}=\frac{\frac{0.372}{x}}{\frac{0.372}{x}+0.0495}$

This would make the vapour pressure of the solution as: $VP_{(solution)}=mf_{(ether)}*VP{(pure\;ether)}$

$0.601=\frac{ 0.0495}{\frac{0.372}{x}+0.0495}*0.631$

Where x would be 150.1g