Raoult’s law Question 2

What is the vapour pressure at 298K of a 20% solution of KNO₃ by mass. Vapour pressure of water at 298K is 3.2kPa.

In order to be able to find the vapour pressure of the solution the mole fraction of both the water and the KNO₃.

1 mole water = 18g and 1 mole KNO₃ = 101g

Lets assume that there is 100g of solution, therefore 20g is equal to the potassium nitrate and 80g is equal to the water.

$moles\;water = \frac{80}{18}=4.44\;moles$

$moles\;KNO_3 = \frac{20}{101}=0.198\;moles$

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

$mf_{(water)}=\frac{moles_{(water)}}{total\;moles}=\frac{4.44}{4.44+0.198}=0.957$

This would make the vapour pressure of the solution as: $VP_{(solution)}=mf_{(water)}*VP{(pure\;water)}=00.957*3.52=3.37kPa$