# How do you calculate the vapor pressure of water above a solution prepared by dissolving 39.5 g in glycerin (C_3H_8O_3) to 149 g of water at 343 K?

Feb 6, 2017

The vapour pressure of water at $343 \cdot K$ is $233.8 \cdot m m \cdot H g \ldots . .$The vapour pressure of your solution is $222.3 \cdot m m \cdot H g .$

#### Explanation:

The given vapour pressure was taken from this site here. (I am doing you a favour here by digging for data that you should have supplied yourself!)

Now it is a fact that the vapour pressure of a solution is proportional to the mole fraction of the volatile component(s) of the solution. For this problem, we assume that glycerine is involatile (this is not entirely true but for the purposes of this problem I am not going to pursue more info).

So we work out the $\text{mole fraction}$ of each component:

${\chi}_{\text{water"="Moles of water"/"Moles of water + moles of glycerol}}$

chi_"water"=[[149*g)/(18.01*g*mol^-1)]/([149*g)/(18.01*g*mol^-1)+(39.5*g)/(92.09*g*mol^-1]

${\chi}_{\text{water}} = 0.951$. Because there are only the two components in solution, ${\chi}_{\text{glycerol"=1-chi_"water}} = 1 - 0.951 = 0.0490$.

And so (finally), we can calculate the vapour pressure of the solution, which is proportional to its mole fraction of the volatile components, i.e.

$\text{Vapour pressure of solution } =$

${\chi}_{\text{water"xx"vapour pressure of pure water"=0.951xx233.8*mm*Hg=??"mm Hg}} .$

This is, as we would expect, slightly depressed from the vapour pressure of pure water.