# Question fb71b

Feb 17, 2016

$\text{23.2 torr}$

#### Explanation:

The vapor pressure of a solution that contains a non-volatile solute depends exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.

color(blue)(P_"sol" = chi_"solvent" xx P_"solvent"^@)" ", where

${P}_{\text{sol}}$ - the vapor pressure of the solution
${\chi}_{\text{solvent}}$ - the mole fraction of the solute
${P}_{\text{solvent}}^{\circ}$ - the vapor pressure of the pure solvent

Now, glycerin is not a non-volatile solute, but you can assume it to be because it has a very, very low vapor pressure at room temperature.

https://en.wikipedia.org/wiki/Glycerol_%28data_page%29

The idea here is that you need to use the density and the volume of glycerin to determine the mass of glycerin you're adding to the solution.

You would then do the same for water. Once you know the masses of the solute and of the solvent, determine how many moles of each you have present in solution.

So, density tells you the mass of one unit of volume. In your case, the density of glycerin tells you that every $\text{1 mL}$ of glycerin has a mass of "1.26 g".

This means that you will have

50 color(red)(cancel(color(black)("mL"))) * overbrace("1.26 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density of glycerin")) = "63 g C"_3"H"_8"O"_3

present in your solution. Water's density if equal to ${\text{0.997 g mL}}^{- 1}$ at ${25}^{\circ} \text{C}$

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

which means that your solution will also contain

500 color(red)(cancel(color(black)("mL"))) * overbrace("0.997 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density of water")) = "498.5 g water"

This will be equivalent to

63color(red)(cancel(color(black)("g"))) * ("1 mole C"_3"H"_8"O"_3)/(92.09color(red)(cancel(color(black)("g")))) = "0.684 moles C"_3"H"_8"O"_3

498.5color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "27.67 moles H"_2"O"

The total number of moles present in solution will be

${n}_{\text{total}} = {n}_{{C}_{3} {H}_{8} {O}_{3}} + {n}_{{H}_{2} O}$

${n}_{\text{total" = "0.684 moles" + "27.67 moles" = "28.35 moles}}$

The mole fraction of water will be

$\textcolor{b l u e}{{\chi}_{\text{water" = n_"water"/n_"total}}}$

chi_"water" = (27.67color(red)(cancel(color(black)("moles"))))/(38.35color(red)(cancel(color(black)("moles")))) = 0.976

The vapor pressure of the solution will thus be

${P}_{\text{sol" = 0.976 * "23.8 torr" = "23.23 torr}}$

The number of sig figs you have for the two volumes would only justify using one sig fig for the answer, but I'll leave it rounded to three sig figs, just for good measure

P_"sol" = color(green)("23.2 torr")