[munoz_m1]

The vapor pressure of a solution containing 53.6g glycerin (C3H8O3) in 133.7g ethanol (C2H5OH) is 113 torr at 40 degrees celsius. Calculate the vapor pressure of pure ethanol at 40 degrees celsius assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.


Any tips, hints, or advice would be greatly appreciated!
Thank you!

[Lysandor]

Ok.  In order to solve this problem we need to understand a few things.  First of all we need to understand the formula for pressures of solution, Which is:

Psol'n = P(pure substance) * Mole fraction of ethanol

We also need to understand what mole fraction is defined as, which is:

             molesA
X =        -------
       molesA + molesB

Now that we have our two base formulas, we need to know how many moles of each substance we have.  This is easy using the molar mass of each substance.

The molar mass of ethanol is 46.08 g/mol

The molar mass of glycerin is 92.11 g/mol


Given that we know we have 133.7g of glycerin and 53.6g of ethanol, we simply divide out our mass by our molar mass:

                      133.7g              1 mol
moles ethanol = ----------  *   -----------  =  2.901 moles
                                             46.08g

                          53.6g                1 mol
moles glycerin = -----------  *   -------------   =  .582 moles
                                                 92.11g 



Now that we know our moles we can figure out our mole fraction:

                      2.901 moles
Xethanol = ---------------                   =      .834 is our mole fraction of ethanol
                 2.901 moles + 0.582 moles



Now that we have our mole fraction we simply plug it into our formula from the beginning:

                     
113 torr = P(pure ethanol) * 0.834

Divide out and you end up with 135.5 torr.

This checks because we know that due to intermolecular forces the vapor pressure of a solution should be lower than the vapor pressure of pure substance.  113 torr is lower than 135.5 torr..so our answer makes sense.