Question

The solvent for an organic reaction is prepared by mixing 70.0mL of acetone

(C3H6O) with 69.0mL of ethyl acetate (C4H8O2). This mixture is stored at 25.0 ?C. The vapor pressure and the densities for the two pure components at 25.0 ?C are given in the following table. What is the vapor pressure of the stored mixture?

Compound	Vapor pressure (mmHg)	Density (g/mL)
acetone	230.0	0.791
ethyl acetate	95.38	0.900

Answer 1

Lets suppose that the liquid was pure acetone, if it were the vapor pressure would be 230 torr (mm of Hg). If it were pure ethyl acetate it would be 95.4 mm. If you combine the two, then the vapor pressure will be less than pure acetone and greater than pure ethyl acetate. What you've got to determine is what the mole fraction of each of the two components are and multiply that fraction times the vapor pressure of the pure liquid. For instance, if the mole fraction of each of the solvents was 50% of the total, then 50% of the total pressure would be due to one of them and 50% to the other one. In this case it would be 0.5 x 230 torr for acetone and 0.5 times 95.4 torr for ethyl acetate. That would make the total pressure 162.7 torr. Well that's a little easier because the mole fractions aren't 50:50, we must find out what they are.

for acetone: 50ml x 0.791g/ml = 39.55g and 39.55g x 1 mole/58g = 0.68moles

for ethyl acetate: 58ml x 0.9g/ml = 52.2g and 52.2g x 1mole/88g = 0.59moles

So the total number of moles is 0.68 + 0.59 = 1.27moles, and the vapor pressure for acetone:

0.68/1.27 x 230 torr = 123.2 torr and for ethyl acetate = 0.59/1.27 = 0.46 x 95.4torr = 43.9torr, and the total vapor pressure is 123.2 + 43.9 or 167 torr.