Question

A solution is made by mixing 49.0 mL of ethanol, C2H6O, and 51.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 °C?

Answer 1

volume of ethanol = 49.0 mL

mass of ethanol = 49 x 0.789 = 38.661 g

moles of ethanol = 38.661 / 46 = 0.84 mol

mass of water = 51 x 0.998 = 50.898 g

moles of water = 50.898 / 18.02 = 2.825 mol

total moles = 3.665 mol

mole fraction of ethanol = 0.84 / 3.665 = 0.229

mole fraction of water = 0.771

vapor pressure of solution = Xethanol x Pethanol + Xwater + Pwater

                                          = 0.229 x 43.9 + 0.771 x 17.5

vapor pressure = 23.5 torr