Question

A solution is made by mixing 33.0 ml of ethanol, C2H6O and 67.0 ml of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 degree C? Ethanol= .789 g/ml 43.9 torr. water= .998 g/ml 17.5 torr

Answer 1

Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms 

Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles 

Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles 

total moles = .57 + 3.72 = 4.29 moles 

Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13

Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87

Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr 

partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr 

Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr