Use Raoult’s law to answer this question, with A = H 2 O and B = ethyl alcohol.
Since we are told that XB = 0.2, then XA = 1 – XB = 1 – 0.2 = 0.8.
Thus the vapor pressure due to water is 0.016 atm and the vapor pressure due to the alcohol is
0.10 atm.