The vapor pressures of pure propyl alcohol and isopropyl alcohol are 21.0 mmHg and 45.2 mmHg,respectively,at 25 °C.What is the composition of the vapor in equilibrium with a propyl alcohol - isopropyl alcohol solution in which the mole fraction of propyl alcohol is 0.250?

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Quizplus · Accepted Answer

The composition of the vapor in equilibrium can be calculated using Raoult's Law, which states that the partial pressure of each component in the vapor phase is proportional to its mole fraction in the liquid phase times its vapor pressure as a pure substance. For propyl alcohol (P_propyl) with a mole fraction of 0.250 and a vapor pressure of 21.0 mmHg, its contribution to the vapor phase is 0.250 * 21.0 = 5.25 mmHg. For isopropyl alcohol (P_isopropyl), with a mole fraction of 0.750 (since the total must equal 1 and 1 - 0.250 = 0.750) and a vapor pressure of 45.2 mmHg, its contribution to the vapor phase is 0.750 * 45.2 = 33.9 mmHg. The total pressure of the mixture is the sum of these, 5.25 mmHg + 33.9 mmHg = 39.15 mmHg. The mole fraction of each component in the vapor phase is then the partial pressure of that component divided by the total pressure. For isopropyl alcohol, it's 33.9 / 39.15 = 0.866, and for propyl alcohol, it's 5.25 / 39.15 = 0.134.

The Vapor Pressures of Pure Propyl Alcohol and Isopropyl Alcohol