On the afternoon of the first day of the professional exams in 2014, 35 questions
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At 20 degrees Celsius, the vapor pressures of methanol and ethanol are 12.5 kPa and 5.9 kPa respectively, in accordance with Raoult’s law. What is the vapor pressure of a mixture composed of 50 kilograms of methanol and 100 kilograms of ethanol at 20 degrees Celsius?
To calculate the mole fractions of methanol and ethanol in the mixture, simply multiply each by its saturated vapor pressure
The answer is: x(methanol, mol%) = 50/32/(50/32 + 100/46) = 0.418, so x(ethanol, mol%) = 0.582. The vapor pressure of the mixture at 20 degrees is P=0.418x12.5+0.582x5.9=8.66 KPa.