Please explain! The liquid boiling in a simple distillation flask consists of 65
ID: 888965 • Letter: P
Question
Please explain! The liquid boiling in a simple distillation flask consists of 65 mol% toluene and 35 mol% hexane. At 75 degree C, a small amount of distillate is collected. The standard vapor pressure of toluene is 136 mmHg and the standard vapor pressure of benzene is 925 mmHg at 75 degree C. Calculate the percentage of each compound in the distillate collected at this temperature, showing all your calculations. Why is it important to allow the hot vapor in a microscale fractional distillation to climb up the column slowly?Explanation / Answer
We know that the vapour pressure = mole fraction X totoal pressure
the vapor pressure of toluene = 0.65 X 136 mmHg = 88 mmHg
vapor pressure of benzene = 0.35 X 925 mmHg = 324 mmHg
So vapour composition will be
% of toluene = 88 / (88 + 324 X 100% = 21% toluene
% of benzene = 324 / (88 + 324) * 100% = 79% benzene.
2) We allow the vapours to climb up slowly so that the vapours of the less volatile components can condense. It will not separate the different liquids if the process is done rapidly.