Gases Liquids and Solids 4.27.20

Question

A solution of ethanol CH3CH2OH and water that is 10.% ethanol by mass is boiling at 98.7°C. The vapor is collected and cooled until it condenses to form a new solution.

Calculate the percent by mass of ethanol in the new solution. Here's some data you may need:

normal boiling point density vapor pressure at 98.7°C

ethanol 78.°C 0.79gmL 1563.torr

water 100.°C 1.00gmL 725.torr

Answer 1

In 100 g solution you have 10 g ethanol (etoh) and 90 g H2O

moles C2H5OH = 10 g x 1 mol/46 g = 0.217 moles C2H5OH

moles H2O = 90 g x 1 mol/18 g = 5.0 moles H2O

mole fraction X etoh = 0.217 / (0.217+5.0) = 0.0416

X_H2O = 5/5.217 = 0.958

P_soln = Pº_etoh x X_etoh + Pº_H2O x X_H2O = (1563 torr x 0.0416) + (725 torr x 0.958) = 65.0 torr + 695 torr

P_soln = 760 torr (1 atm)

P_etoh = 65 torr

P_H2O = 695 torr

X_etoh in vapor = Petoh/Psoln = 65 torr/760 torr = 0.0855

X_H2O in vapor = PH2O/Psoln = 695 torr/760 torr = 0.914

mass C2H5OH = 0.0855 mol x 46 g/mol = 3.93 g C2H5OH

mass H2O = 0.914 mol x 18 g/mol = 16.45 g H2O

mass % etoh = 3.93 g /3.93 g + 16.45 g = 19.3%