At 63.5 C the vapor pressure of H2O is 175 torr and that of ethanol is 400 torr. A solution is made by adding equal masses of H2O and C2H5OH. What is the mole fraction of ethanol in this solution? Sdduming ideal-solution behavior, what is the vapor pressure of the solution at 63.5 C? What is the mole fraction of ethanol in the vapor above the solution?

Moel fraction of ethanol in the solution = 0.28

Vapor pressure of the solution = 238 torr

Mole fraction of ethanol in the vapor = 0.47

Explanation:

Let's use 100 g of each substance as a calculus basis. Knowing that the molar mass of water is 18 g/mol and the molar mass of ethanol is 46 g/mol, the number of moles (n = mass/molar mass) of each one is:

nw = 100/18 = 5.56 mol

ne = 100/46 = 2.17 mol

The total number of moles is 7.73 mol, so the mole fraction of ethanol is

2.17/7.73 = 0.28

The mole fraction of water must be 0.72, so if we assume that the solution is ideal, by the Raoult's law, the solution vapor pressure is the sum of the multiplication of the mole fraction by the vapor pressure of each substance, thus:

P = 0.28*400 + 0.72*175

P = 238 torr

The partial pressure of each substance can be found by the multiplication of the molar fraction by the vapor pressure, thus:

Pw = 0.72*175 = 126 torr

Pe = 0.28*400 = 112 torr

To know the number of moles that is vaporized above the solution, we may use the ideal gas law:

PV = nRT

P/n = RT/V

R is the gas constant, T is the temperature and V is the volume, so they are the same for both water and ethanol, thus

Pw/nw = Pe/ne

126/nw = 112/ne

ne = (112/126) * nw

ne = 0.89nw

So, the mole fraction of ethanol is:

ne / (ne + nw) = 0.89nw / (0.89nw + nw) = 0.89/1.89 = 0.47