Answer:
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
