When two volatile liquids mix to form an ideal solution, both components appear

Question

When two volatile liquids mix to form an ideal solution, both components appear in the vapor phase over the solution. Each liquid can be viewed as either the solute or the solvent, and each lowers the vapor pressure of the other. When the solution behaves ideally, Raoult's law can be used to calculate the partial pressure of each of the components over the solution. This defines the composition of the gas phase in equilibrium with the solution. At 313 K the vapor pressure of pure methanol (CH_3 OH) is 0.345 atm and the vapor pressure of pure ethanol (C_2 H_5 OH) is 0.169 atm. Equal chemical amounts (equal numbers of moles) of methanol and ethanol are mixed and form an ideal solution. Compute the mole fraction of ethanol in the vapor in equilibrium with this solution.

Explanation / Answer

Assume this is ideal so

raoult law:

x1*P°1 = y1*PT

x2*P°2 = y2*PT

x1,x2 liquid fractions

y1,y2 vapor fractions

P°1, P°2, vapor pressures

PT = Total pressure

Assume total P = 1 atm

P°1 = 0.345 atm

P°2 = 0.169 atm

x1 = x2 = 0.5

y1+y2 = 1

y1 = 1-y2

note that 2 is ethanol!

Pt = 1 atm

so

0.5*0.345 = (1-y2)*1

0.5*0.169 = y2*1

0.1725 = 1-y2

0.0845 = y2

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