Chemistry / Unit 8 Chem 1558 worked example, 60 to 90 seconds
Calculating a Solution's Vapor Pressure Under Raoult's Law
The idea
Calculating a solution's vapor pressure under Raoult's law means finding the solvent's mole fraction from the moles of solvent and solute present, then multiplying that mole fraction by the pure solvent's known vapor pressure.
In practice
A solution of 92.1 g glycerin (nonvolatile, molar mass 92.1 g/mol) and 184.4 g ethanol (molar mass 46.1 g/mol) at 40 °C contains n(glycerin) = 92.1 g ÷ 92.1 g/mol = 1.00 mol and n(ethanol) = 184.4 g ÷ 46.1 g/mol = 4.00 mol, giving ethanol a mole fraction X(ethanol) = 4.00 mol ÷ (4.00 mol + 1.00 mol) = 0.800, a pure number with the moles canceling. Substituting into P(solution) = X(ethanol) × P°(ethanol), with pure ethanol's vapor pressure at 40 °C, P°(ethanol) = 0.178 atm, gives P(solution) = (0.800)(0.178 atm) = about 0.142 atm — a pressure lower than pure ethanol's, in atmospheres as expected, and within the range Raoult's law predicts for a mole fraction near 0.8.
Where people slip
Using the solute's mole fraction instead of the solvent's in Raoult's law inverts which fraction multiplies the pure vapor pressure and produces a vapor pressure that is far too low.
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