Chemistry · General chemistry II · Worked example
Vapor pressure of a glucose solution
50.0 g of glucose, C₆H₁₂O₆, is dissolved in 250.0 g of water at 25 °C, where pure water’s vapor pressure is 23.8 torr. Find the solution’s vapor pressure and the lowering.
Moles of each component
Glucose is 180.16 g/mol and water is 18.015 g/mol. Glucose is nonvolatile and does not ionize.
Mole fraction of the solvent
Mole fraction counts particles, so use moles, not grams.
Apply Raoult’s law
Multiply by the pure-water vapor pressure at the same temperature. The lowering is the difference, or equivalently χ_glucose × P°.
Result
The vapor pressure is 23.3 torr, lowered by about 0.47 torr.
Your turn
Would 0.2775 mol of NaCl in the same water lower the vapor pressure more or less than the glucose, and by about how much?
Show the answer and explanation
More: about 0.92 torr, nearly twice as much.
Ideally NaCl gives 2 × 0.2775 = 0.555 mol of ions. χ_solute = 0.555 ÷ (13.88 + 0.555) = 0.0385, so ΔP = 0.0385 × 23.8 ≈ 0.92 torr.
Keep exploring
Check the answer in the Colligative studio’s Raoult mode, then double the glucose.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
- Tro, Chemistry: A Molecular Approach, 4th ed., §13.5 Expressing Solution Concentration, pp. 585–592 (molality, p. 588)
- Tro, Chemistry: A Molecular Approach, 4th ed., §13.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure, pp. 593–605 (volatile solutes, p. 597; osmotic pressure, pp. 603–604)
- Tro, Chemistry: A Molecular Approach, 4th ed., §13.7 Colligative Properties of Strong Electrolyte Solutions, pp. 605–607
- OpenStax Chemistry 2e — Colligative properties