Solutions · Lesson 6 of 11
Vapour-Pressure Lowering and Molar Mass
“Derive relative vapour-pressure lowering and use it to determine solute amounts without confusing exact and dilute forms.”
• Explain why a non-volatile solute lowers solvent vapour pressure. • Derive exact and dilute relative-lowering equations for an ideal solution. • Determine solute molar mass from measured pressure changes. • Recognise when finite-concentration corrections are significant.
Adding particles changes solvent escaping tendency
A non-volatile solute contributes negligible vapour at the measurement temperature. Dissolving it in a volatile solvent changes the solvent’s molecular environment and decreases its escaping tendency in an ideal solution. The equilibrium vapour pressure then comes almost entirely from the solvent and is lower than that of the pure solvent at the same temperature. This is not a permanent cap that seals the surface: solvent molecules still evaporate and condense.
A surface picture can show fewer solvent molecules among all particles near an interface, but the equilibrium effect belongs to the whole solution, not simply to mechanical obstruction of the surface. The reliable quantitative statement is the ideal-solution relation p₁=x₁p₁°. More generally, real solutions require an effective solvent activity rather than treating every mixture as ideal.
Derive the relative lowering before making an approximation
Substitute p₁=x₁p₁° into the pressure difference: Δp₁=p₁°−x₁p₁°=p₁°(1−x₁). In a binary mixture, 1−x₁=x₂. Dividing both sides by p₁° removes pressure units and gives the solute mole fraction. This equality is exact within the ideal, non-volatile, non-associating binary model; it does not yet require a dilute approximation.
For a dilute solution, n₂ is much smaller than n₁, so n₁+n₂ can be approximated by n₁. This simplifies calculations but slightly overestimates x₂ for a given n₂. A relative lowering of 20% is not dilute enough to discard n₂: its exact amount ratio n₂/n₁ is 0.20/0.80=0.25, rather than 0.20.
To derive the exact mass expression, first solve r(n₁+n₂)=n₂, giving rn₁=n₂(1−r). Substitute n₁=w₁/M₁ and n₂=w₂/M₂. Rearranging isolates M₂. As r tends to zero, the factor 1−r tends to one, so the exact expression approaches the dilute formula. If several non-volatile solutes are present in an ideal mixture, the lowering depends on their combined mole fraction, not the name of any one solute.
Problem
Pure benzene has pressure 0.850 bar. Adding 0.500 g non-electrolyte to 39.0 g benzene gives 0.845 bar. Use benzene molar mass 78 g mol⁻¹. Find solute molar mass.
- 1.Relative lowering r=(0.850−0.845)/0.850=1/170=0.00588235. Benzene amount n₁=39/78=0.500 mol.
- 2.The dilute formula gives M₂=0.500×78/(39×r)=170 g mol⁻¹.
- 3.Keeping the finite-composition factor gives M₂=170×(1−1/170)=169 g mol⁻¹.
- 4.The difference is about 0.6%, small for the pressure precision. State which approximation is used rather than treating both values as contradictory results.
Problem
What mass of a non-volatile solute of molar mass 40 g mol⁻¹ reduces the vapour pressure of 114 g octane to 80% of its pure value? Use octane 114 g mol⁻¹ and ideal behaviour.
- 1.Octane amount n₁=114/114=1 mol. Since p/p°=x₁=0.80, solute fraction r=0.20.
- 2.Use the exact ratio n₂=rn₁/(1−r)=0.20×1/0.80=0.25 mol.
- 3.Solute mass=0.25×40=10.0 g. The dilute approximation would give 8.0 g and is inadequate here.
- 4.This example also illustrates that the ideality assumption and the dilution assumption are separate questions.
The common thread of colligative properties
Relative vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure all reflect the concentration of dissolved particles in an ideal dilute model. Equal amounts of different non-reacting solutes in the same solvent can therefore produce nearly equal effects. The solvent identity and temperature still matter. Later we will correct particle counts for association and dissociation, and recognise non-ideal interactions.
In r=(p°−p)/p°, the denominator is pure-solvent pressure. Do not substitute atmospheric pressure except when it is known to equal p° at the stated temperature. A normal boiling point corresponds to 1 atm, approximately 1.01325 bar, not 1.000 bar.
Quick Quiz
Quiz
For an ideal binary solution with a non-volatile solute, relative lowering equals:
Explanation: Subtracting p=x₁p° from p° gives (p°−p)/p°=1−x₁=x₂.
If p/p°=0.80, what is n₂/n₁ in the exact ideal model?
Explanation: x₁=n₁/(n₁+n₂)=0.80, so n₂/n₁=(1−0.80)/0.80=0.25.
Which condition allows r≈n₂/n₁?
Explanation: Only when n₂ is negligible compared with n₁ can n₁+n₂ be approximated by n₁.
Which statement is correct for a non-volatile solute at the stated temperature?
Explanation: Non-volatile means its vapour pressure contribution is negligible under the conditions; solvent exchange continues.
What pressure appears in the denominator of relative vapour-pressure lowering?
Explanation: Relative lowering is defined as Δp/p°, so the reference is pure solvent at the same temperature.
Practice Problems
- At 298 K, water pressure is 23.8 mm Hg. Dissolve 50 g urea in 850 g water. Use molar masses 60 and 18 g mol⁻¹. Find solution pressure and relative lowering under ideal behaviour.
- Pure water pressure at 300 K is 12.3 kPa. Find the pressure of an ideal 1 molal solution of a non-volatile non-electrolyte.
- A solution has 30 g solute in 90 g water and pressure 2.8 kPa. Adding 18 g water gives pressure 2.9 kPa at the same temperature. Find solute molar mass and pure-water pressure using the exact ideal model.
- At 293 K, p°(water)=17.535 mm Hg. Find p after dissolving 25 g glucose in 450 g water. Use glucose 180 and water 18 g mol⁻¹.
- A 2% by mass aqueous solution has pressure 1.004 bar at water’s normal boiling point. Estimate solute molar mass using p°=1.01325 bar and water molar mass 18 g mol⁻¹.
- Two non-reacting, non-volatile solutes have combined amount 0.2 mol in 9.8 mol solvent. What is relative lowering in the ideal model?
n(urea)=50/60=0.833333 mol; n(water)=850/18=47.222222 mol. x(urea)=0.833333/48.055556=0.01734104. Thus r=0.01734104 and p=23.8(1−r)=23.387283 mm Hg≈23.4 mm Hg. Use the pure pressure, not solution pressure, as the relative-lowering denominator.
Key Takeaways
• A non-volatile solute lowers ideal solvent vapour pressure at the same temperature. • Relative lowering equals solute mole fraction in the finite-composition ideal model. • The relation r≈n₂/n₁ is a dilute approximation. • Exact inversion gives n₂=rn₁/(1−r). • Molar-mass calculations require separate solute and solvent masses. • Normal boiling pressure is 1 atm, approximately 1.01325 bar. • Colligative effects count dissolved particles under their stated model assumptions.