Relative Lowering of VP

When a non-volatile solute is dissolved in a solvent, one of the most fundamental changes observed is a reduction in the vapour pressure of the solution compared to that of the pure solvent. This phenomenon is known as vapour pressure lowering, and the fractional decrease relative to the pure solvent's vapour pressure is called the relative lowering of vapour pressure (RLVP). It is a colligative property, meaning it depends only on the number of solute particles present, not on the chemical identity of the solute. The theoretical foundation for this property was established by the French chemist François-Marie Raoult in the 1880s. Raoult's Law states that the vapour pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent. As a consequence, the relative lowering of vapour pressure is numerically equal to the mole fraction of the solute. This elegant relationship makes RLVP an extremely useful experimental tool: by measuring the drop in vapour pressure, chemists can calculate the mole fraction of the solute and from there determine the molar mass of an unknown substance. At the molecular level, the lowering of vapour pressure occurs because solute particles occupy surface sites on the liquid, reducing the rate at which solvent molecules can escape into the vapour phase. Since the solute is non-volatile, it does not contribute to the vapour pressure itself, so the net effect is always a reduction. This article explores the mathematical derivation, practical calculations, and applications of relative lowering of vapour pressure for high school students preparing for advanced examinations.

Key takeaways
  • Raoult's Law for a Non-Volatile Solute
  • Calculating the Lowering of Vapour Pressure
  • Defining Relative Lowering of Vapour Pressure
  • Expressing Mole Fraction in Terms of Masses
  • Determining Molar Mass from Vapour Pressure Measurements
  • Colligative Nature and Electrolyte Solutions

Raoult's Law for a Non-Volatile Solute

For a solution containing a non-volatile solute, Raoult's Law states that the vapour pressure of the solution (p) equals the mole fraction of the solvent (χ₁) multiplied by the vapour pressure of the pure solvent (p°): p = χ₁ × p°. Here, χ₁ = n₁ / (n₁ + n₂), where n₁ is the moles of solvent and n₂ is the moles of solute.

Calculating the Lowering of Vapour Pressure

The absolute lowering of vapour pressure is Δp = p° − p. Substituting Raoult's Law: Δp = p° − χ₁·p° = p°(1 − χ₁). Since the sum of mole fractions must equal 1 (χ₁ + χ₂ = 1, where χ₂ is the mole fraction of the solute), we get 1 − χ₁ = χ₂. Therefore, Δp = χ₂ × p°.

Defining Relative Lowering of Vapour Pressure

Dividing both sides by p° gives the relative lowering: Δp / p° = χ₂. This is the key result — the relative lowering of vapour pressure is exactly equal to the mole fraction of the solute. This quantity is dimensionless and always lies between 0 and 1 for a real solution.

Expressing Mole Fraction in Terms of Masses

For a solution where w₂ grams of solute (molar mass M₂) is dissolved in w₁ grams of solvent (molar mass M₁), the moles are n₂ = w₂/M₂ and n₁ = w₁/M₁. Substituting: Δp/p° = (w₂/M₂) / [(w₁/M₁) + (w₂/M₂)]. For dilute solutions, n₁ >> n₂, so the denominator ≈ n₁, and the expression simplifies to Δp/p° ≈ (w₂ × M₁) / (w₁ × M₂). This approximation is very useful for determining unknown molar masses.

Determining Molar Mass from Vapour Pressure Measurements

Rearranging the dilute-solution approximation: M₂ = (w₂ × M₁ × p°) / (w₁ × Δp). By experimentally measuring p° (pure solvent vapour pressure) and p (solution vapour pressure) at a fixed temperature, and knowing the masses and molar mass of the solvent, one can calculate the molar mass of the solute. This technique has historically been used to determine molar masses of polymers and large organic molecules.

Colligative Nature and Electrolyte Solutions

Because RLVP depends only on particle count, electrolytes that dissociate in solution produce a greater effect than expected from the formula mass alone. For an electrolyte AxBy that dissociates into (x+y) ions, the van't Hoff factor i = x + y (for complete dissociation) must be incorporated: Δp/p° = i × χ₂(formula). For example, NaCl (i ≈ 2) dissolved in water lowers vapour pressure nearly twice as much as a non-electrolyte at the same molality. Always check whether the solute is an electrolyte before applying the simple formula.

The Key Equation

Δpp∘=χ2=n2n1+n2\frac{\Delta p}{p^\circ} = \chi_2 = \frac{n_2}{n_1 + n_2}
  • \Delta p — Lowering of vapour pressure (Pa or mmHg): Difference between vapour pressure of pure solvent (p°) and that of the solution (p): Δp = p° − p
  • p^\circ — Vapour pressure of pure solvent (Pa or mmHg): The equilibrium vapour pressure exerted by the pure solvent at the given temperature
  • \chi_2 — Mole fraction of solute (dimensionless): Ratio of moles of solute to the total moles of all components in the solution
  • n_2 — Moles of solute (mol): Number of moles of the dissolved substance (solute)
  • n_1 — Moles of solvent (mol): Number of moles of the dissolving substance (solvent)

Worked Example

Problem. A solution is prepared by dissolving 6.0 g of a non-volatile, non-electrolyte solute in 180 g of water at 25 °C. The vapour pressure of pure water at 25 °C is 23.8 mmHg. The measured vapour pressure of the solution is 23.4 mmHg. Calculate: (a) the relative lowering of vapour pressure, (b) the mole fraction of the solute, and (c) the molar mass of the solute.

Δpp∘=0.0168,χ2=0.0168,M2≈35 g/mol\frac{\Delta p}{p^\circ} = 0.0168, \quad \chi_2 = 0.0168, \quad M_2 \approx 35\,\text{g/mol}

Answer. Relative lowering of VP = 0.0168; Mole fraction of solute χ₂ = 0.0168; Molar mass of solute ≈ 35 g/mol

Figure
Diagram
Relative Lowering of Vapour Pressure: Molecular View & GraphPure SolventLiquid Surface (solvent only)Many moleculesescape → high VPSolution (solute added)Liquid Surface (blocked sites)SSSFewer moleculesescape → low VPVapour Pressure vs Mole Fraction of Solvent (χ₁)χ₁ (mole fraction of solvent) → 0 to 1p (VP)p°p = χ₁ · p° (Raoult's Law)χ₁ = 0.77p (solution)
Origins · 1 minFrançois-Marie Raoult and how this was discovered
FR
François-Marie Raoult
1830–1901 · French

Raoult formulated the law bearing his name in 1882–1887 after meticulous vapour pressure experiments on hundreds of solutions. He showed that the relative lowering of vapour pressure is equal to the mole fraction of the solute — a purely colligative relationship. His work laid the groundwork for cryoscopy, ebullioscopy, and osmometry as methods for molar mass determination, fundamentally changing physical chemistry. Raoult conducted much of his groundbreaking research while a professor at the University of Grenoble, using simple but precise mercury manometers. He tested so many solute-solvent pairs that his notebooks contained data for over 150 different substances, leading him to discover the generality of his law almost by accident.

  • Raoult's Law — Relative lowering of vapour pressure is derived directly from Raoult's Law; understanding the law is prerequisite to this concept.
  • Boiling Point Elevation — Boiling point elevation is caused by the same vapour pressure lowering effect — the solution's reduced VP requires a higher temperature to equal atmospheric pressure.
  • Freezing Point Depression — Another colligative property linked to vapour pressure lowering; the lowered VP of the solution shifts the solid-liquid equilibrium to a lower temperature.
  • Osmotic Pressure — Osmotic pressure is the fourth colligative property; like RLVP, it depends on mole fraction (or molarity) of solute particles.
  • Mole Fraction and Concentration Units — Mole fraction is the central quantity in the RLVP equation; students must be comfortable converting between mass, moles, and mole fractions.