Relative Lowering of Vapour Pressure

Introduction

Adding a nonvolatile solute to an ideal liquid solution reduces the solvent's mole fraction below one. The solvent vapour pressure falls from p° to p. Dividing this drop by p° gives a particularly simple relation: relative lowering equals the solute mole fraction in a binary ideal mixture. The algebra is short, but its meaning depends on using the correct pressure and counting the appropriate dissolved particles.

Core explanation

Raoult's law for the solvent gives p = xsolvent p°. Rearrange to (p° − p)/p° = 1 − xsolvent. In a binary mixture, xsolvent + xsolute = 1, so (p° − p)/p° = xsolute. The left side is dimensionless because a pressure difference is divided by a pressure. It can be expressed as a decimal or as a percentage after multiplication by 100. A relative lowering of 0.020 is a 2.0% lowering, not a 0.020 kPa lowering.

For example, pure solvent has p° = 50.0 kPa and the solution gives p = 49.0 kPa at the same temperature. The absolute lowering is 1.0 kPa. The relative lowering is 1.0/50.0 = 0.020, so ideal xsolute = 0.020. A different pure solvent with p° = 10.0 kPa and a 1.0 kPa drop would have a relative lowering of 0.10, a much larger composition implication. The absolute difference alone cannot reveal mole fraction without a reference p°.

If nsolute and nsolvent are moles of the named chemical components, xsolute = nsolute/(nsolute + nsolvent). At low solute concentration, nsolute ≪ nsolvent and xsolute ≈ nsolute/nsolvent. This dilute approximation is useful but should not replace the exact denominator in a concentrated solution. For 0.20 mol solute with 9.80 mol solvent, exact xsolute = 0.20/10.00 = 0.0200; the approximation 0.20/9.80 ≈ 0.0204 is close but not identical.

The solute must be effectively nonvolatile for its vapour pressure not to add to the measured total. If the solute is volatile, a measurement of total vapour pressure contains both components and the simple pressure-lowering equation for total pressure is invalid. One may still use a component's solvent partial pressure in a suitable ideal model, but a pressure gauge reading that does not separate vapours cannot be inserted blindly. Temperature must also be the same for p and p°.

Dissociation complicates particle counting. A dissolved electrolyte can produce multiple ions per formula unit, and a basic ideal calculation based only on intact salt formula units can underpredict a colligative effect. Association can reduce the effective number of particles. The relative-pressure relation is most straightforward for a nonelectrolyte in a dilute ideal solution. If a salt is involved, clarify whether the provided mole fraction is of analytical formula units or actual effective particles and whether a van 't Hoff correction is expected.

This pressure relation can estimate a solute's molar mass if its mass is known and the solvent amount is known, provided the solute does not associate, dissociate, react or contribute significant vapour. Such inference is only as good as the pressure measurement and ideal approximation. At very small lowering, a tiny pressure error can produce a large relative uncertainty in calculated solute amount.

Step-by-step reasoning

1. Verify that p° and p refer to the same solvent at the same temperature. 2. Subtract to find p° − p, then divide by p°. 3. Under ideal, binary, nonvolatile-solute conditions, set the result equal to xsolute. 4. Convert xsolute to a mole ratio using the exact total-moles denominator. 5. Check for volatility, dissociation, association and measurement uncertainty before drawing a precise conclusion.

Visual explanation

Draw a pure-solvent bar reaching p° and a solution bar reaching lower p. Mark the vertical gap Δp = p° − p. Next to it, write Δp/p° = 1 − xsolvent = xsolute. A second box shows 98 solvent dots and 2 solute dots, making xsolute = 0.02 and an ideal 2% relative lowering.

Real-world analogy

A five-unit drop in a value of 100 is a 5% change, but a five-unit drop in a value of 20 is 25%. Relative vapour-pressure lowering likewise requires the original pure-solvent pressure as its denominator. The absolute drop is not a composition measure by itself.

Real-world example

Suppose an analyst compares the equilibrium vapour pressure of pure solvent and a dilute nonelectrolyte solution at one controlled temperature. If the pure value is 100 kPa and the solution's solvent vapour pressure is 98 kPa, the ideal relative lowering is 2%, suggesting an effective solute mole fraction of 0.02. The analyst would check temperature control and whether any other gas was included in the pressure measurement.

Why?

Why does xsolute appear even though only solvent enters the vapour? Adding solute reduces xsolvent. Since the ideal solvent partial pressure scales as xsolvent p°, the fraction of pressure lost is exactly the fraction by which xsolvent falls below one in the binary model.

Common misconception

“A one-kilopascal lowering means xsolute = 0.01.” That conclusion needs p° = 100 kPa. For any other pure pressure, divide the one-kilopascal change by the actual p° before interpreting composition.

Worked example

At 298 K, a pure solvent has vapour pressure 40.0 kPa. A binary ideal solution of a nonvolatile nonelectrolyte has solvent pressure 38.4 kPa. Then Δp = 1.6 kPa and Δp/p° = 1.6/40.0 = 0.0400, so xsolute = 0.0400 and xsolvent = 0.9600. If the liquid has 24.0 mol solvent, then nsolute/(nsolute + 24.0) = 0.0400. Solving gives 0.9600 nsolute = 0.960, so nsolute = 1.00 mol. Direct check: 1/(1 + 24) = 0.0400.

Quick check

1. Which denominator defines relative lowering: solution pressure p or pure-solvent pressure p°? Answer: Pure-solvent pressure p°; the relation is (p° − p)/p° under the stated model.

Exam focus

Derive the equation rather than memorising only its last line. Distinguish absolute Δp from relative Δp/p°, and use the exact mole-fraction denominator unless dilute approximation is explicitly justified. State ideal and nonvolatile assumptions.

Advanced insight

For real solutions, solvent activity replaces its mole fraction, so measured relative lowering reflects 1 − asolvent rather than simply the analytical solute mole fraction. Electrolyte solutions can deviate because ions interact even when nominally dilute. Thus vapour-pressure measurements can also probe nonideal solution behaviour rather than merely count added formula units.

Summary

For an ideal binary solution with nonvolatile solute, Δp/p° = (p° − p)/p° = xsolute. Both pressures must refer to the same temperature and solvent, and the result is dimensionless. Volatility or changes in effective particle number require more careful treatment.

Practice questions

1. A pure solvent has p° = 80 kPa and a solution has p = 76 kPa. Find relative lowering. Answer: (80 − 76)/80 = 0.050, or 5.0%. 2. Under ideal nonvolatile-solute assumptions, what is xsolute for that solution? Answer: xsolute = 0.050 and xsolvent = 0.950. 3. Why can total pressure above an ethanol–water mixture not be inserted into this one-nonvolatile-solute formula? Answer: Both components contribute vapour, so total pressure is the sum of their partial pressures.