Chemistry 2 • Solutions & Colligative Properties

Raoult's Law — Apply Raoult's law qualitatively and quantitatively

Predict how dissolved solutes lower a solvent's vapor pressure using mole fractions.

An atomic orbit with a reaction beaker and changing energy trace.
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

Throughout the 1800s, chemists noticed something curious: dissolving a substance like sugar or salt in water caused the water to boil at a slightly higher temperature and freeze at a slightly lower one. These observations hinted that the dissolved particles were interfering with the solvent's normal behavior at the surface, where liquid molecules escape into the gas phase. The French chemist François-Marie Raoult set out to quantify exactly how much a solute lowers the vapor pressure of a solvent, and his careful experiments led to one of the foundational laws of solution chemistry.

1840s • Early Solution Studies

Chemists like Michael Faraday and Thomas Graham investigated how dissolved substances change the physical properties of liquids, setting the stage for quantitative colligative-property research.

1882 • Raoult Publishes His Law

François-Marie Raoult, after studying hundreds of solutions, published the relationship showing that a solvent's vapor pressure is proportional to its mole fraction in solution.

1886 • Van 't Hoff Connects Osmotic Pressure

Jacobus van 't Hoff linked osmotic pressure to other colligative properties, reinforcing the theoretical framework behind Raoult's experimental findings.

1920s • Ideal vs. Non-Ideal Solutions

Researchers distinguished ideal solutions (which obey Raoult's law perfectly) from non-ideal solutions that show positive or negative deviations, refining the law's range of applicability.

The central question Raoult answered was deceptively simple: if you dissolve a non-volatile solute in a liquid, by exactly how much does the liquid's vapor pressure drop? His answer—now called Raoult's law—ties the vapor pressure directly to mole fractions, giving chemists a powerful predictive tool that remains essential in modern chemistry.

SECTION 2

Core Principles & Definitions

Before diving into Raoult's law itself, you need a solid grip on a few key ideas. Vapor pressure is the pressure exerted by the gas-phase molecules above a liquid in a closed container when evaporation and condensation reach equilibrium. A mole fraction (χ) is a way to express concentration: the number of moles of one component divided by the total moles in the mixture. An ideal solution is one in which the intermolecular forces between different molecules are essentially the same as those between like molecules, so the components mix without releasing or absorbing heat.

1

Vapor Pressure (P°)

The pressure of a pure solvent's vapor at a given temperature. Each liquid has a characteristic P° that increases with temperature. A higher P° means the liquid evaporates more readily.
2

Mole Fraction (χ)

χA = moles of A ÷ total moles. Mole fractions are dimensionless and always sum to 1.00 for all components in a solution.
3

Ideal Solution

A solution where solute–solvent interactions equal solvent–solvent interactions. Both components obey Raoult's law across all concentrations. Examples include benzene–toluene mixtures.
4

Non-Volatile Solute

A solute with negligible vapor pressure at the temperature of interest—like sugar or salt dissolved in water. Only the solvent contributes meaningful vapor above the solution.
5

Colligative Property

A property that depends on the number of solute particles, not their identity. Vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure are the classic four.
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KEY TAKEAWAY

Think of vapor pressure like the number of people who can exit a crowded room through a door. In pure water, every molecule at the surface has a chance to escape into the gas phase. Now imagine you scatter a bunch of large, stationary obstacles (solute particles) across the exit. Fewer water molecules can reach the door at any given moment, so the escape rate—and thus the vapor pressure—drops. The more obstacles you add, the greater the reduction.

SECTION 3

Visual Explanation — Vapor Pressure Lowering

Vapor Pressure: Pure Solvent vs. SolutionPure WaterH₂O molecules only↑ H₂O↑ H₂O↑ H₂O↑ H₂O↑ H₂O↑ H₂OP° = highSolution (solute added)H₂O + solute particles↑ H₂O↑ H₂O↑ H₂OP = lowerSSSPink circles (S) = solute particles occupying surface sites
On the left, pure water has maximum surface exposure, so many molecules escape into the vapor phase (high P°). On the right, solute particles (pink, labeled S) block some surface sites, reducing the number of solvent molecules that can evaporate and lowering the observed vapor pressure.

The diagram above captures the qualitative heart of Raoult's law. When a non-volatile solute dissolves in a solvent, solute particles take up space at the liquid's surface. Because those solute particles have essentially zero vapor pressure of their own, only the remaining solvent molecules contribute to the vapor. The fraction of surface sites occupied by solvent molecules is directly related to the solvent's mole fraction in the solution. This is why adding more solute always lowers the vapor pressure further—it steadily reduces the solvent's mole fraction.

SECTION 4

Mathematical Framework

Raoult's law gives us a direct, algebraic way to calculate the vapor pressure of a solvent in a solution. The math is straightforward once you know the mole fraction of each component.

RAOULT'S LAW

Psolution=χsolvent×P°solventP_solution = χ_solvent × P°_solvent

Psolution = vapor pressure of the solvent above the solution (in mmHg, atm, or kPa); χsolvent = mole fraction of the solvent; P°solvent = vapor pressure of the pure solvent at the same temperature.

Because mole fractions must add to 1 in a two-component solution (χsolvent + χsolute = 1), we can rewrite Raoult's law to express the vapor-pressure lowering directly.

VAPOR-PRESSURE LOWERING

ΔP=P°solvent−Psolution=χsolute×P°solventΔP = P°_solvent − P_solution = χ_solute × P°_solvent

ΔP = the decrease in vapor pressure; χsolute = mole fraction of the solute. This form shows that the drop in vapor pressure is proportional to how many mole-fraction units the solute occupies.

MOLE FRACTION FORMULA

χA=nA÷(nA+nB)χ_A = n_A ÷ (n_A + n_B)

nA = moles of component A; nB = moles of component B. For solutions with more than two components, the denominator becomes the sum of all moles present.

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Two Volatile Components

When both components of a solution are volatile (e.g., a benzene–toluene mixture), Raoult's law applies to each one independently: Ptotal = χA × P°A + χB × P°B. Each component's partial pressure follows the same mole-fraction rule.

SECTION 5

Ideal vs. Non-Ideal Solutions — Deviations from Raoult's Law

Raoult's law holds perfectly only for ideal solutions, where the forces between unlike molecules match the forces between like molecules. In reality, many mixtures deviate from this ideal behavior. When solute–solvent attractions are weaker than solvent–solvent attractions, molecules escape the liquid more easily than expected, producing a positive deviation (measured vapor pressure is higher than Raoult's law predicts). When solute–solvent attractions are stronger, molecules are held in the liquid more tightly, producing a negative deviation (measured vapor pressure is lower than predicted).

Raoult's Law: Ideal, Positive, and Negative DeviationsMole fraction of component A (χ_A) →Vapor Pressure (P) →010P°_AIdeal (Raoult's law)Positive deviationNegative deviationPositive: P_measured > P_ideal (weaker solute–solvent forces)Negative: P_measured < P_ideal (stronger solute–solvent forces)
The dashed cyan line shows the ideal Raoult's law prediction—a straight line between zero and P°. The red curve represents a positive deviation (vapor pressure above the line), and the violet curve shows a negative deviation (vapor pressure below the line).
Deviation TypeIntermolecular ForcesExample Mixture
PositiveSolute–solvent attractions are weaker than solvent–solvent or solute–solute attractions.Ethanol + hexane
NegativeSolute–solvent attractions are stronger than those in the pure components.Acetone + chloroform
Ideal (none)Solute–solvent forces roughly equal solvent–solvent forces.Benzene + toluene

Summary of deviation types and their molecular-level causes

SECTION 6

Worked Example — Calculating Vapor Pressure

Let's walk through a complete problem to see how Raoult's law works in practice. Suppose you dissolve 45.0 g of glucose (C6H12O6, molar mass = 180.16 g/mol) in 250.0 g of water at 25 °C. The vapor pressure of pure water at 25 °C is 23.8 mmHg. What is the vapor pressure of the solution?

Vapor Pressure of a Glucose–Water Solution

  1. 1

    Step 1 — Calculate moles of solute and solvent

    Moles of glucose = 45.0 g ÷ 180.16 g/mol = 0.2498 mol. Moles of water = 250.0 g ÷ 18.015 g/mol = 13.88 mol.
    nglucose = 0.250 mol; nwater = 13.88 mol
  2. 2

    Step 2 — Find the mole fraction of water (the solvent)

    χwater = 13.88 ÷ (13.88 + 0.250) = 13.88 ÷ 14.13 = 0.9823.
    χwater = 0.9823
  3. 3

    Step 3 — Apply Raoult's law

    Psolution = χwater × P°water = 0.9823 × 23.8 mmHg = 23.4 mmHg.
    Psolution = 23.4 mmHg
  4. 4

    Step 4 — Calculate the vapor-pressure lowering

    ΔP = P° − Psolution = 23.8 − 23.4 = 0.4 mmHg. This small but measurable decrease comes from the glucose particles occupying about 1.8% of the total mole fraction.
    ΔP = 0.4 mmHg
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Check Your Units

Notice that Raoult's law does not require you to convert pressure units. Whatever unit P° is in, Psolution comes out in the same unit. Just be consistent—if P° is in atm, your answer is in atm.

SECTION 7

Strengths & Limitations of Raoult's Law

Like any scientific model, Raoult's law works beautifully under certain conditions and breaks down under others. Understanding its boundaries helps you know when to trust it and when to look for corrections.

StrengthsLimitations
Mathematically simple—only requires mole fractions and pure-component vapor pressures.Assumes ideal solution behavior, which is rarely exact for real mixtures.
Works well for dilute solutions, where the solvent greatly outnumbers the solute.Breaks down at high solute concentrations where solute–solvent interactions dominate.
Applies to both volatile and non-volatile solutes (non-volatile is the simpler case).Does not account for electrolyte dissociation—NaCl produces two ions per formula unit, doubling the effective particle count.
Serves as the basis for other colligative property equations (boiling-point elevation, freezing-point depression).Cannot predict behavior of strongly interacting pairs (e.g., HCl + water) without activity coefficients.

When Raoult's law works and when it doesn't

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KEY TAKEAWAY

Think of Raoult's law like the speed-equals-distance-over-time formula in physics: it's perfectly accurate in a simplified, frictionless scenario, and it gives excellent approximations in many real-world cases. But just as air resistance makes real projectiles deviate from the simple equation, real intermolecular forces make solutions deviate from ideal behavior. The law is your go-to starting point, and corrections (activity coefficients, the van 't Hoff factor) are added only when precision demands it.

SECTION 8

Connection to Advanced Theory — The Van 't Hoff Factor & Activity

Once you master Raoult's law for molecular (non-electrolyte) solutes, the next step is handling electrolytes—substances that dissociate into ions when they dissolve. Sodium chloride, for instance, splits into Na⁺ and Cl⁻, effectively doubling the number of solute particles. The van 't Hoff factor (i) accounts for this by multiplying the solute's mole count. For NaCl, i ≈ 2; for CaCl2, i ≈ 3.

FeatureRaoult's Law (basic)Modified / Advanced
Solute typeNon-electrolyte (molecular), e.g., glucoseElectrolytes and non-electrolytes; uses van 't Hoff factor i
Vapor-pressure loweringΔP = χsolute × P°ΔP = i × χsolute × P° (approximate for dilute electrolytes)
Accuracy at high concentrationDecreases; deviations growActivity coefficients (γ) replace mole fractions: a = γ × χ
Math complexityAlgebra onlyMay require thermodynamic models (Margules, NRTL) studied in college

Raoult's law vs. advanced modifications

In your current course, you will most often use Raoult's law in its standard form for non-electrolytes, or with the van 't Hoff factor for simple salts in dilute solution. In college-level physical chemistry, you would encounter activity coefficients that correct mole fractions for non-ideal interactions—essentially fine-tuning the model when precision is critical, such as in industrial distillation design.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student adds a spoonful of table sugar to a beaker of water at 40 °C. Without doing any calculations, predict whether the vapor pressure of the solution will be higher than, lower than, or equal to the vapor pressure of pure water at the same temperature. Explain your reasoning.
Show answer
The vapor pressure of the solution will be lower than that of pure water. According to Raoult's law, adding a non-volatile solute (sugar does not evaporate at 40 °C) reduces the mole fraction of water, the solvent. Because Psolution = χwater × P°water, any χwater less than 1 makes Psolution less than P°. Qualitatively, the sugar molecules occupy surface sites and reduce the rate at which water molecules escape into the gas phase.

Question 2: BASIC CALCULATION

At 25 °C, the vapor pressure of pure water is 23.8 mmHg. What is the vapor pressure above a solution made by dissolving 0.500 mol of urea (a non-volatile, non-electrolyte solute) in 10.0 mol of water?
Show answer
First, find the mole fraction of water: χwater = 10.0 ÷ (10.0 + 0.500) = 10.0 ÷ 10.5 = 0.9524. Then apply Raoult's law: Psolution = 0.9524 × 23.8 mmHg = 22.7 mmHg. The vapor pressure dropped by about 1.1 mmHg (ΔP = 0.0476 × 23.8), which equals χsolute × P°, confirming the vapor-pressure lowering form of the equation.

Question 3: INTERMEDIATE

A solution is prepared by dissolving 62.1 g of ethylene glycol (C2H6O2, molar mass 62.07 g/mol) in 500.0 g of water at 30 °C. The vapor pressure of pure water at 30 °C is 31.8 mmHg. (a) Calculate the vapor pressure of the solution. (b) Calculate ΔP.
Show answer
Moles of ethylene glycol = 62.1 ÷ 62.07 = 1.000 mol. Moles of water = 500.0 ÷ 18.015 = 27.75 mol. χwater = 27.75 ÷ (27.75 + 1.000) = 27.75 ÷ 28.75 = 0.9652. (a) Psolution = 0.9652 × 31.8 = 30.7 mmHg. (b) ΔP = 31.8 − 30.7 = 1.1 mmHg. You could also compute ΔP directly: χsolute × P° = 0.0348 × 31.8 = 1.1 mmHg, confirming the result.

Question 4: APPLIED

A car's cooling system contains a 50/50 mixture by mass of ethylene glycol (C2H6O2, 62.07 g/mol) and water. Assume the total mass is 4 000 g. At 100 °C the vapor pressure of pure water is 760 mmHg, and ethylene glycol is essentially non-volatile at this temperature. What is the vapor pressure of the coolant solution, and what does this imply about boiling behavior?
Show answer
Each component contributes 2 000 g. Moles of water = 2 000 ÷ 18.015 = 111.0 mol. Moles of glycol = 2 000 ÷ 62.07 = 32.22 mol. χwater = 111.0 ÷ (111.0 + 32.22) = 111.0 ÷ 143.2 = 0.7750. Psolution = 0.7750 × 760 = 589 mmHg. Because the solution's vapor pressure at 100 °C (589 mmHg) is well below atmospheric pressure (760 mmHg), the coolant will not boil at 100 °C—it requires a higher temperature to reach 760 mmHg. This is exactly why antifreeze also raises the boiling point and protects against engine overheating.

Question 5: CRITICAL THINKING

Two solutions are prepared at 25 °C. Solution A dissolves 0.20 mol of sucrose (C12H22O11) in 10.0 mol of water. Solution B dissolves 0.10 mol of NaCl in 10.0 mol of water. Which solution has the lower vapor pressure? Justify your answer using Raoult's law and the concept of the van 't Hoff factor.
Show answer
For Solution A (sucrose, a non-electrolyte): effective solute particles = 0.20 mol; χwater = 10.0 ÷ 10.2 = 0.9804. For Solution B (NaCl, i ≈ 2): effective solute particles ≈ 0.10 × 2 = 0.20 mol; χwater = 10.0 ÷ 10.2 = 0.9804. With the same effective number of solute particles, both solutions produce essentially the same mole fraction for water and therefore the same vapor pressure. This illustrates that colligative properties depend on the total number of dissolved particles, not their chemical identity. If NaCl fully dissociates, 0.10 mol of NaCl is equivalent to 0.20 mol of a non-electrolyte in its effect on vapor pressure.

SUMMARY

Lesson Summary

Summary

Raoult's law states that the vapor pressure of a solvent above a solution equals the solvent's mole fraction multiplied by the vapor pressure of the pure solvent (P = χ × P°). Qualitatively, adding a non-volatile solute lowers the solvent's vapor pressure because solute particles occupy surface sites and reduce the rate of evaporation. Quantitatively, the vapor-pressure lowering (ΔP) equals χsolute × P°.

The law holds best for ideal solutions and dilute non-electrolyte mixtures. Real solutions may show positive or negative deviations depending on intermolecular forces. For electrolytes, the van 't Hoff factor (i) accounts for ion dissociation, effectively multiplying the number of dissolved particles. Raoult's law is the foundation for all four colligative properties and remains one of the most practical tools in solution chemistry.

Varsity Tutors • Chemistry 2 • Raoult's Law — Apply Raoult's law qualitatively and quantitatively