On this page
Most students meet Raoult's law through colligative properties, where only the solvent is volatile. The harder questions come when both liquids evaporate: what is the total vapour pressure, what is in the vapour, and why can't distillation give pure ethanol? This post covers the two-volatile-liquid case, with worked numbers, the molecular reason for deviations, and the azeotrope facts exams ask for.
Where is this in the syllabus?
It is in sections 1.4 and 1.5 of NCERT Class 12 chemistry, Unit 1, Solutions. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "the vapour pressure of solutions and Raoult's Law - Ideal and non-ideal solutions, vapour pressure - composition, plots for ideal and non-ideal solutions". The JEE (Advanced) 2026 brochure lists Raoult's law and ideal solutions. Check the current year's bulletin before you plan.
What does Raoult's law say for two volatile liquids?
NCERT states that for a solution of volatile liquids, "the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution":
p1 = x1 × p1°
p2 = x2 × p2°
where p° is the vapour pressure of the pure liquid at the same temperature. Dalton's law of partial pressures then gives the total: p(total) = p1 + p2. For an ideal solution, a plot of p(total) against mole fraction is a straight line between the two pure vapour pressures.
The vapour has its own composition. If y1 is the mole fraction of component 1 in the vapour, then y1 = p1 / p(total).
Worked example 1: heptane and octane
NCERT says heptane and octane form an ideal solution and gives their vapour pressures at 373 K as 105.2 kPa and 46.8 kPa. Take a liquid mixture with a heptane mole fraction of 0.400 (for example, 40.08 g of heptane and 68.54 g of octane).
p(heptane) = 0.400 × 105.2 = 42.08 kPa
p(octane) = 0.600 × 46.8 = 28.08 kPa
p(total) = 42.08 + 28.08 = 70.16 kPa
y(heptane) = 42.08 / 70.16 = 0.600
The liquid is 40% heptane, but the vapour is 60% heptane. As NCERT notes, the vapour is always richer in the more volatile component. That is the whole basis of fractional distillation: condense the vapour, and you have a liquid richer in heptane.
Worked example 2: which mixture boils at 373 K?
A liquid boils when its vapour pressure equals the outside pressure. At 1 atm (101.325 kPa), what heptane mole fraction x makes the mixture boil at 373 K?
101.325 = 46.8 + (105.2 − 46.8)x, so x = 54.525 / 58.4 = 0.934.
The vapour then has y = (0.934 × 105.2) / 101.325 = 0.969 heptane. Reading boiling points off vapour pressures like this is the link between Raoult's law and distillation.
Why do some solutions deviate from Raoult's law?
It comes down to how strongly unlike molecules attract each other compared with like molecules. NCERT's examples:
Positive deviation (vapour pressure higher than predicted). A–B attractions are weaker than A–A and B–B. In ethanol and acetone, acetone molecules get between ethanol molecules and break some of their hydrogen bonds, so molecules escape more easily. Carbon disulphide and acetone also deviate positively. Because weaker attractions form on mixing, heat is absorbed (ΔmixH > 0).
Negative deviation (vapour pressure lower than predicted). A–B attractions are stronger. In phenol and aniline, the phenolic hydrogen bonds to nitrogen's lone pair. In chloroform and acetone, chloroform's hydrogen bonds to acetone's oxygen. Heat is released on mixing (ΔmixH < 0).
Ideal (nearly). A–B attractions roughly equal A–A and B–B. NCERT lists n-hexane and n-heptane, bromoethane and chloroethane, and benzene and toluene; for an ideal solution ΔmixH = 0 and ΔmixV = 0.
What is an azeotrope, and why does it stop distillation?
NCERT defines azeotropes as binary mixtures "having the same composition in liquid and vapour phase" that boil at a constant temperature. Once the vapour has the same composition as the liquid, condensing it gains nothing, so fractional distillation cannot separate them further.
Minimum boiling azeotropes form from large positive deviations. Fermentation ethanol, fractionally distilled, stops at about 95% ethanol by volume.
Maximum boiling azeotropes form from large negative deviations. NCERT's example is nitric acid and water, about 68% nitric acid and 32% water by mass, boiling at 393.5 K.
Which mistakes cost marks?
Using mass fractions in Raoult's law. Convert to moles first.
Using liquid mole fractions for the vapour. The vapour composition comes from y = p / p(total).
Adding the two pure vapour pressures instead of weighting each by its mole fraction.
Swapping the azeotrope types. Positive deviation gives a minimum boiling azeotrope; negative gives a maximum.
Calling an azeotrope a compound. It is a mixture whose composition changes with pressure.
For the related case where only the solvent is volatile, see our post on colligative properties and the van 't Hoff factor, and for unit conversions, molarity, molality and mole fraction.
A practice routine
Draw the three vapour pressure plots (ideal, positive, negative) from memory and label p1, p2 and p(total).
Solve NCERT's Example 1.5 and intext question 1.8, then change one number and solve again.
Write one sentence of molecular explanation for each of NCERT's five non-ideal examples.
Close the book and test yourself on these three days later. In one experiment, students who took recall tests remembered more a week later than students who restudied the same material.
If vapour composition still feels abstract, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
This week, compute the vapour composition for a 50:50 mole mixture of heptane and octane at 373 K and check that heptane is above 0.5.
