What does the vapour over a mixture contain?
"What does the vapour over a mixture contain?" is a free, interactive Class 12 Chemistry simulation for NEET and JEE, built on the lesson "Vapour pressure and Raoult's law" in Solutions. Slide the mole fraction of hexane in the liquid. Each liquid adds its own pressure, p = p° × x, and the gauge reads the sum. Watch the vapour: it never matches the liquid. Ideal: p_total = p₁° + (p₂° − p₁°) x₂, a straight line, and y_i = p_i/p_total. Real mixtures bend away from the line: up when unlike molecules attract weakly, down when they attract strongly. JEE Main asked 53 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Vapour pressure and Raoult's law in Solutions. Change the values and watch what happens.
In JEE Main
42 shifts analysed
53 questions from Solutions across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §1.4
- Two volatile liquids in a closed vessel both evaporate until liquid and vapour are in equilibrium. By Dalton's law of partial pressures the total vapour pressure is p_total = p₁ + p₂.
- Raoult's law: in a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the liquid: p₁ = p₁° x₁ and p₂ = p₂° x₂, where p° is the vapour pressure of the pure liquid at the same temperature.
- Combining the two, p_total = p₁° + (p₂° − p₁°) x₂: the total vapour pressure is a straight line in the mole fraction of either component, running from p₁° to p₂°.
- The vapour's composition follows from Dalton's law: y_i = p_i / p_total. The vapour is always richer than the liquid in the more volatile component, the one with the larger p°.
- Raoult's law is a special case of Henry's law: both say a volatile component's partial pressure is proportional to its mole fraction, and for Raoult's law the constant K_H equals p°.
- When the solute is non-volatile, only the solvent contributes vapour: p₁ = x₁ p₁°. Solute particles occupy part of the surface, fewer solvent molecules escape, and the solution's vapour pressure falls below the pure solvent's.
- This lowering depends on how much solute is present, not on what it is: 1.0 mol of sucrose and 1.0 mol of urea in 1 kg of water lower the vapour pressure by nearly the same amount.
Questions
3 answers
- What does the "What does the vapour over a mixture contain?" simulation show?
- Slide the mole fraction of hexane in the liquid. Each liquid adds its own pressure, p = p° × x, and the gauge reads the sum. Watch the vapour: it never matches the liquid.
- Does JEE Main ask questions from Solutions?
- Yes. JEE Main asked 53 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Vapour pressure and Raoult's law" in Solutions (Class 12 Chemistry).
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