JEE (Advanced) 2026, Paper 2, Chemistry, Section 4 — a question stem shared by Q.15 and Q.16. This simulation answers Q.15.
Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.
| Given | Value |
|---|---|
| The gas constant R | 0.08 L atm K−1 mol−1 |
| Molar mass of A | 50 g mol−1 |
| Molar mass of B | 57 g mol−1 |
| Density of liquid B at 300 K | 0.5 g/mL |
| 1 atm | 760 mm Hg |
Q.15 At 300 K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ____ .
Enter the numbers the question gives you into the instrument rail, then run the experiment. The bench works out the mole fraction, boils the solution to find pB°, fills two chambers with one mole of B each — one liquid, one vapour — and only then tells you the ratio.
The container is empty. Fill in the data from the question stem and press RUN VIRTUAL EXPERIMENT.
An ideal solution has a straight-line pressure diagram: each component contributes its own vapour pressure times its own mole fraction. Drag the composition and watch the total move. The measured point is where your 5 molal solution sits.
Every quantity is filled in only when the experiment has established it, and every completed run is logged so two sets of data can be compared side by side.
No runs yet. The log fills itself in as you run the experiment — change a number and run it again to see what moves and what does not.
Each card re-runs the model with one mistaken assumption and on your own numbers, so the wrong answer appears as a measurement rather than as a warning. With the question's data the right answer is a round number and every one of these is not.
Run the experiment and reveal the answer — the derivation opens with it.
The bench has not produced an answer yet. Enter the data, run the experiment through to the end and press REVEAL ANSWER — this opens by itself at the same moment.
A 5 molal solution has 5 mol of B for every 1000 g of A. Nothing else about the amount matters — the ratio of the two components is what the question fixes.
Raoult's law is written in mole fractions, never in molality, so this conversion is the first real step of the problem.
In closed form, xB = m / (m + 1000/MA), which shows that the mass of A you start from cancels out entirely.
For an ideal solution the total pressure is the sum of the two partial pressures, each of them the pure-component vapour pressure scaled by that component's mole fraction. Only pB° is unknown, so the equation solves for it.
A check that costs nothing: 105 × 0.8 = 84 and 80 × 0.2 = 16, and 84 + 16 = 100, the pressure that was measured.
One mole of liquid B weighs 57 g, and 57 g of a liquid of density 0.5 g/mL occupies 57/0.5 mL. Density is the only property of the liquid that enters.
Pure B as a vapour at 300 K sits at pB° = 80 mm Hg. Convert to atmospheres, because R is quoted in L atm K−1 mol−1.
Both volumes are per mole, so the units cancel and the answer is a pure number.
It is worth seeing why the number is so large. A mole of gas at this pressure is spread over roughly two hundred litres while the same mole of liquid is about a tenth of a litre: the molecules are the same, the spacing is not. In each direction the expansion is only ∛2000 ≈ 12.6, which is why the two cubes on the bench are drawn at that edge ratio rather than at 2000:1.
Five ideas carry this question, and the experiment used them in this order. The cards animate in when the answer is revealed — or right now, if you would rather read first.
Molality counts moles of solute per kilogram of solvent; mole fraction counts them against all the moles present. Raoult's law is written in mole fractions, so a molality always has to be converted first — and the conversion needs the molar mass of the solvent, which is why it is given.
Each component escapes into the vapour in proportion to how much of the surface it owns. The partial pressure is the pure-component vapour pressure scaled by that component's mole fraction, and the total is their sum — a straight line between pA° and pB°.
A liquid's molecules are already touching, so its volume is set by how much mass you have and how tightly it packs. One mole is M grams, and M grams of a liquid of density ρ occupies M/ρ — no temperature, no pressure, no gas law.
"Pure B in the vapour phase" is one mole of B alone, at 300 K, at the pressure pure B actually exerts: the pB° the bench measured — not the mixture's total and not 1 atm. Convert to atmospheres, because R is quoted in L atm K−1 mol−1.
Both volumes are per mole, so the units cancel and the answer is a pure number. It rises with temperature, falls as the vapour pressure rises, and rises with the density of the liquid. In each direction the expansion is only the cube root of it.
Notice what is not in the last expression: the mass of A, the molar mass of A and the molality appear only through pB°, and pA° and ptotal only through the same quantity. Sweep the mass of A on the bench and the answer does not move.
| Exam | JEE Advanced 2026 · Paper 2 |
|---|---|
| Subject | Chemistry |
| Section | Section 4 · question stem for Q.15 and Q.16 |
| Question | Q.15 — numerical value |
| Marking | +2 if correct, 0 otherwise |
| Chapter | Solutions and Colligative Properties |
| Topic | Raoult's law for ideal solutions of two volatile liquids |
| Also draws on | Mole concept · Ideal gas equation · Density |
| Difficulty | Moderate |
| Answer | — run the experiment |