Mole fraction of B in the vapour phase — JEE Advanced 2026 Paper 2 Q.16

01Original question — verbatim

JEE (Advanced) 2026, Paper 2, Chemistry, Section 4 — the second question on the stem shared with Q.15.

Question stem for question nos. 15 and 16

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	Needed for Q.16?
Molar mass of A	50 g mol⁻¹	yes — turns the molality into a mole fraction
The gas constant R	0.08 L atm K⁻¹ mol⁻¹	no
Molar mass of B	57 g mol⁻¹	no
Density of liquid B at 300 K	0.5 g/mL	no
1 atm	760 mm Hg	no

Q.16 The mole fraction of B in vapour phase which is in equilibrium with this solution is ____ .

The whole question is in one word: vapour. The liquid's composition, x_B, follows from the molality in two lines. The vapour sitting above it has a different composition, y_B, because the two liquids do not evaporate equally readily. Writing down x_B and stopping is the mistake this question is built to catch.

Three of the five given quantities — R, the molar mass of B and the density of liquid B — belong to Q.15 and are not needed here. They are on the bench anyway, greyed out, because recognising unused data is part of reading a question stem.

02The vapour analyser

Enter the stem's data, press start, and the bench takes it through five stages: make the solution, work out what the liquid is made of, let it come to equilibrium and split the pressure gauge in two, then sample the vapour and count what is actually up there.

00Charge 01Prepare solution 02Liquid composition 03Raoult's law 04Analyse vapour 05Discover answer waiting for your data

Bench — closed equilibrium vessel at 300 K

t = 0.0 s

The vessel is empty. Fill in the data from the question stem and press START EXPERIMENT.

Seven boxes carry this part; three more are on the bench but belong to Q.15.

03Liquid against vapour

The two phases sit on top of each other in the same sealed vessel and they are not made of the same stuff. This is the picture the question is testing, and the curve below is where every composition of this pair ends up.

Liquid phase

—

x(B) — what you made, straight from the molality

25 molecules drawn — count the orange ones

→

Vapour phase

—

y(B) — what is actually above it at equilibrium

25 molecules drawn — count the orange ones

liquid composition → vapour pressures → vapour composition

Explore a different liquid composition — x(B) 0.200

Your own solution stays marked. Drag to see that the vapour never matches the liquid except at the two pure ends.

x(B) explored

0.200

y(B) there

—

total pressure there

— mm Hg

richer phase in B

—

04Record of the run

Each quantity appears only once the experiment has established it, and every completed run is logged so two sets of data can be compared.

What this run established

Trial log

No runs yet. Change a number, run it again, and the log shows you what moves — the mass of A, for one, never does.

05Where this question goes wrong

Each card re-runs the model with one mistaken assumption, on your own numbers, so the wrong answer turns up as a measurement rather than as a warning.

06Step-by-step solution

Run the experiment and reveal the answer — the derivation opens with it.

The solution is still locked

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.

Step 1 — moles of A

Molality is quoted per kilogram of solvent, so take 1000 g of A as the basis. Nothing about the answer depends on that choice; it just gives us something to count.

n_A = 1000 / 50 = 20 mol
Step 2 — the liquid-phase mole fraction of B

A 5 molal solution has 5 mol of B for every kilogram of A, so n_B = 5 mol. Mole fraction counts against everything present, not against the solvent alone.

x_A = 20 / (20 + 5) = 0.8 · x_B = 5 / 25 = 0.2

This is the number the question is waiting for you to hand in by mistake.
Step 3 — partial pressure of A, from Raoult's law

In an ideal solution each component exerts its own pure vapour pressure scaled by its own mole fraction in the liquid.

p_A = x_A p_A° = 0.8 × 105 = 84 mm Hg
Step 4 — partial pressure of B, from the total

The gauge on the vessel reads the sum of the two partial pressures, and A's share is now known, so B's share is whatever is left.

p_B = p_T − p_A = 100 − 84 = 16 mm Hg
Step 5 — the vapour-phase mole fraction

The vapour is an ideal gas mixture, so each component's partial pressure is its mole fraction in the vapour times the total pressure — Dalton's law. Rearranged, the vapour composition is simply the pressure share.

p_i = y_i p_T ⟹ y_B = p_B / p_T = 16 / 100 = 0.16
Why it is smaller than x_B, and what that means

Putting the steps together gives y_B = x_Bp_B°/p_T, and here p_B° = 16/0.2 = 80 mm Hg against p_A° = 105 mm Hg. B is the less volatile of the two, so the vapour is poorer in B than the liquid was: 0.16 against 0.20, an enrichment factor of exactly p_B°/p_T = 0.8.

y_B / x_B = p_B° / p_T = 80/100 = 0.8 · y_A = 84/100 = 0.84 > x_A = 0.80

The vapour is always richer in the more volatile component. That single sentence is the basis of fractional distillation, and it is what this question is really about.

Independent verification

Derived first, checked afterwards. Every step — the molality conversion, Raoult, the subtraction and Dalton — was worked out from the stem alone before any published solution was consulted.
The two phases close separately. x_A + x_B = 0.80 + 0.20 = 1 and y_A + y_B = 0.84 + 0.16 = 1, and the partial pressures add back to the measured total: 84 + 16 = 100 mm Hg.
The answer by a second, independent route. Recovering p_B° = p_B/x_B = 80 mm Hg and using y_B = x_Bp_B°/p_T gives 0.16 again, as does 1 − x_Ap_A°/p_T. The relative-volatility form with α = p_A°/p_B° = 21/16 reproduces y_A = 0.84 exactly.
The sign of the effect predicted before it was computed. p_B° < p_A° forces y_B < x_B, and the computed numbers agree — so the result is not a coincidence of arithmetic.
Limiting cases. Setting p_A° = p_T makes the two liquids equally volatile and the model returns y_B = x_B exactly; setting p_A° below p_T makes B the more volatile one and the model returns y_B > x_B; a vanishing molality returns y_B → 0. The model therefore behaves correctly across its whole range, not only at the paper's numbers.
The basis cancels. Repeating the calculation with 500 g, 2500 g or 137 g of A returns 0.16 every time, which the bench reproduces live — the mass of A moves n(A) and n(B) on the gauges and never moves the answer.
Distractor audit. Handing in x_B gives 0.20; y_A gives 0.84; p_B/p_A gives 0.1905; x_Bp_A°/p_T gives 0.21; 1 − x_B gives 0.80. None is 0.16, and every one is reproduced on the bench as a run.
Headless-browser run of this page. The finished simulation was driven end to end in Chromium and asserted to reproduce 0.16 from the live model, to give different answers on different data, to refuse a total pressure that would make p_B negative, to keep the answer off the page until it is revealed, and to render with no console errors and no horizontal overflow at 390 px and 360 px.
Cross-check. Afterwards the answer was compared against a published worked solution for this paper, which follows the same route and reports the same intermediate values (x_B = 0.2, p_A = 84, p_B = 16 mm Hg) and the same 0.16. IIT Roorkee's official key was not retrievable at the time of writing.

07What did we learn?

Four ideas, in the order the bench used them. The cards fill their numbers in when the answer is revealed.

The four concepts this question turns on

CARD 01

Molality ≠ mole fraction

5 molal does not mean x_B = 5, and it does not mean 5/20 either. Molality counts moles of B per kilogram of A; mole fraction counts them against every mole in the liquid. Converting between them is what the molar mass of A is given for.

x_B = n_B / (n_A + n_B) ⟹ …

CARD 02

Raoult's law — the liquid's side

Each component of an ideal solution pushes molecules into the vapour in proportion to how much of the liquid it owns, scaled by how volatile it is on its own.

p_i = x_i p_i° ⟹ …

CARD 03

Dalton's law — the vapour's side

Once the molecules are in the gas phase they are an ideal gas mixture, and there each component's partial pressure is its own mole fraction times the total. This is the step that turns pressures back into a composition.

p_i = y_i p_T ⟹ …

CARD 04

The key connection

Put the two laws together and the vapour composition is the liquid composition scaled by that component's own volatility relative to the total pressure. Less volatile than the mixture as a whole means the vapour is poorer in it; more volatile means richer.

y_i = x_i p_i° / p_T ⟹ …

Formulas used

n_A = w_A/M_A, n_B = m · w_A/1000

x_B = n_B/(n_A + n_B), x_A = 1 − x_B

p_A = x_A p_A° (Raoult)

p_B = p_T − p_A

y_B = p_B / p_T (Dalton)

combined: y_B = x_B p_B° / p_T, y_B/x_B = p_B°/p_T

Everything here is a ratio of pressures or of moles, so the answer is dimensionless — and the gas constant, the molar mass of B and the density of B, all printed in the stem, never appear. They belong to Q.15.

Classification

Exam	JEE Advanced 2026 · Paper 2
Subject	Chemistry
Section	Section 4 · question stem for Q.15 and Q.16
Question	Q.16 — numerical value
Marking	+2 if correct, 0 otherwise
Chapter	Solutions and Colligative Properties
Topic	Vapour-phase composition over an ideal solution
Also draws on	Mole concept · Raoult's law · Dalton's law of partial pressures
Difficulty	Moderate
Answer	— run the experiment

01Original question — verbatim

Question stem for question nos. 15 and 16

02The vapour analyser

Bench — closed equilibrium vessel at 300 K

03Liquid against vapour

04Record of the run

What this run established

Trial log

05Where this question goes wrong

06Step-by-step solution

The solution is still locked

Step 1 — moles of A

Step 2 — the liquid-phase mole fraction of B

Step 3 — partial pressure of A, from Raoult's law

Step 4 — partial pressure of B, from the total

Step 5 — the vapour-phase mole fraction

Why it is smaller than xB, and what that means

Independent verification

07What did we learn?

The four concepts this question turns on

Molality ≠ mole fraction

Raoult's law — the liquid's side

Dalton's law — the vapour's side

The key connection

Formulas used

Classification

Why it is smaller than x_B, and what that means