Boiling point of a partially dissociated solute — JEE Advanced 2026 Paper 2 Q.12

01The question — verbatim

Section 3, numerical value — +4 for the correct value, 0 otherwise. Read it, then type what it gives you into the bench below.

In a solvent S, a compound B is partially dissociated into C and D as given below:

B ⇌ 2C + 2D

B, C and D are non-volatile in nature. The molar mass of B is 10 times the molar mass of S. The standard boiling point and the standard enthalpy of vaporization of S are 400 K and 10R J mol⁻¹, respectively (R is the gas constant in J K⁻¹ mol⁻¹). A solution of B in S with an initial concentration of B as 0.25% (mass/mass) has a boiling point of 408 K at 1 bar pressure. In this solution, the mole percent of B that has been dissociated is ______.

Not given: the molar mass of the solvent itself. That is deliberate — section 03 shows why the experiment never needs it.

What each number in the question is for

What you are given	Symbol	Value	What it is for
Standard boiling point of the pure solvent	T_b°	400 K	builds K_b, and is the baseline the rise is measured from
Enthalpy of vaporisation of the solvent	ΔH_vap	10R J mol⁻¹	builds K_b — and cancels the R in it
Molar mass of B relative to S	M_B	10 M_S	turns the 0.25% into a molality
Initial concentration of B	—	0.25% (m/m)	0.25 g of B per 100 g of solvent
The dissociation	B ⇌ 2C + 2D	1 → 4	how many particles one B leaves behind
Boiling point of the solution	T_b	408 K	the one measurement — everything is read from it

02The bench

Type the data from the question into the instrument panel, press Run experiment, and watch the bench work. Nothing here is precomputed — every number comes out of the values you enter.

1Enter the data 2Run the experiment 3Read the result waiting for your data

The apparatusempty

The working, written as it is earned

predicted your reading flask now

03Why the solvent’s molar mass never mattered

The question never tells you what S is. This panel, driven by the data you entered, shows that it could not have mattered — and keeps a log of every run you make.

Pick any solvent you like

Molar mass of the solvent S 40.0 g mol⁻¹

g mol⁻¹

K_b rises with it, the molality falls with it.

K_b carries one factor of M_S and the molality carries one factor of 1/M_S, so their product is fixed by the data you entered and by nothing else. That product is what the whole experiment runs on.

Runs you have made

No runs yet — enter the data above and press Run experiment.

Lab notebook

The notebook writes itself as the bench settles things.

04Where it goes wrong

Four answers this question produces every year. Each is a single wrong move, and the bench above will reproduce all of them if you enter the data that way.

100 — one particle instead of four

Reading “B ⇌ 2C + 2D” as though one molecule became one. Enter 1 C and 1 D in the stoichiometry boxes and the bench returns 100 % — complete dissociation, which the word partially rules out. One B leaves four particles, so i = 1 + 3α.

404 K — forgetting dissociation

Step 3 of the run boils the solution with nothing dissociated and lands on 404 K. The thermometer says 408 K, so the extra 4 K has to come from somewhere — and the only thing left is more particles.

102 — T_b instead of ΔT_b

Putting 408 K into ΔT_b = i K_b m gives i = 102. The van’t Hoff factor here can never exceed 4, so any i above 4 says a boiling point was used where a rise was meant. Enter 808 K as the solution’s boiling point and the bench refuses the run for exactly that reason.

stuck — hunting for M_S

The molar mass of the solvent is not in the question and it is not missing. There is no input box for it on the bench because the experiment does not need one — section 03 shows why.

05The record of the run

Filled in from your own run, quantity by quantity, as each one is established.

06Step-by-step solution

Complete a run on the bench and the derivation opens by itself.

The solution is still locked

Enter the data from the question and let the experiment finish — the derivation and the answer box unlock the moment the flask matches your measurement.

Build K_b out of the solvent’s own data

No table of ebullioscopic constants is given, and none is needed. K_b follows from the boiling point and the enthalpy of vaporisation:

K_b = R (T_b°)² M_S / ΔH_vap

with M_S in kg mol⁻¹. Because ΔH_vap is quoted as 10R, the gas constant cancels straight away:

K_b = R(400)²(M_S/1000) / (10R) = 16 M_S K kg mol⁻¹ (M_S in g mol⁻¹)
Turn 0.25% (m/m) into a molality

0.25 g of B in 100 g of solvent, and M_B = 10 M_S:

n_B = 0.25 / (10 M_S) mol in 0.100 kg of S

m = 0.25 / (10 M_S × 0.100) = 0.25 / M_S mol kg⁻¹
The molar mass of the solvent cancels

This is the move the question is really testing. K_b goes up with M_S, the molality goes down with it, and the product is a pure number:

K_b m = (16 M_S) × (0.25 / M_S) = 4.000 K

So the solvent’s identity never enters the answer, which is why it was never given and why the bench has no input box for it.
Read the van’t Hoff factor off the thermometer

ΔT_b = T_b − T_b° = 408 − 400 = 8 K

i = ΔT_b / (K_b m) = 8 / 4 = 2

Two particles now exist for every one that was dissolved. That is exactly what step 4 of the run exposes: an undissociated solution boils at 404 K, and the second 4 K is entirely the work of the dissociation.
Count the particles the dissociation makes

Start with 1 mol of B and let a fraction α break up. Each one that goes produces 2 C and 2 D — four particles from one:

B ⇌ 2C + 2D → (1 − α) + 2α + 2α = 1 + 3α

The limits check out: α = 0 gives i = 1, and complete dissociation gives i = 4.
Solve for α

1 + 3α = 2 ⇒ α = 1/3 = 0.3333

mole percent dissociated = 100 α = 33.33 %

Which is what the flask shows at the end of the run: 4 of the 12 drawn units broken up, 8 whole B plus 16 fragments — 24 particles where there were 12.

Independent verification

Derived before anything was looked up. K_b from thermodynamics, the molality from the mass percent, i from the measured rise and α from the stoichiometry — all fixed independently, and only then compared with a published solution.
Back-substitution. Taking any M_S at all — 12, 20, 33.3, 58.5, 180 g mol⁻¹ — and putting α = 1/3 back into ΔT_b = i K_b m returns 8.000000 K and a boiling point of exactly 408 K.
The cancellation checked symbolically, not just numerically: K_b m reduces to the constant 4 with M_S carried as a free symbol, so the result holds for every solvent, not merely for the values tried. The bench proves the same thing by construction — it never asks for M_S and still reaches the answer.
Limiting cases. i = 1 + 3α gives i = 1 at α = 0 and i = 4 at α = 1, bracketing boiling points of 404 K and 416 K. The measured 408 K sits inside that window, so a partially dissociated solution is consistent with the data — as the question states. The bench enforces the same bracket and refuses any measurement outside it, saying why.
The distractors reproduced deliberately. One C and one D in the stoichiometry boxes gives 100 %; ignoring dissociation gives 404 K; using T_b in place of ΔT_b gives i = 102, above the maximum of 4; ignoring the factor of ten in M_B multiplies the molality by ten and gives a rise of 80 K. None of them is 33.33.
Units. K_b in K kg mol⁻¹ times m in mol kg⁻¹ leaves kelvin, and i is dimensionless — so ΔT_b comes out in kelvin as it must.
Answer key cross-check. A published worked solution for this paper follows the same route — K_b = 16 M_S, K_bm = 4, i = 2, i = 1 + 3α — and reports the same 33.33. IIT Roorkee’s official key was not retrievable at the time of writing.
Headless-browser run of this page, driving the form with the paper’s data and with deliberately wrong data, asserting that the run refuses impossible measurements, that the answer is nowhere on the page until the run finishes, that the scene, gauges, working, chart and thermometer all move together, and that the page never scrolls sideways at 390 px.

07Concepts and classification

What the question is made of, and where it sits in the syllabus.

Formulas used

K_b = R (T_b°)² M_S / ΔH_vap

ΔT_b = i K_b m

m = n_solute / (mass of solvent in kg)

i = (particles after) / (particles before)

B ⇌ 2C + 2D → i = 1 + 3α

The bench applies these in exactly this order, one per step of the run.

Classification

Exam	JEE (Advanced) 2026, Paper 2
Question	Q.12, Section 3 — numerical value
Marking	+4 for the correct value, 0 otherwise, no negative marking
Subject	Physical Chemistry
Chapter	Solutions and Colligative Properties
Topic	Elevation of boiling point with a partially dissociated solute
Also needs	Thermodynamic origin of K_b; concentration terms; van’t Hoff factor from stoichiometry
Difficulty	Moderate — the arithmetic is short, the trap is structural
Answer	— run the experiment

01The question — verbatim

02The bench

03Why the solvent’s molar mass never mattered

Pick any solvent you like

Runs you have made

Lab notebook

04Where it goes wrong

05The record of the run

06Step-by-step solution

The solution is still locked

Build Kb out of the solvent’s own data

Turn 0.25% (m/m) into a molality

The molar mass of the solvent cancels

Read the van’t Hoff factor off the thermometer

Count the particles the dissociation makes

Solve for α

Independent verification

07Concepts and classification

Formulas used

Classification

Build K_b out of the solvent’s own data