PYQ explainedSolutions · colligative properties14 November 2026

Abnormal molar mass: the van 't Hoff factor for benzoic acid in benzene and K₂SO₄ in water

By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS

One freezing point, two surprises. Benzoic acid in benzene behaves as if it weighs almost twice as much, and K₂SO₄, a strong electrolyte, gives fewer particles than it should. Both are the same idea: colligative properties count particles. In this post I will show you how often UPSC asks it, which book to use, how I plan the answer, and how it should look in your booklet.

Day 09 question (Daily Answer Writing, 12 November 2026), 10 marks

(a) 1.00 g of benzoic acid dissolved in 50.0 g of benzene lowers the freezing point by 0.48 K. Calculate the apparent molar mass and the degree of dimerisation of benzoic acid (K_f of benzene = 5.12 K kg mol⁻¹; C₆H₅COOH = 122.12 g mol⁻¹).
(b) A 0.100 mol kg⁻¹ aqueous solution of K₂SO₄ freezes at −0.43 °C. Calculate the van 't Hoff factor and the apparent degree of dissociation (K_f of water = 1.86 K kg mol⁻¹). Why is this less than 100% though K₂SO₄ is a strong electrolyte?

Built on UPSC Combined Geo-Scientist (Main) 2017, Chemistry Paper II, Q5, 10 marks: two aqueous 1:1 electrolyte solutions of 0.1 mol kg⁻¹ freeze at −0.360 °C and −0.208 °C; calculate their van 't Hoff factors, degrees of ionisation and likely nature (K_f = 1.86 K kg mol⁻¹).

Short answer

Colligative properties depend on the number of particles: ΔT_f = i K_f m. (a) M_app = K_fw₂ × 1000/(ΔT_fw₁) = 5.12 × 1.00 × 1000/(0.48 × 50.0) = 213 g mol⁻¹; i = 122.12/213.3 = 0.572. For dimerisation i = 1 − α/2, so α = 0.855 (85.5%): benzoic acid forms an H-bonded cyclic dimer in benzene. (b) i = 0.43/(1.86 × 0.100) = 2.31; for K₂SO₄, ν = 3, so α = (i − 1)/(ν − 1) = 0.66. The salt is fully ionised; the shortfall is due to interionic attraction (ionic atmosphere, ion pairs), so 66% is only an apparent degree and i → 3 on dilution.

Watch the video

Day 09 of my Daily Answer Writing series on YouTube: I solve both parts the way they should go into your booklet, with the particle count that gives i = 1 − α/2.

Video coming soon. It goes up on the VKC YouTube channel and is announced first on Telegram.

A word from me

Most students can plug numbers into ΔT_f = K_fm. Where they lose marks is the step in between: they write i = 1 − α/2 without showing where it comes from, or they say K₂SO₄ is “66% dissociated”. Both are red flags to an examiner. The first one costs you the derivation marks; the second one is simply wrong chemistry, because K₂SO₄ is a strong electrolyte.

So always count particles in one line before you write the formula for i: start with 1 mol, write what is left and what is formed, and add. It works for dimers, trimers, ternary salts and weak acids alike. And for part (b), use the words ionic atmosphere and apparent degree of dissociation; that is the line that separates a 6 from a 9. If something does not add up, write to me on Telegram or at [email protected].

A Geo-Scientist favourite

In Geo-Scientist Mains Paper II, colligative properties came 4 times, for 36 marks, in 2017 and 2018: the van 't Hoff factor and its relation with the degree of ionisation (2017 Q1(g), 6 marks), the van 't Hoff factor of two electrolytes from their freezing points (2017 Q5, 10 marks), the derivation of ΔT_f = K_fm from chemical potential (2017 Q11(a), 10 marks) and a molar mass from vapour-pressure lowering (2018 Q5, 10 marks).

The Geo-Scientist Prelims Chemistry paper asks it every year, usually two or three questions: from 2021 to 2026 there are questions on Raoult's law, elevation of boiling point, osmotic pressure, the van 't Hoff factor and why colligative results deviate (for example 2023 Q64, 2024 Q59, 2026 Q63). In Civil Services and IFoS, part (b) of this question is the Debye–Hückel theory of strong electrolytes from the Electrochemistry unit.

Syllabus and books

For Geo-Scientist, colligative properties are part of the Prelims Chemistry paper and have been asked in Mains Paper II. Part (b) is common to all three exams through this line of the Civil Services and IFoS Paper I syllabus (Geo-Scientist Mains Paper II has the same Debye–Hückel limiting law):

“Debye-Huckel theory of strong electrolytes and Debye-Huckel limiting law for various equilibrium and transport properties …”

And the derivation of ΔT_f = K_fm from chemical potential, which Geo-Scientist asked in 2017, belongs with “phase equilibria and solutions” in Civil Services and IFoS Paper I.

You do not need all of these books. Any one of them is enough for this question.

Book 1: Puri, Sharma and Pathania, Principles of Physical Chemistry, the chapter on dilute solutions and colligative properties, the part on abnormal molar masses and the van 't Hoff factor. Plenty of numericals of exactly this type; if you have this book, this is the one I would use.
Book 2: Atkins' Physical Chemistry, the chapter “Simple mixtures”, colligative properties, for the chemical-potential derivation; and the Debye–Hückel section of the electrolyte chapter for part (b).
Book 3: NCERT Chemistry, Class XII, Part I, Chapter 2 “Solutions”, the section on abnormal molar masses. Enough for the Prelims questions.

How to build the answer

Read the tree from left to right. Every leaf is a point to write, in that order. The highlighted words are the ones the examiner looks for.

Abnormal molar mass: benzoic acid dimerises in benzene; K₂SO₄ dissociates in water

Introduction

Given

w₂ 1.00 g, w₁ 50.0 g, ΔT_f 0.48 K
0.100 m K₂SO₄, ΔT_f 0.43 K

Assumptions

dilute, ideal solution
non-volatile solute
only solvent freezes out

Key law

ΔT_f = i K_f m

Body

(a) Association

M_app = 213 g mol⁻¹
i = 122.12/213 = 0.572
i = 1 − α/2 → α = 0.855
sketch: H-bonded dimer

(b) Dissociation

i = 0.43/0.186 = 2.31
α = (i − 1)/(ν − 1) = 0.66

Why not 100%

interionic attraction, ion pairs
apparent α; g = i/3 = 0.77

Conclusion

Principle

association: i < 1
dissociation: i > 1
i → ν on dilution (Debye–Hückel)

Application

molar mass by cryoscopy
antifreeze, de-icing salt, isotonic saline

Final answer

213 g mol⁻¹, 85.5%; i 2.31, 66%

Pictures that earn marks

Figure 1. The van 't Hoff factor on one line. Association pulls i below 1 (benzoic acid in benzene, 0.572; full dimerisation would be 0.5). Dissociation pushes i above 1, but interionic attraction keeps K₂SO₄ at 2.31 instead of the ideal 3.

Figure 2. Benzoic acid in benzene: two molecules join through two O–H···O hydrogen bonds into a cyclic dimer. Benzene cannot hydrogen-bond to the acid, so the dimer survives.

Figure 3. Why a solute lowers the freezing point. The solution's vapour pressure lies below that of the pure liquid, so it meets the curve of the solid solvent at a lower temperature. The gap is ΔT_f, proportional to the number of solute particles.

Figure 4. Each sulfate ion is surrounded by an ionic atmosphere of K⁺ ions, and some K⁺ and SO₄²⁻ stay together as ion pairs. The ions are all there, but they do not act as independent particles.

What gains marks

The law ΔT_f = i K_f m stated, with the assumptions (dilute, ideal, non-volatile solute).
M_app = 213 g mol⁻¹ worked with units (2 marks).
The particle count for 2A ⇌ A₂ giving i = 1 − α/2, then α = 0.855 (3 marks); the H-bonded dimer drawn (1 mark).
(b) i = 2.31 and α = (i − 1)/(ν − 1) = 0.66 (3 marks).
The reason: interionic attraction, ionic atmosphere, ion pairs; “apparent” degree; i → 3 on dilution (1 mark).

Match the length to the marks

For 5 marks, as in a single part, write the formula, one particle count and the number, about half a page. For 10 marks, as in Geo-Scientist 2017 Q5 and this question, write both parts with the derivation of i and the reason in (b): the two pages below. For 15 marks, add the derivation of ΔT_f = K_fm from the chemical potential of the solvent, as asked in Geo-Scientist 2017 Q11(a).

How it looks in the booklet

This is the full answer in the order of the tree above. Page 1 states the law and solves part (a) with the particle count; page 2 solves part (b), gives the reason, and closes with the principle and one application.

Page 1 of 2 · Law and part (a)ΔTf = iKf m, apparent molar mass, degree of dimerisation

Q.)(a) 1.00 g benzoic acid in 50.0 g benzene: ΔT_f = 0.48 K. Find M_app and degree of dimerisation. (b) 0.100 m K₂SO₄ freezes at −0.43 °C: find i and α; why < 100%? [Day 09], 10M

Ans.Assume a dilute, ideal solution of a non-volatile solute; only pure solvent freezes out. Then

ΔT_f = i K_f m— — (i)

(a) Taking i = 1 gives the apparent molar mass:

M_app = K_f w₂ × 1000 / (ΔT_f w₁) = 5.12 × 1.00 × 1000 / (0.48 × 50.0) = 213 g mol⁻¹ — — (ii)

i = M_normal/M_app = 122.12/213.3 = 0.572

2C₆H₅COOH ⇌ (C₆H₅COOH)₂: from 1 mol, (1 − α) + α/2 mol of particles, so i = 1 − α/2 — — (iii)

α = 2(1 − i) = 2(1 − 0.572) = 0.855, i.e. 85.5% dimerised

Page 2 of 2 · Part (b) and conclusioni of K₂SO₄, apparent α, ionic atmosphere, principle

(b) From (i): i = ΔT_f/(K_f m) = 0.43/(1.86 × 0.100) = 2.31

K₂SO₄ → 2K⁺ + SO₄²⁻, ν = 3; i = 1 + (ν − 1)α:

α = (i − 1)/(ν − 1) = 1.31/2 = 0.66 (66%)

Why not 100%: K₂SO₄ is fully dissociated, but at 0.1 m oppositely charged ions attract each other: each ion has an ionic atmosphere, some form ion pairs KSO₄⁻.

So the ions do not act as fully independent particles; 66% is only an apparent degree (osmotic coefficient g = i/ν = 0.77). On dilution i → 3 (Debye–Hückel).

Conclusion: association lowers the number of particles (i < 1), dissociation raises it (i > 1), and interionic attraction keeps i below ν. Application: molar mass by cryoscopy; ethylene glycol antifreeze, salt on icy roads.

So, M_app = 213 g mol⁻¹, 85.5% dimerised; K₂SO₄: i = 2.31, α(app) = 66%

The same question in other exams

Geo-Scientist 2017, Paper II, Q5, 10 marks: 0.1 mol kg⁻¹ solutions of two 1:1 electrolytes freeze at −0.360 °C and −0.208 °C. i = 0.360/0.186 = 1.94 and 0.208/0.186 = 1.12; α ≈ 94% and 12%: A behaves as a strong electrolyte, B as a weak one.

Geo-Scientist 2017, Paper II, Q1(g), 6 marks: define the van 't Hoff factor and relate it to the degree of ionisation: from 1 mol, (1 − α) + να particles, so i = 1 + (ν − 1)α.

Geo-Scientist 2018, Paper II, Q5, 10 marks: vapour pressure of benzene falls from 53.3 to 51.5 kPa when 19.0 g of solute is dissolved in 500 g. Raoult's law gives x₂ = 1.8/53.3 and M ≈ 85 g mol⁻¹. Same particle-counting idea, another colligative property.

Geo-Scientist Prelims 2026, Q63: order of osmotic pressure of 0.01 mol dm⁻³ acetic acid in benzene, acetic acid in water, NaCl and glucose: the dimerised acid in benzene has the fewest particles and NaCl the most.

Geo-Scientist Prelims 2023, Q64: the largest elevation of boiling point comes from 2 mol Ca(OH)₂ (6 mol of ions), ahead of 2 mol MgSO₄ or NaCl (4 mol) and 1 mol CaCl₂ (3 mol).

Every version is the same two steps: count the particles that one formula unit really gives (association, dissociation or neither), then put i into the colligative formula. If the solute is a strong electrolyte and i still falls short, the reason is interionic attraction, never incomplete dissociation.

Questions students ask

What is the van 't Hoff factor?

The van 't Hoff factor i is the ratio of the observed colligative effect to the effect calculated for the same amount of a solute that neither associates nor dissociates. Equivalently, i = normal molar mass / apparent molar mass. For a solute that dissociates into ν ions, i = 1 + (ν − 1)α; for association of n molecules into one, i = 1 − α(1 − 1/n).

Why does benzoic acid show a higher molar mass in benzene?

In a non-polar solvent like benzene, two benzoic acid molecules join through two O–H···O hydrogen bonds into a cyclic dimer. Fewer particles means a smaller freezing-point depression, so the apparent molar mass comes out higher than 122. Complete dimerisation would give 244 g/mol; the data in this question give 213 g/mol, that is 85.5% dimerised.

How do you find the degree of dimerisation from the van 't Hoff factor?

For 2A ⇌ A2, one mole of A gives (1 − α) mol of monomer and α/2 mol of dimer, so i = 1 − α/2 and α = 2(1 − i). With i = 122.12/213.3 = 0.572, α = 0.855.

Why is the apparent degree of dissociation of K2SO4 less than 100%?

K2SO4 is a strong electrolyte and is completely ionised, but at 0.1 mol/kg the K+ and SO4 2− ions attract each other. Each ion is surrounded by an ionic atmosphere of opposite charge and some form ion pairs such as KSO4−, so the ions do not behave as fully independent particles. The value from freezing point, i = 2.31, gives an apparent α of 66%; i approaches 3 on dilution, as Debye–Hückel theory predicts.

What is the formula for apparent molar mass from freezing point depression?

M(apparent) = Kf × w2 × 1000 / (ΔTf × w1), with w2 the mass of solute and w1 the mass of solvent in grams, and Kf in K kg/mol.

How often does UPSC ask colligative properties and the van 't Hoff factor?

Geo-Scientist Mains Paper II asked it 4 times for 36 marks in 2017 and 2018, including the van 't Hoff factor from freezing points. The Geo-Scientist Prelims Chemistry paper asks two to three questions on solutions and colligative properties every year from 2021 to 2026.

Read the complete answers

Geo-Scientist 2017 Q5 solution Geo-Scientist 2017 Q1(g) solution Notes: solutions and colligative properties Prelims notes: solutions Day 09 question and model answer Join on Telegram

By VKC, your mentor. Integrated M.Sc. Chemistry, IIT Roorkee. CSIR NET Chemical Sciences, All India Rank 4. New posts are announced first on Telegram.