ChemistryISC Class 12 · Solutions

Abnormal Molar Mass and van't Hoff Factor for ISC Class 12

Updated 15 October 2026 · Fact-checked

A molar mass found from a colligative property is abnormal when the solute dissociates or associates in solution. The van't Hoff factor i is normal molar mass divided by observed molar mass. Dissociation gives i greater than 1 and a low molar mass; association gives i less than 1 and a high molar mass.

More particles (dissociation) give i > 1 and a lower observed molar mass; fewer particles (association) give i < 1 and a higher one.

Illustration drawn for QuestionBuddy.

Understand Abnormal Molar Mass and van't Hoff Factor

Colligative properties count particles. The formulas $Δ T_{f} = K_{f} m$ , $Δ T_{b} = K_{b} m$ and $π = C R T$ assume that every formula unit of solute stays as one particle. When that is true, the molar mass found from the measurement is the normal (true) molar mass. When it is not, the molar mass that comes out is abnormal.

Dissociation. An electrolyte such as KCl splits into K⁺ and Cl⁻ in water. One mole of KCl added gives nearly two moles of particles, so the depression of freezing point is nearly double the value expected for one mole of particles. A student who assumes no dissociation divides the mass by roughly twice the true number of moles, and gets a molar mass of nearly 37.25 g mol⁻¹ instead of 74.5 g mol⁻¹. Dissociation always makes the observed molar mass lower than the true value.

Association. Ethanoic acid and benzoic acid form hydrogen-bonded dimers in benzene, a solvent of low dielectric constant: $2 C H_{3} C O O H ⇌ (C H_{3} C O O H)_{2}$ . Two molecules become one particle, so the number of particles falls. If dimerisation were complete, $Δ T_{f}$ would be half the normal value and the observed molar mass would be double. Association always makes the observed molar mass higher than the true value.

The van't Hoff factor. In 1880 van't Hoff introduced $i$ to correct for this. It can be written three equivalent ways:

$i = \frac{normal molar mass}{observed molar mass} = \frac{observed colligative property}{calculated colligative property} = \frac{moles of particles after}{moles of particles before}$ .

For dissociation $i > 1$ (aqueous KCl is close to 2); for association $i < 1$ (ethanoic acid in benzene is close to 0.5); for a non-electrolyte such as glucose or urea, $i = 1$ . The corrected formulas are $Δ T_{f} = i K_{f} m$ , $Δ T_{b} = i K_{b} m$ , $π = i C R T$ , and for relative lowering of vapour pressure $\frac{p ^{0} - p}{p ^{0}} = i \frac{n _{2}}{n _{1}}$ (dilute).

Degree of dissociation. If one formula unit gives $n$ ions and a fraction $α$ dissociates, then from 1 mol we have $(1 - α) + n α$ mol of particles. So $i = 1 + (n - 1) α$ and $α = \frac{i - 1}{n - 1}$ . Count $n$ carefully: NaCl 2, BaCl₂ and K₂SO₄ 3, AlCl₃ and K₃[Fe(CN)₆] 4, K₄[Fe(CN)₆] and Al₂(SO₄)₃ 5.

Degree of association. If $n$ molecules join to form one particle and a fraction $α$ associates, then from 1 mol we have $(1 - α) + \frac{α}{n}$ mol of particles. So $i = 1 - α (1 - \frac{1}{n})$ and $α = \frac{1 - i}{1 - \frac{1}{n}}$ . For a dimer ( $n = 2$ ), $α = 2 (1 - i)$ .

A practical route for numericals: find the observed molar mass (or the observed colligative property), divide to get $i$ , then use the right degree formula. The same solute can behave both ways: ethanoic acid dissociates slightly in water ( $i$ just above 1) but dimerises in benzene ( $i$ near 0.5).

Abnormal Molar Mass and van't Hoff Factor: formulas and key facts

van't Hoff factor: $i = \frac{M _{n or ma l}}{M _{o b ser v e d}} = \frac{observed colligative property}{calculated colligative property}$; Calculated values assume no dissociation or association.
Modified colligative formulas: $Δ T_{f} = i K_{f} m, Δ T_{b} = i K_{b} m, π = i C R T$; Relative lowering: $\frac{p ^{0} - p}{p ^{0}} = i \frac{n _{2}}{n _{1}}$ for dilute solutions.
Degree of dissociation: $α = \frac{i - 1}{n - 1}$; $n$ = number of ions from one formula unit (NaCl 2, CaCl₂ 3, K₄[Fe(CN)₆] 5).
Degree of association: $α = \frac{1 - i}{1 - \frac{1}{n}}$; $n$ = number of molecules that join; for a dimer $α = 2 (1 - i)$ .
Observed molar mass from freezing point: $M_{o b ser v e d} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}}$; $w_{1}$ is the mass of solvent in grams.

How to solve Abnormal Molar Mass and van't Hoff Factor questions

Every van't Hoff factor question in ISC runs through the same three moves: get $i$ , choose dissociation or association, then apply the matching degree formula.

1Find $i$ . Either compute the observed molar mass and divide: $i = M_{n or ma l} / M_{o b ser v e d}$ ; or compute the 'normal' colligative value (with $i = 1$ ) and divide the observed value by it.
2Check the sign: $i > 1$ means dissociation; $i < 1$ means association. If the answer contradicts the chemistry (a salt with $i < 1$ ), recheck your arithmetic.
3For dissociation, write the ionisation equation and count $n$ , the ions per formula unit. Then $α = \frac{i - 1}{n - 1}$ .
4For association, write the equilibrium ( $2 A ⇌ A_{2}$ ) and note $n$ . Then $α = \frac{1 - i}{1 - 1/ n}$ .
5When predicting a property, use $Δ T = i K m$ or $π = i C R T$ with $i = 1 + (n - 1) α$ (dissociation).
6Give $α$ as a fraction and as a percentage; state the unit of any molar mass (g mol⁻¹).

Abnormal Molar Mass and van't Hoff Factor in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

More particles per gram give a bigger freezing-point drop

Suppose a hill-town council must choose between NaCl and CaCl₂ to keep the water on a road from freezing, and wants the larger drop in freezing point for each 100 g of salt dissolved in 1 kg of water. Assume both salts dissociate completely and the dilute-solution formula holds ( $K_{f} = 1.86$ K kg mol⁻¹; NaCl 58.5, CaCl₂ 111 g mol⁻¹).

Calculate $Δ T_{f}$ for 100 g of each salt in 1 kg of water, and say which salt gives more effect per gram.

Try it first, then show the solutionSolution

NaCl: $m = \frac{100}{58.5} = 1.71$ mol kg⁻¹, $i = 2$ , $Δ T_{f} = 2 \times 1.86 \times 1.71 = 6.36$ K.
CaCl₂: $m = \frac{100}{111} = 0.901$ mol kg⁻¹, $i = 3$ , $Δ T_{f} = 3 \times 1.86 \times 0.901 = 5.03$ K.
Per gram, NaCl gives $\frac{2}{58.5} = 0.0342$ mol of particles, CaCl₂ only $\frac{3}{111} = 0.0270$ mol.
So NaCl lowers the freezing point more for the same mass, even though each CaCl₂ unit gives more ions.

Answer: NaCl about 6.4 K, CaCl₂ about 5.0 K. NaCl gives more particles per gram, so a larger drop per 100 g (in this ideal model).

Memory hook: Colligative properties count particles, so compare moles of ions per gram, not ions per formula.

Tricks to remember Abnormal Molar Mass and van't Hoff Factor

Short memory aids. Say each one aloud once, then use it on a question the same day.

Break up, mass goes down
Dissociation breaks a solute into more particles, so the observed molar mass goes DOWN and $i$ goes UP. Association does the reverse.
Use it when: Deciding which degree formula to use.
i minus one over n minus one
Dissociation: ' $i$ minus one over $n$ minus one'. Association: 'one minus $i$ over one minus one-by- $n$ '. The '1 − i' keeps $α$ positive when $i < 1$ .
Use it when: Writing the formula line correctly.
Dimer means double, half
Complete dimerisation: molar mass doubles, $i$ halves to 0.5, and $α = 2 (1 - i)$ .
Use it when: Benzoic or ethanoic acid in benzene.

Abnormal Molar Mass and van't Hoff Factor: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Largest van't Hoff factor · 1 mark · MCQBoard level

Assuming complete dissociation, which solute has the largest van't Hoff factor in dilute aqueous solution?
(a) K₃[Fe(CN)₆]
(b) Al₂(SO₄)₃
(c) MgSO₄
(d) urea

Try it first, then show the solutionSolution

K₃[Fe(CN)₆] → 3K⁺ + [Fe(CN)₆]³⁻: 4 ions.
Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻: 5 ions.
MgSO₄: 2 ions; urea: non-electrolyte, $i = 1$ .

Answer: (b) Al₂(SO₄)₃, $i = 5$ .

Example 2: Degree of dissociation of KCl · 2 marks · Short answerBoard level

A 0.10 mol kg⁻¹ aqueous solution of KCl freezes at −0.344 °C. Calculate the van't Hoff factor and the degree of dissociation of KCl. ( $K_{f}$ for water $= 1.86$ K kg mol⁻¹)

Try it first, then show the solutionSolution

Calculated $Δ T_{f}$ (no dissociation) $= K_{f} m = 1.86 \times 0.10 = 0.186$ K; observed $= 0.344$ K (½ mark).
$i = \frac{0.344}{0.186} = 1.85$ (½ mark).
KCl → K⁺ + Cl⁻, $n = 2$ : $α = \frac{i - 1}{n - 1} = \frac{0.85}{1} = 0.85$ (1 mark).

Answer: $i \approx 1.85$ ; $α \approx 0.85$ (85% dissociated).

Example 3: Association of benzoic acid in benzene · 3 marks · Short answerHarder board level

0.976 g of benzoic acid ( $C_{6} H_{5} C O O H$ , $M = 122$ g mol⁻¹) is dissolved in 25.0 g of benzene. The freezing point falls by 0.90 K. Calculate the observed molar mass, the van't Hoff factor and the percentage of benzoic acid that has dimerised. ( $K_{f}$ for benzene $= 5.12$ K kg mol⁻¹)

Try it first, then show the solutionSolution

$M_{o b s} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}} = \frac{1000 \times 5.12 \times 0.976}{0.90 \times 25.0} = 222$ g mol⁻¹ (1 mark).
$i = \frac{122}{222} = 0.549$ (½ mark).
$2 C_{6} H_{5} C O O H ⇌ (C_{6} H_{5} C O O H)_{2}$ , $n = 2$ : $α = \frac{1 - i}{1 - \frac{1}{2}} = 2 (1 - 0.549)$ (1 mark).
$α = 0.901$ , i.e. about 90% dimerised (½ mark).

Answer: $M_{o b s} \approx 222$ g mol⁻¹; $i \approx 0.549$ ; about 90% of the acid is dimerised.

Example 4: From boiling point to freezing point · 3 marks · Short answerHarder board level

A salt of formula MX₂ dissociates partly in water. A 0.020 mol kg⁻¹ aqueous solution boils 0.026 K above pure water. (a) Calculate the van't Hoff factor and the degree of dissociation. (b) Predict the freezing point of the same solution. ( $K_{b} = 0.52$ , $K_{f} = 1.86$ K kg mol⁻¹)

Try it first, then show the solutionSolution

(a) $i = \frac{Δ T _{b}}{K _{b} m} = \frac{0.026}{0.52 \times 0.020} = 2.5$ (1 mark).
MX₂ → M²⁺ + 2X⁻, $n = 3$ : $α = \frac{2.5 - 1}{3 - 1} = 0.75$ (1 mark).
(b) $Δ T_{f} = i K_{f} m = 2.5 \times 1.86 \times 0.020 = 0.093$ K, so the solution freezes at −0.093 °C (1 mark).

Answer: (a) $i = 2.5$ , $α = 0.75$ (75%). (b) About −0.093 °C.

Example 5: One acid, two solvents · 5 marks · Long answerHOTS / competency

0.60 g of ethanoic acid ( $C H_{3} C O O H$ , $M = 60$ g mol⁻¹) is dissolved separately in 50.0 g of water and in 50.0 g of benzene.
(a) The aqueous solution freezes 0.379 K below pure water. Calculate $i$ and the degree of dissociation. ( $K_{f}$ water $= 1.86$ K kg mol⁻¹) [2]
(b) The benzene solution shows a freezing-point depression of 0.55 K. Calculate $i$ , the observed molar mass and the degree of association. ( $K_{f}$ benzene $= 5.12$ K kg mol⁻¹) [2]
(c) Explain why the same acid gives $i > 1$ in one solvent and $i < 1$ in the other. [1]

Try it first, then show the solutionSolution

Molality in both solvents: $m = \frac{0.60/60}{0.0500} = 0.200$ mol kg⁻¹.
(a) Calculated $Δ T_{f} = 1.86 \times 0.200 = 0.372$ K, so $i = \frac{0.379}{0.372} = 1.019$ ; $α = \frac{i - 1}{2 - 1} = 0.019$ , about 1.9% dissociated (2 marks).
(b) Calculated $Δ T_{f} = 5.12 \times 0.200 = 1.024$ K, so $i = \frac{0.55}{1.024} = 0.537$ ; $M_{o b s} = \frac{60}{0.537} = 112$ g mol⁻¹; $α = 2 (1 - 0.537) = 0.926$ , about 93% dimerised (2 marks).
(c) Water has a high dielectric constant and solvates ions, so the weak acid ionises slightly; benzene has a low dielectric constant, cannot support ions, and the acid molecules hydrogen-bond in pairs to form dimers (1 mark).

Answer: (a) $i \approx 1.02$ , $α \approx 0.019$ . (b) $i \approx 0.537$ , $M_{o b s} \approx 112$ g mol⁻¹, about 93% associated. (c) Slight ionisation in polar water; hydrogen-bonded dimers in non-polar benzene.

Where students lose marks in Abnormal Molar Mass and van't Hoff Factor

The slips examiners see most often, and what to write instead.

Mistake: Wrong number of ions, $n$
ISC examiners reported wrong values of $n$ for BaCl₂ in a degree-of-dissociation question (2024).
Wrong working
BaCl₂: $n = 2$ , $α = \frac{i - 1}{1}$
Right working
BaCl₂ → Ba²⁺ + 2Cl⁻, $n = 3$ , $α = \frac{i - 1}{2}$
Do this: Always write the dissociation equation first and count every ion on the right.
Usually costs: 1 to 2 marks.
Mistake: Using the dissociation formula for association
Examiners noted wrong formulas for degree of association, or direct answers with no formula (2025).
Wrong working
Benzoic acid dimer, $i = 0.5$ : $α = \frac{i - 1}{n - 1} = - 0.5$
Right working
$α = \frac{1 - i}{1 - \frac{1}{n}} = \frac{0.5}{0.5} = 1$
Do this: If $i < 1$ , switch to the association formula. A negative $α$ is a signal you used the wrong one.
Usually costs: The method mark and the answer.
Mistake: Assuming an associated solute is 'normal' or 'dissociated'
In a 2024 MCQ, candidates ignored that a smaller depression for an equimolal solute means $i < 1$ .
Do this: Compare the observed colligative effect with the expected one: smaller than expected means association; larger means dissociation.
Usually costs: 1 mark.
Mistake: Arithmetic slips in $i$
Examiners asked candidates to recheck calculations; many had the right formula but the wrong substitution.
Do this: Write $i$ to three significant figures and check that $α$ lies between 0 and 1.
Usually costs: 1 mark.

Exam tips

Start every answer by writing the dissociation or association equation; it fixes $n$ and earns credit.
Show $i$ as a separate line before $α$ ; ISC answer keys award a mark for $i$ itself.
For complex salts, count the complex ion as one particle: K₄[Fe(CN)₆] gives 4K⁺ + [Fe(CN)₆]⁴⁻, so $n = 5$ .
If a question gives the percentage dissociation, convert it to $α$ (25% = 0.25) and use $i = 1 + (n - 1) α$ .
Remember the direction: dissociation lowers the observed molar mass; association raises it.

Practice questions on Abnormal Molar Mass and van't Hoff Factor

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A solute trimerises completely in a solvent ( $3 A ⇌ A_{3}$ ). Its van't Hoff factor is:
(a) 3
(b) 1
(c) $\frac{1}{3}$
(d) $\frac{2}{3}$
Hint
Three particles become one.
Answer
(c) $\frac{1}{3}$ .
Q2 · 2 marks · Short answerBoard level
What is meant by abnormal molar mass? Give one example of a solute that shows a lower and one that shows a higher molar mass than normal.
Hint
Think of KCl in water and ethanoic acid in benzene.
Answer
A molar mass found from a colligative property that differs from the true value because the solute dissociates or associates in solution. Lower: KCl in water (dissociates into K⁺ and Cl⁻). Higher: ethanoic acid (or benzoic acid) in benzene (forms dimers).
Q3 · 2 marks · Short answerHarder board level
0.87 g of K₂SO₄ ( $M = 174$ g mol⁻¹) is dissolved in 250 g of water. Assuming complete dissociation, calculate the depression in freezing point. ( $K_{f} = 1.86$ K kg mol⁻¹)
Hint
K₂SO₄ gives three ions.
Answer
$m = \frac{0.87/174}{0.250} = 0.020$ mol kg⁻¹; $i = 3$ ; $Δ T_{f} = 3 \times 1.86 \times 0.020 = 0.112$ K.
Q4 · 3 marks · Short answerHarder board level
The molar mass of calcium nitrate, Ca(NO₃)₂ (normal value 164 g mol⁻¹), found from the elevation of boiling point of its dilute aqueous solution is 70.0 g mol⁻¹. Calculate the van't Hoff factor and the degree of dissociation.
Hint
Ca(NO₃)₂ → Ca²⁺ + 2NO₃⁻.
Answer
$i = \frac{164}{70.0} = 2.34$ . $n = 3$ , so $α = \frac{2.34 - 1}{2} = 0.671$ , i.e. about 67% dissociated.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The molar mass of KCl determined by measuring the depression in freezing point of its aqueous solution is less than 74.5 g mol⁻¹.
Reason (R): KCl dissociates in water, so the number of solute particles is greater than the number of formula units dissolved.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
More particles mean a bigger depression and a smaller calculated molar mass.
Answer
(a).
Q6 · 2 marks · Short answerHarder board level
Arrange the following aqueous solutions in decreasing order of depression in freezing point, assuming complete dissociation of electrolytes: 0.05 m urea, 0.02 m AlCl₃, 0.03 m MgCl₂, 0.035 m KCl. Show your working.
Hint
Compare $i \times m$ .
Answer
$im$ : urea 0.05; AlCl₃ $4 \times 0.02 = 0.08$ ; MgCl₂ $3 \times 0.03 = 0.09$ ; KCl $2 \times 0.035 = 0.07$ . Order: MgCl₂ > AlCl₃ > KCl > urea.
Q7 · 3 marks · Short answerHOTS / competency
Equimolal solutions of a non-electrolyte A and a salt of type BC₂ in water show depressions of freezing point in the ratio 1 : 2.6. Calculate the degree of dissociation of BC₂. Would the ratio ever exceed 1 : 3? Explain.
Hint
For equal molality, the ratio of $Δ T_{f}$ equals the ratio of $i$ .
Answer
$i_{A} = 1$ , so $i_{B C_{2}} = 2.6$ . BC₂ → B²⁺ + 2C⁻, $n = 3$ : $α = \frac{2.6 - 1}{2} = 0.8$ (80%). The ratio cannot exceed 1 : 3 because complete dissociation gives at most 3 particles per formula unit (in this ideal model).
Q8 · 3 marks · Short answerHOTS / competency
A 0.10 mol kg⁻¹ aqueous solution of a weak monobasic acid HA freezes at −0.2046 °C. (a) Calculate $i$ and the degree of dissociation. (b) Taking molarity equal to molality, estimate its osmotic pressure at 300 K. ( $K_{f} = 1.86$ K kg mol⁻¹, $R = 0.0821$ L atm K⁻¹ mol⁻¹)
Hint
HA ⇌ H⁺ + A⁻ gives $n = 2$ .
Answer
(a) $i = \frac{0.2046}{1.86 \times 0.10} = 1.10$ ; $α = 0.10$ (10%). (b) $π = i C R T = 1.10 \times 0.10 \times 0.0821 \times 300 = 2.71$ atm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define the van't Hoff factor in terms of molar mass.Show answer
$i = \frac{normal molar mass}{observed (abnormal) molar mass}$ .
2.Value of $i$ for ethanoic acid fully dimerised in benzene?Show answer
0.5.
3.Formula for degree of dissociation?Show answer
$α = \frac{i - 1}{n - 1}$ .
4.Formula for degree of association?Show answer
$α = \frac{1 - i}{1 - 1/ n}$ .
5.What is $n$ for K₄[Fe(CN)₆]?Show answer
5 (four K⁺ and one [Fe(CN)₆]⁴⁻).
6.Why do carboxylic acids associate in benzene?Show answer
They form hydrogen-bonded dimers; benzene has a low dielectric constant and does not separate them.

Abnormal Molar Mass and van't Hoff Factor: frequently asked questions

What is the van't Hoff factor in ISC Class 12 Chemistry?

It is the ratio of the normal molar mass to the observed molar mass, or of the observed colligative property to the value calculated assuming no dissociation or association.

Why is the observed molar mass of NaCl less than 58.5 g mol⁻¹?

NaCl dissociates into Na⁺ and Cl⁻, so there are more particles than formula units. The colligative effect is larger, and dividing the mass by the larger apparent number of moles gives a smaller molar mass.

Can the van't Hoff factor be less than 1?

Yes. When solute molecules associate, as ethanoic acid does in benzene, the number of particles falls and i is less than 1.

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

Use \alpha = \frac{i - 1}{n - 1}, where n is the number of ions one formula unit gives on complete dissociation.

What is i for a non-electrolyte such as glucose?

1, because glucose neither dissociates nor associates in water.