ChemistryCBSE Class 12 · Solutions

Abnormal Molar Mass and the van't Hoff Factor

Q: What is abnormal molar mass in Class 12 Chemistry?

It is a molar mass, found from a colligative property, that is higher or lower than the formula value because the solute associates or dissociates in solution.

Q: What is the van't Hoff factor?

It is the ratio of the normal molar mass to the abnormal molar mass, or of the observed colligative property to the calculated one. It tells how many particles each formula unit effectively gives in solution.

Q: Why is $i$ for a strong electrolyte slightly less than the number of ions?

At ordinary concentrations, oppositely charged ions attract each other and some move together, so the solution behaves as if it had slightly fewer particles. The value approaches n on dilution.

Q: How do I find the degree of dissociation from $i$?

Use \alpha = \frac{i-1}{n-1}, where n is the number of ions produced by one formula unit.

Q: Does association increase or decrease the observed molar mass?

It increases it, because fewer particles give a smaller colligative effect.

Updated 15 October 2026 · Fact-checked

When a solute dissociates into ions or associates into dimers, the number of particles changes, so colligative properties and the molar mass worked out from them come out abnormal. The van't Hoff factor $i$ corrects this: $i$ = normal molar mass ÷ abnormal molar mass. Dissociation gives $i > 1$; association gives $i < 1$.

Understand Abnormal Molar Mass and van't Hoff Factor

Colligative properties count particles. The formulas $Δ T_{b} = K_{b} m$ , $Δ T_{f} = K_{f} m$ and $π = C R T$ assume that every formula unit of solute stays as one particle. That is true for glucose or urea in water, but not for every solute.

Dissociation. Potassium chloride in water splits into $K^{+}$ and $C l^{-}$ . One mole of KCl gives up to two moles of particles, so the freezing point falls almost twice as much as for the same molality of glucose. If you then use the ordinary formula to calculate molar mass, the larger $Δ T_{f}$ makes the molar mass come out about half the true value (around 37 instead of 74.5 g mol⁻¹).

Association. Ethanoic acid or benzoic acid in benzene forms dimers held by two hydrogen bonds. Two molecules behave as one particle, so the depression is roughly halved and the calculated molar mass is about double the true value.

A molar mass that is higher or lower than the formula value is called an abnormal molar mass. In 1880 van't Hoff introduced a factor $i$ to account for it:

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

The corrected formulas are $\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = i x_{2}$ , $Δ T_{b} = i K_{b} m$ , $Δ T_{f} = i K_{f} m$ and $π = i C R T$ .

Getting the degree of dissociation. If one formula unit can give $n$ ions and a fraction $α$ dissociates, then for each mole $(1 - α)$ mol stays whole and $n α$ mol of ions form, so $i = 1 + (n - 1) α$ and $α = \frac{i - 1}{n - 1}$ . For KCl, $n = 2$ ; for $K_{2} S O_{4}$ , $n = 3$ ; for $K_{4} [F e (C N)_{6}]$ , $n = 5$ .

Getting the degree of association. If $n$ molecules join into one, $i = 1 - α + \frac{α}{n}$ ; for dimerisation ( $n = 2$ ), $i = 1 - \frac{α}{2}$ and $α = 2 (1 - i)$ .

For strong electrolytes in dilute solution, $i$ is close to the number of ions; at higher concentrations inter-ionic attraction makes it a little smaller. A weak electrolyte such as ethanoic acid in water has $i$ only slightly above 1. The same idea lets you count ions in a complex: the freezing point of a dilute solution tells you how many particles each formula unit releases.

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

van't Hoff factor: $i = \frac{normal molar mass}{abnormal molar mass} = \frac{observed colligative property}{calculated colligative property}$; Also equals total moles of particles after association or dissociation divided by moles before.
Modified colligative formulas: $Δ T_{b} = i K_{b} m, Δ T_{f} = i K_{f} m, π = i C R T, \frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = i x_{2}$; Use $i = 1$ for non-electrolytes.
Degree of dissociation: $i = 1 + (n - 1) α \Rightarrow α = \frac{i - 1}{n - 1}$; $n$ = number of ions from one formula unit (KCl 2, $K_{2} S O_{4}$ 3, $A l C l_{3}$ 4, $A l_{2} (S O_{4})_{3}$ 5).
Degree of association: $i = 1 - α + \frac{α}{n} \Rightarrow α = \frac{1 - i}{1 - \frac{1}{n}}$; For dimerisation ( $n = 2$ ): $α = 2 (1 - i)$ .
Limits of i: $complete dimerisation: i = 0.5; complete dissociation: i = n$; Real strong electrolytes give $i$ a little below $n$ except at very low concentration.

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

Every numerical in this topic asks you to go from a measured colligative property to $i$ , and then to $α$ (or the other way). Keep the steps separate so each earns its mark.

1Write the dissociation or association equation and note $n$ (ions per unit, or molecules per associated particle).
2Calculate the 'normal' value of the property for $i = 1$ , e.g. $Δ T_{f} (calc) = K_{f} m$ , or the normal molar mass from the formula.
3Find the observed value, or the abnormal molar mass $M_{o b s} = \frac{1000 K w _{2}}{Δ T w _{1}}$ .
4Divide: $i = \frac{observed property}{calculated property}$ or $i = \frac{M _{n or ma l}}{M _{o b s}}$ .
5Convert $i$ to $α$ with $α = \frac{i - 1}{n - 1}$ (dissociation) or $α = \frac{1 - i}{1 - 1/ n}$ (association); give it as a fraction or percentage.
6If $i$ is given (or assumed for complete dissociation), go forward instead: $Δ T = i K m$ or $π = i C R T$ .
7Sense-check: dissociation must give $i > 1$ and a smaller observed molar mass; association must give $i < 1$ and a larger one.

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.

Salt raises the boiling point twice as much as an equal number of moles of sugar

Suppose a cook adds 58.5 g of common salt (NaCl, 58.5 g mol⁻¹) to 2.0 kg of water before boiling pasta and claims this makes the water 'much hotter'.

Assuming complete dissociation, calculate the rise in boiling point ( $K_{b}$ = 0.52 K kg mol⁻¹). Would 342 g of sugar (1 mol) in the same water give the same rise?

Try it first, then show the solutionSolution

Moles NaCl $= 58.5/58.5 = 1.0$ mol; $m = 1.0/2.0 = 0.50$ mol kg⁻¹.
NaCl gives $N a^{+} + C l^{-}$ , so $i = 2$ .
$Δ T_{b} = i K_{b} m = 2 \times 0.52 \times 0.50 = 0.52$ K.
One mole of sugar gives one mole of particles ( $i = 1$ ), so $Δ T_{b} = 0.52 \times 0.50 = 0.26$ K, half as much.

Answer: About 0.52 K, so the water boils at about 100.5 °C; the same number of moles of sugar would give only 0.26 K.

Memory hook: Salt splits into two particles, so it counts twice: that doubling is the van't Hoff factor.

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.

Split up, i goes up
Dissociation splits particles: $i > 1$ , molar mass looks smaller. Association sticks them together: $i < 1$ , molar mass looks bigger.
Use it when: Predicts the direction before you calculate.
Normal over abnormal
'N over A' like a fraction in the alphabet order: $i = M_{n or ma l} / M_{abn or ma l}$ .
Use it when: Avoids inverting the ratio.
Count the ions with charges
Write the ions with their charges; the number of ions is $n$ . $K_{2} S O_{4} \to 2 K^{+} + S O_{4}^{2 -}$ gives $n = 3$ .
Use it when: Prevents counting atoms instead of ions.

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: Lowest freezing point · 1 mark · MCQBoard level

Which of the following 0.1 m aqueous solutions has the lowest freezing point (assume complete dissociation of electrolytes)?
(a) Glucose (b) NaCl (c) $C a C l_{2}$ (d) $A l_{2} (S O_{4})_{3}$

Try it first, then show the solutionSolution

$i$ : glucose 1, NaCl 2, $C a C l_{2}$ 3, $A l_{2} (S O_{4})_{3} \to 2 A l^{3 +} + 3 S O_{4}^{2 -}$ gives 5.
Largest $i$ gives the largest $Δ T_{f}$ and the lowest freezing point.

Answer: (d) $A l_{2} (S O_{4})_{3}$ ( $i = 5$ ).

Example 2: Freezing point of a salt solution · 2 marks · Very short answerBoard level

Calculate the freezing point of a 0.10 m aqueous solution of $K_{2} S O_{4}$ , assuming complete dissociation. ( $K_{f}$ for water = 1.86 K kg mol⁻¹)

Try it first, then show the solutionSolution

$K_{2} S O_{4} \to 2 K^{+} + S O_{4}^{2 -}$ , so $i = 3$ (½ mark).
$Δ T_{f} = i K_{f} m = 3 \times 1.86 \times 0.10 = 0.558$ K (1 mark).
Freezing point $= 0 - 0.558 = - 0.558$ °C (272.59 K) (½ mark).

Answer: −0.558 °C (about 272.59 K).

Example 3: Degree of dissociation of KCl · 3 marks · Short answerHarder board level

A solution of 0.50 g of KCl (74.5 g mol⁻¹) in 100 g of water freezes at −0.24 °C. Calculate the van't Hoff factor and the degree of dissociation of KCl. ( $K_{f}$ = 1.86 K kg mol⁻¹)

Try it first, then show the solutionSolution

$m = \frac{0.50/74.5}{0.100} = 0.0671$ mol kg⁻¹; calculated $Δ T_{f} = 1.86 \times 0.0671 = 0.125$ K (1 mark).
$i = \frac{0.24}{0.125} = 1.92$ (1 mark).
$K C l \to K^{+} + C l^{-}$ , $n = 2$ : $α = \frac{i - 1}{n - 1} = \frac{0.92}{1} = 0.92$ , i.e. 92% (1 mark).

Answer: $i \approx 1.92$ ; degree of dissociation about 0.92 (92%).

Example 4: Association of benzoic acid in benzene · 3 marks · Short answerHarder board level

3.05 g of benzoic acid ( $C_{6} H_{5} C O O H$ , 122 g mol⁻¹) dissolved in 50 g of benzene lowers the freezing point by 1.40 K. $K_{f}$ for benzene is 5.12 K kg mol⁻¹. Calculate (a) the observed molar mass, (b) the van't Hoff factor, and (c) the percentage of benzoic acid that has dimerised.

Try it first, then show the solutionSolution

(a) $M_{o b s} = \frac{1000 \times 5.12 \times 3.05}{1.40 \times 50} = 223$ g mol⁻¹ (1 mark).
(b) $i = \frac{122}{223} = 0.547$ (1 mark).
(c) $2 C_{6} H_{5} C O O H ⇌ (C_{6} H_{5} C O O H)_{2}$ , $i = 1 - \frac{α}{2}$ , so $α = 2 (1 - 0.547) = 0.906$ : about 91% dimerised (1 mark).

Answer: (a) About 223 g mol⁻¹. (b) $i \approx 0.547$ . (c) About 91% associated.

Example 5: Counting ions in a hydrate isomer · 3 marks · Short answerHOTS / competency

Chromium(III) chloride hexahydrate, $C r C l_{3} \cdot 6 H_{2} O$ , can exist as $[C r (H_{2} O)_{6}] C l_{3}$ , $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$ or $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O$ . A 0.010 m aqueous solution of one sample freezes at −0.0558 °C. Treating the complex ion as stable and the outer chlorides as fully released, identify the sample. ( $K_{f}$ = 1.86 K kg mol⁻¹)

Try it first, then show the solutionSolution

Calculated $Δ T_{f}$ for $i = 1$ : $1.86 \times 0.010 = 0.0186$ K (½ mark).
$i = \frac{0.0558}{0.0186} = 3.0$ (1 mark).
Particles per formula unit: $[C r (H_{2} O)_{6}] C l_{3}$ gives 4, $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$ gives 3 ( $[C r (H_{2} O)_{5} C l]^{2 +} + 2 C l^{-}$ ), and $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O$ gives 2 (1 mark).
So the sample is $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$ (½ mark).

Answer: $[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O$ , which releases three ions ( $i = 3$ ).

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: Inverting the definition of $i$
Students divide abnormal by normal molar mass.
Wrong working
$i = \frac{223}{122} = 1.83$ for benzoic acid in benzene
Right working
$i = \frac{122}{223} = 0.547$
Do this: Association must give $i < 1$ ; if your answer is above 1, you have inverted it.
Usually costs: 1-2 marks, since $\alpha$ also comes out wrong.
Mistake: Using $α = i - 1$ for every electrolyte
The KCl case is memorised and applied to $K_{2} S O_{4}$ .
Wrong working
$K_{2} S O_{4}$ with $i = 2.6$ : $α = 1.6$
Right working
$α = \frac{i - 1}{n - 1} = \frac{1.6}{2} = 0.80$
Do this: Always write $n$ from the ionisation equation before using the formula.
Usually costs: 1 mark.
Mistake: Counting atoms instead of ions
$K_{4} [F e (C N)_{6}]$ is read as many particles.
Wrong working
$K_{4} [F e (C N)_{6}]$ : $i = 4 + 1 + 6 = 11$
Right working
$K_{4} [F e (C N)_{6}] \to 4 K^{+} + [F e (C N)_{6}]^{4 -}$ , so $i = 5$
Do this: A complex ion inside square brackets stays as one particle.
Usually costs: 1 mark in MCQs and in numericals.
Mistake: Leaving out $i$ in osmotic pressure or boiling point of a salt
The plain formula is used out of habit.
Do this: Ask first: is the solute an electrolyte? If yes, include $i$ .
Usually costs: 1-2 marks.

Exam tips

Write the ionisation or association equation in the first line; examiners give a step mark for it and it fixes $n$ .
If the question says 'assume complete dissociation', use $i = n$ directly.
For weak acids in water, expect $i$ just above 1 and $α$ as a small fraction; an $α$ above 1 means an error.
Give $α$ both as a fraction and as a percentage in the final line.
In assertion-reason items on abnormal molar mass, check 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
If ethanoic acid dimerises completely in benzene, the van't Hoff factor is:
(a) 2 (b) 1 (c) 0.5 (d) 0
Hint
Two molecules make one particle.
Answer
(c) 0.5.
Q2 · 2 marks · Very short answerBoard level
Define the van't Hoff factor. What is its value (greater or less than 1) when the solute (a) dissociates and (b) associates in solution? Give one example of each.
Hint
Ratio of normal to abnormal molar mass.
Answer
$i$ = normal molar mass ÷ abnormal (observed) molar mass, or observed ÷ calculated colligative property. (a) Dissociation, $i > 1$ : KCl in water. (b) Association, $i < 1$ : ethanoic acid in benzene.
Q3 · 3 marks · Short answerHarder board level
A 0.040 m aqueous solution of a weak monobasic acid HX freezes at −0.0893 °C. Calculate the van't Hoff factor and the degree of dissociation of HX. ( $K_{f}$ = 1.86 K kg mol⁻¹)
Hint
$H X ⇌ H^{+} + X^{-}$ , $n = 2$ .
Answer
Calculated $Δ T_{f} = 1.86 \times 0.040 = 0.0744$ K. $i = \frac{0.0893}{0.0744} = 1.20$ . $α = i - 1 = 0.20$ (20%).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The molar mass of NaCl found from the depression of freezing point of its aqueous solution is close to half its formula mass.
Reason (R): NaCl dissociates in water, nearly doubling the number of solute particles.
(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, bigger $Δ T_{f}$ , smaller calculated $M$ .
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
Calculate the osmotic pressure of a 0.10 M aqueous solution of $C a C l_{2}$ at 300 K, assuming complete dissociation. ( $R$ = 0.083 L bar K⁻¹ mol⁻¹)
Hint
$i = 3$ .
Answer
$π = i C R T = 3 \times 0.10 \times 0.083 \times 300 = 7.47$ bar.
Q6 · 3 marks · Short answerHarder board level
14.2 g of sodium sulphate ( $N a_{2} S O_{4}$ , 142 g mol⁻¹) is dissolved in 500 g of water. If the salt is 90% dissociated, find the boiling point of the solution at 1 atm. ( $K_{b}$ = 0.52 K kg mol⁻¹)
Hint
$n = 3$ , so $i = 1 + 2 α$ .
Answer
$i = 1 + 2 (0.90) = 2.8$ ; $m = \frac{0.10}{0.50} = 0.20$ mol kg⁻¹; $Δ T_{b} = 2.8 \times 0.52 \times 0.20 = 0.291$ K. Boiling point ≈ 100.29 °C (373.44 K).
Q7 · 3 marks · Short answerHOTS / competency
Arrange the following 0.1 m aqueous solutions in order of increasing freezing point and justify: ethanoic acid, glucose, $B a C l_{2}$ , NaCl.
Hint
Lowest freezing point = largest $i$ .
Answer
$B a C l_{2}$ ( $i \approx 3$ ) < NaCl ( $i \approx 2$ ) < ethanoic acid ( $i$ slightly above 1, weak electrolyte) < glucose ( $i = 1$ ). Freezing point depression is proportional to $i$ , so more particles means a lower freezing point.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a class prepares 'model sea water' by dissolving 3.5 g of NaCl (58.5 g mol⁻¹) in 96.5 g of water, and measures its freezing point. The teacher points out that the measured depression is a little smaller than the value calculated with $i = 2$ .
(a) Calculate the expected freezing point with $i = 2$ . ( $K_{f}$ = 1.86 K kg mol⁻¹) (2 marks)
(b) Why is the measured depression slightly smaller? (1 mark)
(c) Why does spreading salt on an icy road help to clear it? (1 mark)
Hint
Molality is about 0.62 mol kg⁻¹.
Answer
(a) $m = \frac{3.5/58.5}{0.0965} = 0.62$ mol kg⁻¹; $Δ T_{f} = 2 \times 1.86 \times 0.62 = 2.31$ K; freezing point ≈ −2.3 °C. (b) At this concentration oppositely charged ions attract each other and some move as pairs, so the effective number of particles is less than 2 per NaCl ( $i < 2$ ). (c) Dissolved salt lowers the freezing point of the water film, so the ice melts at temperatures a little below 0 °C.

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.Value of $i$ for $A l_{2} (S O_{4})_{3}$ on complete dissociation?Show answer
5.
2.Relation between $i$ and $α$ for dissociation?Show answer
$i = 1 + (n - 1) α$ .
3.Why is ethanoic acid's molar mass nearly 120 g mol⁻¹ in benzene?Show answer
It dimerises through hydrogen bonding, halving the number of particles.
4.Osmotic pressure formula with $i$ ?Show answer
$π = i C R T$ .
5.Is $i$ for glucose in water greater than, equal to or less than 1?Show answer
Equal to 1 (no association or dissociation).
6.Degree of association for dimerisation in terms of $i$ ?Show answer
$α = 2 (1 - i)$ .

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

What is abnormal molar mass in Class 12 Chemistry?

It is a molar mass, found from a colligative property, that is higher or lower than the formula value because the solute associates or dissociates in solution.

What is the van't Hoff factor?

It is the ratio of the normal molar mass to the abnormal molar mass, or of the observed colligative property to the calculated one. It tells how many particles each formula unit effectively gives in solution.

Why is $i$ for a strong electrolyte slightly less than the number of ions?

At ordinary concentrations, oppositely charged ions attract each other and some move together, so the solution behaves as if it had slightly fewer particles. The value approaches n on dilution.

How do I find the degree of dissociation from $i$?

Use \alpha = \frac{i-1}{n-1}, where n is the number of ions produced by one formula unit.

Does association increase or decrease the observed molar mass?

It increases it, because fewer particles give a smaller colligative effect.