Abnormal Molar Masses

Q: Why does 1 mole of NaCl in water depress the freezing point twice as much as 1 mole of glucose?

NaCl dissociates completely into sodium and chloride ions, giving 2 moles of particles per mole dissolved. Glucose does not dissociate, giving 1 mole of particles. Colligative properties depend on particle count, so NaCl produces twice the freezing point depression.

Q: Why is the observed molar mass of acetic acid higher in benzene than its actual molar mass?

In benzene, acetic acid molecules pair into cyclic dimers via two hydrogen bonds between hydroxyl H and carbonyl O. This halves the effective particle count, so the calculated molar mass from the standard formula comes out about twice the true value.

Q: Can Van't Hoff factor be non-integer?

Yes. Non-integer i values reflect partial dissociation or association. A weak acid at 30 percent ionisation has i equal to 1.3. Only for strong electrolytes with complete dissociation does i equal the integer number of ions.

Q: Why does acetic acid dimerise in benzene but not in water?

In water, both hydrogen bond donor and acceptor sites of acetic acid bond with water molecules which are in vast excess. Water outcompetes any acid-acid pairing. In benzene, no hydrogen bonding with solvent is possible, so acetic acid molecules preferentially bond with each other forming stable cyclic dimers.

Q: What is the Van't Hoff factor for a solute that neither associates nor dissociates?

i equals 1 exactly. Non-electrolytes like glucose, urea, and sucrose give i approximately 1. The standard colligative formulas without an i multiplier apply directly.

Q: Why does the Van't Hoff factor for real electrolytes sometimes come out slightly less than the ideal value?

In real solutions ions attract each other via Coulombic interactions. Each cation is surrounded by a diffuse cloud of anions, making the ions behave partly like paired species. The effective number of independent particles is slightly less than the fully-dissociated count, giving i a bit below the ideal integer value.

Q: How is the degree of dissociation calculated from Van't Hoff factor?

If a solute dissociates into n particles per formula unit and alpha is the degree of dissociation, then i equals 1 plus n minus 1 times alpha. Rearranging gives alpha equals i minus 1 divided by n minus 1.

Q: Which colligative property is best for detecting dissociation or association?

Freezing point depression is usually most convenient because K f is large, so the temperature shift is easy to measure. The observed depression divided by the theoretical value gives Van't Hoff factor directly. Osmotic pressure works well for polymer aggregation studies.

ChemistrySolutionsFor NEET aspirants

The colligative-property formulas assume the solute stays as separate molecules in solution. When a solute dissociates (like NaCl into Na $^{+}$ + Cl $^{-}$ ) or associates (like benzoic acid forming dimers in benzene), the experimentally observed molar mass differs from the theoretical value - this is called the abnormal molar mass. The Van't Hoff factor $i$ corrects for this: $i = 1$ for no change, $i > 1$ for dissociation, $i < 1$ for association. All four colligative-property expressions get multiplied by $i$ .

Key Formulas - Quick Reference

Van't Hoff factor: $i = \frac{observed colligative property}{calculated colligative property} = \frac{theoretical molar mass}{observed molar mass}$
Also: $i = \frac{no. of particles after dissociation/association}{no. of particles initially taken}$
Modified relative VP lowering: $\frac{P ^{\circ} - P}{P ^{\circ}} = i \cdot x_{2}$
Modified BP elevation: $Δ T_{b} = i \cdot K_{b} \cdot m$
Modified FP depression: $Δ T_{f} = i \cdot K_{f} \cdot m$
Modified osmotic pressure: $π = i \cdot M R T$
Dissociation into $n$ ions with degree $α$ : $i = 1 + (n - 1) α$
Association of $n$ molecules into one aggregate, degree $α$ : $i = 1 - α (1 - \frac{1}{n})$

1. Why Molar Masses Come Out Abnormal

All colligative properties count particles, not moles-as-written on paper. The relationship $Δ T_{f} = K_{f} m$ or $π = M R T$ implicitly assumes that dissolving $n$ moles of solute puts $n$ moles of particles into solution. This is only true when the solute:

is a non-electrolyte (so it does not dissociate), and
does not associate into dimers, trimers or larger aggregates in the given solvent

Real solutes often violate these assumptions:

Electrolytes dissociate. 1 mole of NaCl in water becomes 2 moles of ions ( $Na^{+}$ + $Cl^{-}$ ). The measured colligative effect is roughly twice what the formula predicts if you use the formula weight of NaCl.
Some solutes associate. Carboxylic acids like acetic acid or benzoic acid form cyclic dimers in non-polar solvents like benzene via two hydrogen bonds. 2 moles of acid molecules become 1 mole of dimer particles. The measured effect is half of what the formula predicts.

When the calculated colligative effect does not match the observation, using it to compute molar mass gives a value that is either too low (dissociation - because number of particles is higher, calculated $M$ from $M = \frac{R T W}{π V}$ etc. is lower) or too high (association). These non-matching values are the "abnormal molar masses".

2. The Van't Hoff Factor

The Van't Hoff factor $i$ is a correction ratio that turns the observed abnormal behaviour back into a form compatible with the standard formulas.

$i = \frac{observed colligative property}{calculated colligative property (assuming no dissociation or association)}$

Equivalently:

$i = \frac{normal (theoretical) molar mass}{observed molar mass} = \frac{number of particles in solution}{number of particles initially dissolved}$

Three cases:

$i = 1$ : solute neither dissociates nor associates (all non-electrolytes like glucose, urea, sucrose).
$i > 1$ : solute dissociates into more particles (NaCl, KCl, $K_{2} SO_{4}$ , all salts and strong electrolytes).
$i < 1$ : solute associates into fewer, larger particles (carboxylic acids in benzene, phenols in non-polar solvents).

Figure 1: The Van't Hoff factor on a number line.

i < 1

means association (fewer particles than were dissolved),

i = 1

means the solute stays intact, and

i > 1

means dissociation. Because observed molar mass

=

normal molar mass

\div i

, dissociation makes

M

come out too small while association makes it too large.

3. Modified Colligative Property Expressions

Every colligative-property formula gets multiplied by $i$ when the solute is not a pure non-electrolyte:

Property	Ideal formula	Modified formula (with $i$ )
Relative VP lowering	$\frac{Δ P}{P ^{\circ}} = x_{2}$	$\frac{Δ P}{P ^{\circ}} = i \cdot x_{2}$
Boiling point elevation	$Δ T_{b} = K_{b} m$	$Δ T_{b} = i \cdot K_{b} \cdot m$
Freezing point depression	$Δ T_{f} = K_{f} m$	$Δ T_{f} = i \cdot K_{f} \cdot m$
Osmotic pressure	$π = M R T$	$π = i \cdot M R T$

4. Case 1 - Dissociation

Figure 2: Dissociation of NaCl in water. One formula unit gives

Na^{+}

and

Cl^{-}

, each wrapped in a shell of water molecules that turn their oxygen ends towards the cation and their hydrogen ends towards the anion. Two particles from one, so

i = 2

and the observed molar mass is about 29 g/mol instead of 58.5.

Suppose 1 mole of an electrolyte $A_{x} B_{y}$ dissolves and dissociates as:

$A_{x} B_{y} \to x A^{y +} + y B^{x -}$

producing $n = x + y$ ions per formula unit if fully dissociated. Let $α$ be the degree of dissociation (fraction that actually dissociates).

Out of 1 mole taken:

$α$ mole dissociates, producing $n α$ moles of ions
$(1 - α)$ mole stays as intact molecules
Total particles $= (1 - α) + n α = 1 + (n - 1) α$

$i = 1 + (n - 1) α$ (for dissociation into $n$ particles)

Rearranging: $α = \frac{i - 1}{n - 1}$

Examples:

NaCl (dissociates into 2 ions, complete): $i = 1 + (2 - 1) (1) = 2$
$BaCl_{2}$ (3 ions, complete): $i = 1 + (3 - 1) (1) = 3$
$K_{3} [Fe(CN)_{6}]$ (4 ions, complete): $i = 4$
$Al_{2} (SO_{4})_{3}$ (5 ions, complete): $i = 5$
A weak acid HA at 30% dissociation: $i = 1 + (2 - 1) (0.3) = 1.3$

Solved Example 1

The observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. Calculate the degree of dissociation of silver nitrate.

Solution:

$i = \frac{normal mol. wt.}{observed mol. wt.} = \frac{170}{92.64} = 1.835$

$AgNO_{3} \to Ag^{+} + NO_{3}^{-}$ , so $n = 2$ .

Using $i = 1 + (n - 1) α$ : $1.835 = 1 + (2 - 1) α \Rightarrow α = 0.835 = 83.5%$ .

Solved Example 2

A storage battery contains a 38% by mass solution of

H_{2} SO_{4}

. At this concentration the Van't Hoff factor is 2.50. At what temperature will the battery contents freeze? (

K_{f}

(water) = 1.86 K kg mol

^{- 1}

; molar mass of

H_{2} SO_{4}

= 98.)

Solution:

Consider 100 g of solution: 38 g $H_{2} SO_{4}$ + 62 g water.

Molality $m = \frac{38/98}{62/1000} = \frac{0.388}{0.062} = 6.25$ mol/kg.

$Δ T_{f} = i \cdot K_{f} \cdot m = 2.50 \times 1.86 \times 6.25 = 29.06$ K.

Freezing point $= 273 - 29.06 = 243.94$ K $\approx - 29.1$ °C.

This is why lead-acid batteries do not freeze even in bitterly cold weather.

Solved Example 3

Calculate the osmotic pressure at 27 °C of a 1 millimolar solution of potassium ferricyanide,

K_{3} [Fe(CN)_{6}]

, that is 70% dissociated.

Solution:

$K_{3} [Fe(CN)_{6}] \to 3 K^{+} + [Fe(CN)_{6}]^{3 -}$ : $n = 4$ ions.

$i = 1 + (n - 1) α = 1 + (4 - 1) (0.7) = 1 + 2.1 = 3.1$

$π = i \cdot M R T = 3.1 \times 1 0^{- 3} \times 0.0821 \times 300 = 0.0764$ atm.

5. Case 2 - Association

Figure 3: Association of acetic acid in benzene. Two molecules pair head-to-head into an eight-membered ring closed by two

O-H \dots O=C

hydrogen bonds. Two particles become one, so

i = 0.5

and the observed molar mass comes out near 120 g/mol instead of 60. In water the same molecule does the opposite and ionises, giving

i > 1

Suppose $n$ molecules of solute associate into a single aggregate:

$n A ⇌ A_{n}$

Let $α$ be the degree of association (fraction of molecules that associate). Out of 1 mole taken:

$α$ mole associates, forming $\frac{α}{n}$ mole of aggregates
$(1 - α)$ mole stays as monomers
Total particles $= (1 - α) + \frac{α}{n} = 1 - α (1 - \frac{1}{n})$

$i = 1 - α (1 - \frac{1}{n})$ (for association of $n$ molecules)

Common case - dimerisation ( $n = 2$ ): $i = 1 - \frac{α}{2}$ . If dimerisation is complete ( $α = 1$ ), $i = 0.5$ and the observed molar mass is double the theoretical value.

Solved Example 4

Acetic acid (

CH_{3} COOH

) associates into cyclic dimers in benzene. When 1.65 g of acetic acid is dissolved in 100 g of benzene, the boiling point rises by 0.36 °C. Calculate the Van't Hoff factor and the degree of association of acetic acid in benzene. (

K_{b}

for benzene = 2.57 K kg mol

^{- 1}

; molar mass of acetic acid = 60.)

Solution:

Observed molar mass from $Δ T_{b}$ :

$M_{obs} = \frac{1000 \cdot K _{b} \cdot W _{solute}}{Δ T _{b} \cdot W _{solvent}} = \frac{1000 \times 2.57 \times 1.65}{0.36 \times 100} = 118$ g/mol.

Van't Hoff factor: $i = \frac{M _{theor}}{M _{obs}} = \frac{60}{118} = 0.508$ .

Degree of association (dimerisation, $n = 2$ ):

$i = 1 - \frac{α}{2} \Rightarrow 0.508 = 1 - \frac{α}{2} \Rightarrow α = 2 (1 - 0.508) = 0.984$ .

So acetic acid is 98.4% associated as dimers in benzene.

Solved Example 5

Benzoic acid,

C_{6} H_{5} COOH

(M = 122), dimerises in benzene. A solution containing 2.44 g of benzoic acid in 50 g of benzene shows a freezing point depression of 0.98 K. Calculate the degree of association. (

K_{f}

for benzene = 4.90 K kg mol

^{- 1}

Solution:

Molality (theoretical, no association): $m = \frac{2.44/122}{50/1000} = \frac{0.02}{0.05} = 0.40$ mol/kg.

Theoretical $Δ T_{f} = K_{f} \cdot m = 4.90 \times 0.40 = 1.96$ K.

Observed $Δ T_{f} = 0.98$ K.

$i = \frac{observed}{calculated} = \frac{0.98}{1.96} = 0.50$ .

For dimerisation ( $n = 2$ ): $0.50 = 1 - \frac{α}{2} \Rightarrow α = 1.00$ (100% dimerisation).

This confirms the strong tendency of benzoic acid to form cyclic dimers via two O-H...O=C hydrogen bonds in non-polar solvents.

6. Combined and Complex Cases

Solved Example 6

Rank the following equimolar aqueous solutions by their observed freezing point depression, assuming complete dissociation for electrolytes: (a) urea, (b) NaCl, (c)

CaCl_{2}

, (d)

Al_{2} (SO_{4})_{3}

Solution:

$Δ T_{f} = i \cdot K_{f} \cdot m$ . For equimolar solutions, $Δ T_{f} \propto i$ .

Urea (non-electrolyte): $i = 1$
NaCl (2 ions): $i = 2$
$CaCl_{2}$ (3 ions): $i = 3$
$Al_{2} (SO_{4})_{3} \to 2 Al^{3 +} + 3 SO_{4}^{2 -}$ (5 ions): $i = 5$

Order of $Δ T_{f}$ : urea $<$ NaCl $<$ $CaCl_{2}$ $<$ $Al_{2} (SO_{4})_{3}$ .

Correspondingly, the freezing point itself is highest for urea and lowest for $Al_{2} (SO_{4})_{3}$ .

Solved Example 7

River water contains 11.7% NaCl, 9.5%

MgCl_{2}

, and 8.4%

NaHCO_{3}

by mass. Assuming 90% ionisation of NaCl, 70% of

MgCl_{2}

, and 50% of

NaHCO_{3}

, calculate the normal boiling point. (

K_{b}

for water = 0.52 K kg mol

^{- 1}

Solution:

Consider 100 g of solution. Water content = 100 - (11.7 + 9.5 + 8.4) = 70.4 g.

Moles of each solute per 100 g solution:

$n (NaCl) = 11.7/58.5 = 0.200$

$n (MgCl_{2}) = 9.5/95 = 0.100$

$n (NaHCO_{3}) = 8.4/84 = 0.100$

Van't Hoff factors:

NaCl (2 ions): $i = 1 + (2 - 1) (0.9) = 1.9$

$MgCl_{2}$ (3 ions): $i = 1 + (3 - 1) (0.7) = 2.4$

$NaHCO_{3}$ (2 ions): $i = 1 + (2 - 1) (0.5) = 1.5$

Effective moles of particles:

$\sum i \cdot n = 1.9 (0.2) + 2.4 (0.1) + 1.5 (0.1) = 0.38 + 0.24 + 0.15 = 0.77$

Effective molality $= \frac{0.77}{70.4/1000} = 10.94$ mol/kg.

$Δ T_{b} = K_{b} \cdot m_{eff} = 0.52 \times 10.94 = 5.69$ K.

Boiling point $= 100 + 5.69 = 105.69$ °C.

Solved Example 8

The freezing point of an aqueous solution containing 0.02 mole fraction of acetic acid in benzene is 277.4 K. Acetic acid exists partly as dimer. Calculate the equilibrium constant

K_{c}

for dimerisation. (Freezing point of pure benzene

= 278.4

K; heat of fusion of benzene

= 10.042

kJ/mol;

M

of benzene

= 78

;

R = 8.314

J K

^{- 1}

mol

^{- 1}

; assume molarity

\approx

molality.)

Solution:

Cryoscopic constant of benzene:

$K_{f} = \frac{R \cdot T _{f}^{2} \cdot M _{1}}{1000 \cdot Δ H _{f u s}} = \frac{8.314 \times ( 278.4 ) ^{2} \times 78}{1000 \times 10042} = 5.00$ K kg mol $^{- 1}$ .

Molality of acetic acid: with mole fraction $x_{1} = 0.02$ of acetic acid and $x_{2} = 0.98$ of benzene,

$m = \frac{n _{acid}}{w _{benz} ( k g )} = \frac{0.02/0.98 \times 1000}{78} = 0.262$ mol/kg.

Given: $Δ T_{f} = 278.4 - 277.4 = 1.0$ K.

$Δ T_{f} = i \cdot K_{f} \cdot m \Rightarrow 1.0 = i \times 5.00 \times 0.262 \Rightarrow i = 0.763$ .

Degree of association: $i = 1 - α /2 \Rightarrow α = 2 (1 - 0.763) = 0.474$ .

Equilibrium constant for dimerisation: $2 CH_{3} COOH ⇌ (CH_{3} COOH)_{2}$

If initial concentration is $c$ and degree $α$ : $[monomer] = c (1 - α)$ , $[dimer] = c α /2$ .

$K_{c} = \frac{[ dimer ]}{[ monomer ] ^{2}} = \frac{c α /2}{[ c ( 1 - α ) ] ^{2}} = \frac{α}{2 c ( 1 - α ) ^{2}}$

$K_{c} = \frac{0.474}{2 \times 0.262 \times ( 0.526 ) ^{2}} = \frac{0.474}{0.145} = 3.27$ M $^{- 1}$ .

Common Mistakes to Avoid

Watch out

Do not forget to include $i$ when the solute is an electrolyte, associates, or dissociates. Skipping $i$ underestimates the effect for electrolytes and overestimates it for associating solutes.
Number of particles $n$ counts total ions, not formula units. For $Al_{2} (SO_{4})_{3} \to 2 Al^{3 +} + 3 SO_{4}^{2 -}$ , $n = 5$ , not 2 or 3.
For dissociation: $i = 1 + (n - 1) α$ . For association: $i = 1 - α (1 - 1/ n)$ . Sign of the second term is different.
Carboxylic acids associate in non-polar solvents (like benzene) via H-bonded dimers, but dissociate in water as weak acids. Same molecule, opposite behaviour depending on solvent.
$i < 1$ implies association (fewer particles); $i > 1$ implies dissociation (more particles). Do not confuse the direction.
Observed molar mass $= \frac{normal molar mass}{i}$ . For a dissociating electrolyte, observed $M$ is smaller than the true formula weight. For an associating solute, it is larger.
When both dissociation and hydrolysis or protonation occur (as in $NH_{4} Cl$ hydrolysis), carefully account for particles at each equilibrium stage before computing $i$ .

Frequently Asked Questions

Q1. Why does 1 mole of NaCl in water depress the freezing point twice as much as 1 mole of glucose?

NaCl completely dissociates in water into $Na^{+}$ and $Cl^{-}$ , producing 2 moles of particles per mole of NaCl dissolved. Glucose does not dissociate, giving 1 mole of particles. Colligative properties depend on particle count, so NaCl produces twice the freezing point depression.

Q2. Why is the observed molar mass of acetic acid higher in benzene than its actual molar mass?

In benzene (a non-polar solvent), acetic acid molecules pair up into cyclic dimers held together by two hydrogen bonds between the carbonyl oxygen and hydroxyl hydrogen. 2 moles of monomers become 1 mole of dimer, halving the effective particle count. The colligative property is therefore only half of what would be expected, and the calculated molar mass (using the standard formula) comes out about twice the true value.

Q3. Can Van't Hoff factor be non-integer?

Yes. Non-integer $i$ values reflect partial dissociation or association. A weak acid at 30% ionisation has $i = 1.3$ . A carboxylic acid at 80% dimerisation has $i = 0.6$ . Only for strong electrolytes with complete dissociation does $i$ become an integer equal to the number of ions.

Q4. Why does acetic acid dimerise in benzene but not in water?

In water, both the H-bond donor (O-H) and acceptor (C=O) of acetic acid can hydrogen-bond with water molecules, which are in vast excess. Water molecules essentially "compete" for the H-bond sites, so pairing between two acetic acid molecules is not favourable. In benzene, no hydrogen bonding is possible with the solvent, so acetic acid molecules preferentially bond with each other, forming stable cyclic dimers.

Q5. What is the Van't Hoff factor for a solute that neither associates nor dissociates?

$i = 1$ exactly. This is the reference case - non-electrolytes like glucose, urea, sucrose, and most organic molecules dissolved in water give $i \approx 1$ . The standard colligative formulas (without an $i$ multiplier) apply directly.

Q6. Why does the Van't Hoff factor for real electrolytes sometimes come out slightly less than the ideal value?

In real solutions, ions attract each other (Coulombic interactions), and each cation is surrounded by a diffuse cloud of anions and vice versa (Debye-Huckel ion atmosphere). These interactions make the ions behave partly like paired species, so the effective number of independent particles is a bit less than the fully-dissociated count. For example, NaCl at real concentrations gives $i \approx 1.9$ instead of exactly 2.

Q7. How is the degree of dissociation calculated from Van't Hoff factor?

If a solute dissociates into $n$ particles per formula unit and $α$ is the degree of dissociation, then $i = 1 + (n - 1) α$ . Rearranging: $α = \frac{i - 1}{n - 1}$ . For example, if AgNO $_{3}$ ( $n = 2$ ) has $i = 1.835$ , then $α = 0.835$ or 83.5%.

Q8. Which colligative property is best for detecting dissociation or association?

Freezing point depression is usually the most convenient because $K_{f}$ is large (1.86 for water, 5.12 for benzene) so the shift is easy to measure. The observed $Δ T_{f}$ divided by the theoretical value (assuming no dissociation/association) gives $i$ directly. Osmotic pressure works well for polymer aggregation studies; boiling point elevation is used less frequently because heat-sensitive solutes may decompose.

Previous year questions on Abnormal Molar Masses

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