Benzoic acid in benzene gives an apparent molar mass of about 244 instead of 122. This is because:

It associates into dimers through hydrogen bonding

The experiment has a systematic error

A mixture of chloroform and acetone shows a vapour pressure lower than Raoult's law predicts because:

A new hydrogen bond forms between the two, strengthening A–B attraction

A–B attraction is weaker than A–A and B–B

The solubility of a gas in water on raising the temperature:

First increases and then decreases

Ethanol and water form a minimum-boiling azeotrope. This indicates:

Positive deviation from Raoult's law

Negative deviation from Raoult's law

Positive deviation from Raoult's law

That the mixture can be fully separated by distillation

Solutions and Colligative Properties

A colligative property depends on how many solute particles are dissolved and not at all on what they are. That is a strange and powerful idea: a mole of urea and a mole of glucose depress the freezing point of water by exactly the same amount, despite having nothing chemically in common. It is also the source of the one thing that goes wrong in this chapter — the moment the solute dissociates or associates, the number of particles is no longer the number of formula units you dissolved, and every answer must be multiplied by the van't Hoff factor i. Almost all lost marks here come from setting i = 1 without checking, so treat "how many particles does one formula unit become" as the first question you ask, before touching a formula.

Raoult's Law and Non-Ideal Solutions

For a solution of two volatile liquids, p_A = x_A p°_A and the total vapour pressure is p_A + p_B — each component contributes in proportion to its mole fraction. For a non-volatile solute the law reduces to the relative lowering of vapour pressure, (p° − p)/p° = x_solute, which is the form used in molar-mass determinations. An ideal solution is one where A–B interactions are identical in strength to A–A and B–B, so ΔH_mix = 0 and ΔV_mix = 0.

The two kinds of deviation from Raoult's law
Feature	Positive deviation	Negative deviation
A–B interaction	Weaker than A–A and B–B	Stronger than A–A and B–B
Vapour pressure	Higher than Raoult predicts	Lower than Raoult predicts
ΔH_mix	Positive — mixing absorbs heat	Negative — mixing releases heat
Azeotrope formed	Minimum boiling, e.g. ethanol–water	Maximum boiling, e.g. nitric acid–water
Example	Ethanol + acetone; benzene + methanol	Chloroform + acetone; HCl + water

🎯 Why chloroform and acetone show negative deviation

Neither liquid can hydrogen bond with itself — chloroform has no lone pair to accept a bond and acetone has no acidic hydrogen. But mixed together, the electron-poor hydrogen of CHCl₃ forms a genuine hydrogen bond with the carbonyl oxygen of acetone. So A–B attraction is stronger than either pure liquid manages alone, molecules are held back from escaping, vapour pressure falls below the Raoult prediction, and mixing releases heat. Once you can see the new interaction being created, you never have to memorise which pairs deviate which way — you look for whether mixing makes a new attraction possible (negative) or breaks an existing one such as ethanol's own hydrogen-bond network (positive).

Henry's law governs the solubility of a gas: p = K_H·x, so solubility rises with pressure and falls with temperature. A larger K_H means a less soluble gas, which is the reverse of the intuitive reading and is regularly tested. Three familiar consequences: a fizzy drink loses gas when opened because p drops; divers develop the bends because nitrogen dissolved at depth comes out of solution on ascent, which is why breathing mixtures use helium with its much larger K_H; and aquatic life is stressed in warm water because oxygen is less soluble there.

The Four Colligative Properties

The four properties and their working formulae
Property	Formula	Constant
Relative lowering of vapour pressure	(p° − p)/p° = i·x_solute	None needed
Elevation of boiling point	ΔT_b = i·K_b·m	K_b = 0.52 K·kg/mol for water
Depression of freezing point	ΔT_f = i·K_f·m	K_f = 1.86 K·kg/mol for water
Osmotic pressure	π = i·CRT	Uses molarity, not molality

🧠 Osmotic pressure is the method of choice for macromolecules

A protein of molar mass 60 000 dissolved at 1 g/L gives a freezing-point depression of about 0.00003 K — unmeasurable. The same solution gives an osmotic pressure of roughly 0.4 kPa, which a manometer reads comfortably. The reason is that π = CRT carries the factor RT ≈ 2500 J/mol, an enormous multiplier compared with K_f = 1.86. So whenever a question involves a polymer, protein or any solute of very high molar mass, osmotic pressure is the expected technique, and the reason asked for is its far greater sensitivity at low concentration.

🎯 Why a solute raises the boiling point but lowers the freezing point

Both follow from the same fact — a non-volatile solute lowers the vapour pressure of the solvent — yet they move in opposite directions, which is why the pair is confusing until you see the mechanism. Boiling happens when vapour pressure reaches atmospheric; if the solute has pushed vapour pressure down, you must heat further to get back up to 1 atm, so the boiling point rises. Freezing happens when the liquid and solid have equal vapour pressure; the solute lowers the liquid's vapour pressure but not the pure ice's, so the two curves now meet at a lower temperature. Hence salt on icy roads and antifreeze in radiators — the same additive does both jobs at once.

The van't Hoff Factor

The factor i is the ratio of the number of particles actually present to the number dissolved. For a non-electrolyte such as glucose or urea, i = 1. For a strong electrolyte it equals the number of ions: 2 for NaCl, 3 for CaCl₂ or Na₂SO₄, 4 for AlCl₃ or K₃[Fe(CN)₆]. For a weak electrolyte dissociating into n particles with degree α, i = 1 + (n − 1)α. For a solute that associates into groups of n, i = 1/n, so i falls below 1.

⚠️ Benzoic acid in benzene has i = 0.5, not 1

This is the standard association question and it catches almost everyone the first time. In water, benzoic acid ionises slightly and i is a little above 1. In benzene, however, there is no solvent to stabilise ions, so two molecules pair up through hydrogen bonding into a stable dimer. The number of particles is halved, i = 0.5, and the observed freezing-point depression is only half what you would predict — which means the apparent molar mass comes out double the true value, 244 instead of 122. So an experiment reporting a molar mass of about 244 for benzoic acid has not failed; it has detected dimerisation. Whenever a question names a non-polar solvent, check for association before assuming i = 1.

The relation between i and apparent molar mass is worth stating explicitly because it is the form most questions take: i = M_theoretical / M_observed. Dissociation makes more particles, so M_observed comes out smaller than the true value; association makes fewer, so M_observed comes out larger. If an experiment on NaCl reports a molar mass near 29 rather than 58.5, nothing is wrong — the solute has split into two ions and the calculation simply did not account for it.

ℹ️ Isotonic, hypertonic and reverse osmosis

Two solutions with the same osmotic pressure are isotonic, which is why intravenous fluid must be 0.9 % saline — pure water would burst red cells by osmosis inwards, and a concentrated solution would shrivel them by drawing water out. A more concentrated solution is hypertonic, and it is why salting or sugaring preserves food: bacteria in a hypertonic medium lose water and cannot function. Applying pressure greater than π on the concentrated side reverses the flow, and that is reverse osmosis, the basis of desalination. Note carefully that isotonic means equal particle concentration, so 0.1 M NaCl (i = 2) is isotonic with 0.2 M glucose, not with 0.1 M glucose.

Key takeaways

Raoult: p_A = x_A p°_A; for a non-volatile solute, (p° − p)/p° = x_solute.
Positive deviation means weaker A–B forces and a minimum-boiling azeotrope; negative means the reverse.
Henry: p = K_H·x. A larger K_H means a less soluble gas, and solubility always falls on heating.
ΔT_b = iK_b m and ΔT_f = iK_f m use molality; π = iCRT uses molarity.
A solute raises the boiling point and lowers the freezing point — both from the same drop in vapour pressure.
Osmotic pressure is the method for macromolecules, because RT is a far larger multiplier than K_f.
i = 2 for NaCl, 3 for CaCl₂, 4 for AlCl₃; for weak electrolytes i = 1 + (n − 1)α.
Association gives i < 1: benzoic acid in benzene dimerises, so its apparent molar mass doubles to 244.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Equimolal aqueous solutions of glucose, NaCl and CaCl₂ are compared. The order of freezing-point depression is:
1. All three are equal
2. glucose < NaCl < CaCl₂
3. CaCl₂ < NaCl < glucose
4. NaCl < glucose < CaCl₂
Show answer
Answer: B — glucose < NaCl < CaCl₂
ΔT_f = iK_f m, and at equal molality the only difference is i: 1 for glucose, 2 for NaCl and 3 for CaCl₂. So the depressions stand in the ratio 1 : 2 : 3. The first option is the trap for candidates who remember that colligative properties ignore the identity of the solute but forget that they count particles, and an electrolyte supplies several particles per formula unit.
Benzoic acid in benzene gives an apparent molar mass of about 244 instead of 122. This is because:
1. It dissociates into two ions
2. It associates into dimers through hydrogen bonding
3. Benzene is a non-ideal solvent
4. The experiment has a systematic error
Show answer
Answer: B — It associates into dimers through hydrogen bonding
Benzene cannot stabilise ions, so instead of ionising, two benzoic acid molecules hydrogen bond into a dimer through their carboxyl groups. That halves the particle count, giving i = 0.5, and since i = M_theoretical/M_observed the apparent molar mass doubles. Dissociation would have the opposite effect and report a molar mass near 61, so the first option points the wrong way.
A mixture of chloroform and acetone shows a vapour pressure lower than Raoult's law predicts because:
1. A new hydrogen bond forms between the two, strengthening A–B attraction
2. A–B attraction is weaker than A–A and B–B
3. Both liquids are non-polar
4. Mixing absorbs heat
Show answer
Answer: A — A new hydrogen bond forms between the two, strengthening A–B attraction
The acidic hydrogen of CHCl₃ hydrogen bonds to the carbonyl oxygen of acetone, an interaction neither pure liquid can form with itself. Stronger A–B attraction holds molecules in the liquid, so vapour pressure falls below the ideal prediction and mixing is exothermic — the definition of negative deviation. Weaker A–B forces and endothermic mixing describe positive deviation instead.
Which colligative property is preferred for determining the molar mass of a protein?
1. Elevation of boiling point
2. Depression of freezing point
3. Osmotic pressure
4. Relative lowering of vapour pressure
Show answer
Answer: C — Osmotic pressure
A macromolecule gives such a low molal concentration that ΔT_f and ΔT_b are far too small to measure — of the order of 10⁻⁵ K. Osmotic pressure carries the factor RT, some three orders of magnitude larger than K_f, so it produces a pressure a manometer can read at the same tiny concentration. It also has the practical advantage of being measurable at room temperature, without heating a protein that would denature.
Which solution is isotonic with 0.1 M NaCl?
1. 0.1 M glucose
2. 0.2 M glucose
3. 0.1 M CaCl₂
4. 0.05 M glucose
Show answer
Answer: B — 0.2 M glucose
Isotonic means equal osmotic pressure, and π = iCRT, so what must match is the product iC. NaCl at 0.1 M with i = 2 gives an effective 0.2 M in particles, matching 0.2 M glucose with i = 1. 0.1 M CaCl₂ has i = 3 and so gives 0.3 M in particles, which makes it hypertonic relative to the NaCl solution.
The solubility of a gas in water on raising the temperature:
1. Increases
2. Decreases
3. Is unaffected
4. First increases and then decreases
Show answer
Answer: B — Decreases
Dissolving a gas is exothermic, so by Le Chatelier heating drives the gas back out and solubility falls — visible as bubbles forming in water well before it boils. This is why warm river water holds less dissolved oxygen and thermal discharge from power plants stresses aquatic life. Solids typically behave the opposite way, which is why the two cases must not be generalised together.
Ethanol and water form a minimum-boiling azeotrope. This indicates:
1. Negative deviation from Raoult's law
2. Positive deviation from Raoult's law
3. Ideal behaviour
4. That the mixture can be fully separated by distillation
Show answer
Answer: B — Positive deviation from Raoult's law
A minimum-boiling azeotrope means the mixture has a higher vapour pressure than Raoult predicts, which is positive deviation: ethanol's own hydrogen-bond network is disrupted by water without an equally strong replacement, so molecules escape more easily. The existence of the azeotrope is precisely why ethanol cannot be distilled beyond 95.6 % — the last option states the opposite of what an azeotrope implies.
A weak electrolyte AB is 40 % dissociated in solution. Its van't Hoff factor is:
1. 0.4
2. 1.4
3. 2.0
4. 0.6
Show answer
Answer: B — 1.4
For dissociation into n particles, i = 1 + (n − 1)α, and AB gives n = 2, so i = 1 + (1)(0.4) = 1.4. The answer 2.0 would apply only to complete dissociation, and 0.4 is α itself mistaken for i. Note that i must always exceed 1 for any dissociating solute, so an answer below 1 signals association instead and cannot be right here.