ChemistryCBSE Class 12 · Solutions

Raoult's Law, Ideal and Non-ideal Solutions: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Raoult's law says that in a solution of volatile liquids, the partial vapour pressure of each component equals its mole fraction times its pure vapour pressure. A solution that obeys it at every composition is ideal ($\Delta_{mix}H = 0$, $\Delta_{mix}V = 0$). Stronger or weaker A–B attractions give negative or positive deviations and can form azeotropes.

In an ideal solution each partial pressure is a straight line in x, and the total pressure is their sum, a straight line from p₁⁰ to p₂⁰.

Illustration drawn for QuestionBuddy.

Understand Raoult's Law, Ideal and Non-ideal Solutions

Take two liquids that both evaporate, such as benzene and toluene, and mix them in a closed vessel. Each liquid sends molecules into the vapour, so the vapour above the mixture holds both. Raoult's law describes how much: for each volatile component, the partial vapour pressure is proportional to its mole fraction in the liquid, and the constant of proportionality is the vapour pressure of that liquid when pure. So $p_{1} = x_{1} p_{1}^{0}$ and $p_{2} = x_{2} p_{2}^{0}$ . By Dalton's law of partial pressures the total is $p_{t o t a l} = p_{1} + p_{2} = p_{1}^{0} + (p_{2}^{0} - p_{1}^{0}) x_{2}$ , a straight line between $p_{1}^{0}$ and $p_{2}^{0}$ when plotted against $x_{2}$ .

The vapour does not have the same composition as the liquid. Its mole fractions come from the partial pressures: $y_{1} = p_{1} / p_{t o t a l}$ . Because the more volatile liquid has the larger $p^{0}$ , the vapour is always richer in it than the liquid is. This is the whole basis of fractional distillation.

When the solute is a non-volatile solid (sugar, urea, salt), only the solvent contributes to the vapour, so $p = x_{1} p_{1}^{0}$ . The vapour pressure falls, and the relative lowering $(p_{1}^{0} - p) / p_{1}^{0}$ equals the mole fraction of the solute, $x_{2}$ . Relative lowering depends only on the number of solute particles, which makes it the first colligative property. NCERT also shows that Raoult's law is a special case of Henry's law: both say partial pressure is proportional to mole fraction; Raoult's constant is $p_{1}^{0}$ while Henry's is $K_{H}$ .

An ideal solution obeys Raoult's law over the whole range of composition. This happens when the A–B attractions are about equal to the A–A and B–B attractions, so mixing gives out or takes in no heat ( $Δ_{mi x} H = 0$ ) and causes no change in volume ( $Δ_{mi x} V = 0$ ). Benzene + toluene and n-hexane + n-heptane come close.

A non-ideal solution does not obey the law. In positive deviation the A–B attractions are weaker than the like attractions, molecules escape more easily, and the vapour pressure is higher than Raoult's law predicts. Mixing is endothermic and the volume increases ( $Δ_{mi x} H > 0$ , $Δ_{mi x} V > 0$ ). Ethanol + acetone is the NCERT example: ethanol molecules lose some of their hydrogen bonds. Carbon disulphide + acetone is another. In negative deviation the A–B attractions are stronger, molecules are held back, and the vapour pressure is lower. Mixing is exothermic and the volume decreases ( $Δ_{mi x} H < 0$ , $Δ_{mi x} V < 0$ ). Chloroform + acetone is the classic case: the C–H of chloroform forms a hydrogen bond with the C=O oxygen of acetone. Phenol + aniline and nitric acid + water also deviate negatively.

Large deviations produce azeotropes, binary mixtures that boil at a constant temperature with the same composition in liquid and vapour, so fractional distillation cannot separate them. Large positive deviation gives a minimum boiling azeotrope (ethanol–water at about 95% ethanol by volume). Large negative deviation gives a maximum boiling azeotrope (nitric acid about 68% and water 32% by mass, boiling near 393.5 K).

Raoult's Law, Ideal and Non-ideal Solutions: formulas and key facts

Raoult's law (volatile components): $p_{1} = x_{1} p_{1}^{0}, p_{2} = x_{2} p_{2}^{0}$; Each partial pressure depends on that component's mole fraction in the liquid.
Total vapour pressure (ideal): $p_{t o t a l} = p_{1}^{0} + (p_{2}^{0} - p_{1}^{0}) x_{2}$; Straight line on a $p$ versus $x_{2}$ plot; Dalton's law adds the partial pressures.
Vapour-phase composition: $y_{1} = \frac{p _{1}}{p _{t o t a l}}, y_{2} = \frac{p _{2}}{p _{t o t a l}}$; The vapour is richer in the component with the larger $p^{0}$ .
Relative lowering of vapour pressure (non-volatile solute): $\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = x_{2} = \frac{n _{2}}{n _{1} + n _{2}}$; For finding molar mass, the exact rearrangement $\frac{p _{1}^{0} - p _{1}}{p _{1}} = \frac{n _{2}}{n _{1}}$ avoids the dilute approximation.
Ideal solution conditions: $Δ_{mi x} H = 0, Δ_{mi x} V = 0$; Positive deviation: both greater than zero. Negative deviation: both less than zero.
Raoult's law as a case of Henry's law: $p = K_{H} x becomes p_{1} = p_{1}^{0} x_{1} when K_{H} = p_{1}^{0}$; Both laws: partial pressure proportional to mole fraction.

How to solve Raoult's Law, Ideal and Non-ideal Solutions questions

Board questions on this topic are either numericals (vapour pressure, vapour composition, molar mass from lowering) or reasoning (identify the deviation, give signs, explain an azeotrope). Use this routine.

1Decide the type: two volatile liquids (use $p_{i} = x_{i} p_{i}^{0}$ for both), or a non-volatile solute (only the solvent's term survives).
2Convert masses to moles, then find mole fractions in the liquid. Write $x_{1} + x_{2} = 1$ as a check.
3For volatile pairs, find each partial pressure, add them for $p_{t o t a l}$ , then divide for vapour mole fractions $y_{1}$ , $y_{2}$ .
4For a non-volatile solute, use $\frac{p ^{0} - p}{p} = \frac{n _{2}}{n _{1}}$ when you need the molar mass $M_{2} = \frac{w _{2}}{n _{2}}$ ; state the unit g mol⁻¹.
5For deviation questions, compare the observed vapour pressure with the Raoult (ideal) value. Higher means positive deviation; lower means negative.
6Link the sign to forces: weaker A–B attraction gives positive deviation with $Δ_{mi x} H > 0$ and $Δ_{mi x} V > 0$ ; stronger A–B attraction gives the opposite signs.
7If an azeotrope is asked, match positive deviation with minimum boiling and negative deviation with maximum boiling, and give one example of each.

Raoult's Law, Ideal and Non-ideal Solutions 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.

Vapour above a mixture is richer in the more volatile liquid

Suppose you are helping in a small lab that blends two solvents, A and B, for cleaning electronic parts. At room temperature pure A has a vapour pressure of 120 kPa and pure B 40 kPa, and they form a nearly ideal solution. The blend is made with equal numbers of moles.

Find the mole fraction of A in the vapour above the blend, and explain why the blend left in an open tray slowly becomes richer in B.

Try it first, then show the solutionSolution

Liquid: $x_{A} = x_{B} = 0.5$ .
Partial pressures: $p_{A} = 0.5 \times 120 = 60$ kPa and $p_{B} = 0.5 \times 40 = 20$ kPa, so $p_{t o t a l} = 80$ kPa.
Vapour: $y_{A} = 60/80 = 0.75$ , $y_{B} = 0.25$ .
The vapour that escapes carries three molecules of A for every one of B, so the liquid left behind loses A faster and its share of B rises.

Answer: $y_{A} = 0.75$ . The escaping vapour is 75% A, so the remaining liquid becomes richer in the less volatile B.

Memory hook: The vapour always favours the liquid that escapes more easily: that is Raoult plus Dalton.

Tricks to remember Raoult's Law, Ideal and Non-ideal Solutions

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

Positive = Pushed apart
Positive deviation: molecules are Pushed out more easily (weaker A–B), so Pressure up, heat absorbed (+ΔH), volume up (+ΔV), minimum boiling azeotrope.
Use it when: Write all four consequences together in any 3-mark deviation answer.
Negative = Nailed together
Negative deviation: A–B molecules are held more tightly, so pressure down, heat given out (−ΔH), volume shrinks (−ΔV), maximum boiling azeotrope.
Use it when: Chloroform + acetone, nitric acid + water, phenol + aniline.
Volatile wins the vapour
Whichever liquid has the bigger $p^{0}$ is always over-represented in the vapour: $y > x$ for it.
Use it when: Fast check on any vapour-composition answer.

Raoult's Law, Ideal and Non-ideal Solutions: solved examples

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

Example 1: Spot the deviation · 1 mark · MCQBoard level

Which of the following mixtures shows positive deviation from Raoult's law?
(a) Chloroform + acetone
(b) Ethanol + acetone
(c) Nitric acid + water
(d) Benzene + toluene

Try it first, then show the solutionSolution

Chloroform + acetone and nitric acid + water have stronger A–B attraction: negative deviation.
Benzene + toluene is nearly ideal.
In ethanol + acetone, acetone gets between ethanol molecules and breaks some of their hydrogen bonds, so A–B attraction is weaker: positive deviation.

Answer: (b) Ethanol + acetone.

Example 2: Total pressure and vapour composition · 2 marks · Very short answerBoard level

Liquids A and B form an ideal solution. At 300 K the vapour pressures of pure A and pure B are 500 mm Hg and 800 mm Hg. Calculate the total vapour pressure of a solution with $x_{A} = 0.4$ and the mole fraction of B in the vapour.

Try it first, then show the solutionSolution

$p_{A} = x_{A} p_{A}^{0} = 0.4 \times 500 = 200$ mm Hg; $p_{B} = 0.6 \times 800 = 480$ mm Hg (½ + ½ mark).
$p_{t o t a l} = 200 + 480 = 680$ mm Hg (½ mark).
$y_{B} = \frac{p _{B}}{p _{t o t a l}} = \frac{480}{680} = 0.706$ (½ mark).

Answer: $p_{t o t a l} = 680$ mm Hg; $y_{B} = 0.706$ .

Example 3: Molar mass from lowering of vapour pressure · 3 marks · Short answerHarder board level

At 298 K the vapour pressure of pure water is 23.8 mm Hg. When 10 g of a non-volatile, non-electrolyte solute is dissolved in 90 g of water, the vapour pressure falls to 23.2 mm Hg. Calculate the molar mass of the solute.

Try it first, then show the solutionSolution

Use the exact form $\frac{p ^{0} - p}{p} = \frac{n _{2}}{n _{1}}$ (1 mark).
$n_{1} = \frac{90}{18} = 5.0$ mol; $\frac{23.8 - 23.2}{23.2} = \frac{0.6}{23.2} = 0.02586$ (½ mark).
$n_{2} = 0.02586 \times 5.0 = 0.1293$ mol (½ mark).
$M_{2} = \frac{10}{0.1293} = 77.3$ g mol⁻¹ (1 mark).

Answer: About 77 g mol⁻¹ (77.3 g mol⁻¹).

Example 4: Finding both pure vapour pressures · 3 marks · Short answerHarder board level

An ideal solution contains 2 mol of liquid A and 3 mol of liquid B and has a vapour pressure of 480 mm Hg at a certain temperature. When 1 mol more of B is added, the vapour pressure rises to 500 mm Hg at the same temperature. Find the vapour pressures of pure A and pure B.

Try it first, then show the solutionSolution

First solution: $x_{A} = 0.4$ , $x_{B} = 0.6$ , so $0.4 p_{A}^{0} + 0.6 p_{B}^{0} = 480$ … (i) (1 mark).
Second solution: 2 mol A and 4 mol B, $x_{A} = \frac{1}{3}$ , $x_{B} = \frac{2}{3}$ , so $\frac{1}{3} p_{A}^{0} + \frac{2}{3} p_{B}^{0} = 500$ … (ii) (½ mark).
From (i): $p_{A}^{0} = 1200 - 1.5 p_{B}^{0}$ . Substituting in (ii): $400 - 0.5 p_{B}^{0} + 0.667 p_{B}^{0} = 500$ , so $p_{B}^{0} = 600$ mm Hg (1 mark).
$p_{A}^{0} = 1200 - 900 = 300$ mm Hg (½ mark). Check: adding B raised the pressure, which fits B being the more volatile liquid.

Answer: $p_{A}^{0} = 300$ mm Hg, $p_{B}^{0} = 600$ mm Hg.

Example 5: From data to deviation, signs and azeotrope · 3 marks · Short answerHOTS / competency

At 320 K pure liquid X has a vapour pressure of 400 mm Hg and pure Y 200 mm Hg. A mixture with $x_{X} = 0.5$ is found to have a total vapour pressure of 340 mm Hg. (a) What type of deviation does it show? (b) Predict the signs of $Δ_{mi x} H$ and $Δ_{mi x} V$ . (c) If the deviation is large, what kind of azeotrope can form, and would the azeotrope boil above or below both pure liquids?

Try it first, then show the solutionSolution

Ideal value: $0.5 \times 400 + 0.5 \times 200 = 300$ mm Hg. Observed 340 mm Hg is higher (½ mark).
(a) Positive deviation: X–Y attractions are weaker than X–X and Y–Y (½ mark).
(b) Breaking stronger like attractions needs energy, so $Δ_{mi x} H > 0$ ; molecules are less tightly packed, so $Δ_{mi x} V > 0$ (1 mark).
(c) Large positive deviation gives a minimum boiling azeotrope. Its vapour pressure is the highest of any composition, so it reaches the outside pressure first and boils below both pure liquids (1 mark).

Answer: (a) Positive deviation. (b) $Δ_{mi x} H > 0$ , $Δ_{mi x} V > 0$ . (c) A minimum boiling azeotrope, boiling below both pure liquids.

Where students lose marks in Raoult's Law, Ideal and Non-ideal Solutions

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

Mistake: Using liquid mole fractions for the vapour
Students assume the vapour has the same composition as the liquid.
Wrong working
$x_{B} = 0.6$ , so the vapour also has $y_{B} = 0.6$ .
Right working
$y_{B} = \frac{p _{B}}{p _{t o t a l}} = \frac{480}{680} = 0.706$
Do this: Always go liquid → partial pressures → vapour. Divide each partial pressure by the total.
Usually costs: 1 mark in a 2- or 3-mark numerical.
Mistake: Mixing up signs for positive and negative deviation
'Positive' is read as 'heat given out'.
Wrong working
Positive deviation: $Δ_{mi x} H < 0$ because the solution warms up.
Right working
Positive deviation: $Δ_{mi x} H > 0$ (heat absorbed, mixture cools); negative deviation: $Δ_{mi x} H < 0$ (mixture warms).
Do this: Remember: positive deviation means vapour pressure, enthalpy of mixing and volume change are all positive.
Usually costs: 1 mark per wrong sign in a 3-mark question.
Mistake: Pairing azeotropes with the wrong deviation
'Maximum' is linked with 'positive' by habit.
Wrong working
Ethanol–water is a maximum boiling azeotrope.
Right working
Ethanol–water: positive deviation, minimum boiling azeotrope.
Do this: Higher vapour pressure boils sooner, so positive deviation gives the minimum boiling point.
Usually costs: 1 mark.
Mistake: Forgetting that only the solvent is volatile in a sugar or urea solution
Students add a term for the solute.
Wrong working
$p = x_{1} p_{1}^{0} + x_{2} p_{2}^{0}$ for urea in water.
Right working
$p = x_{1} p_{1}^{0}$ , because urea has no measurable vapour pressure.
Do this: Read 'non-volatile' in the question and drop the solute's term.
Usually costs: The whole numerical.

Exam tips

In 'identify A and B type' questions, write the observation (vapour pressure higher or lower than ideal), the cause (A–B forces), and the consequences ( $Δ_{mi x} H$ , $Δ_{mi x} V$ , azeotrope) as three separate lines.
Quote the NCERT examples exactly: positive, ethanol + acetone and carbon disulphide + acetone; negative, chloroform + acetone, phenol + aniline, nitric acid + water.
When a question gives masses, show the mole calculation; an answer without moles loses the method mark.
Keep units consistent: vapour pressures in one unit (mm Hg, bar or kPa) throughout.
For molar mass from vapour pressure lowering, the exact form $\frac{p ^{0} - p}{p} = \frac{n _{2}}{n _{1}}$ and the dilute form $\frac{p ^{0} - p}{p ^{0}} \approx \frac{n _{2}}{n _{1}}$ give slightly different answers; say which one you use.

Practice questions on Raoult's Law, Ideal and Non-ideal Solutions

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 mixture of chloroform and acetone:
(a) shows positive deviation and forms a minimum boiling azeotrope
(b) shows negative deviation and forms a maximum boiling azeotrope
(c) is an ideal solution
(d) shows positive deviation and has $Δ_{mi x} H < 0$
Hint
Think about the hydrogen bond between the C–H of chloroform and the oxygen of acetone.
Answer
(b). The new hydrogen bond makes A–B attraction stronger, so the vapour pressure is lower than ideal.
Q2 · 2 marks · Very short answerBoard level
State Raoult's law for a solution of two volatile liquids. Show that it is a special case of Henry's law.
Hint
Write both laws as partial pressure proportional to mole fraction.
Answer
For each volatile component, partial vapour pressure equals its mole fraction in the solution times the vapour pressure of the pure component: $p_{i} = x_{i} p_{i}^{0}$ . Henry's law says $p = K_{H} x$ ; Raoult's law is the same form with the constant $K_{H}$ equal to $p_{i}^{0}$ .
Q3 · 3 marks · Short answerHarder board level
Benzene (78 g mol⁻¹) and toluene (92 g mol⁻¹) form a nearly ideal solution. At a certain temperature their pure vapour pressures are 95 mm Hg and 28 mm Hg. A solution is made from 39 g benzene and 46 g toluene. Calculate (a) the total vapour pressure and (b) the mole fraction of benzene in the vapour.
Hint
Moles: 0.5 of each.
Answer
$x_{b e n z e n e} = x_{t o l u e n e} = 0.5$ . $p_{b e n z e n e} = 47.5$ mm Hg, $p_{t o l u e n e} = 14.0$ mm Hg. (a) $p_{t o t a l} = 61.5$ mm Hg. (b) $y_{b e n z e n e} = 47.5/61.5 = 0.772$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A mixture of phenol and aniline shows negative deviation from Raoult's law.
Reason (R): Hydrogen bonding between phenol and aniline molecules is stronger than the attractions within either pure liquid.
(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
Stronger A–B attraction holds molecules in the liquid.
Answer
(a). The O–H of phenol forms a strong hydrogen bond with the N of aniline, lowering the vapour pressure.
Q5 · 2 marks · Very short answerHarder board level
6.0 g of urea (60 g mol⁻¹) is dissolved in 180 g of water. The vapour pressure of pure water at this temperature is 31.8 mm Hg. Calculate the vapour pressure of the solution and the lowering of vapour pressure.
Hint
$n_{2} = 0.10$ , $n_{1} = 10$ .
Answer
$x_{1} = \frac{10}{10.1} = 0.990$ . $p = 0.990 \times 31.8 = 31.49$ mm Hg. Lowering $= 31.8 - 31.49 = 0.31$ mm Hg.
Q6 · 3 marks · Short answerHarder board level
Explain, with one example each, why some solutions show positive deviation and others negative deviation from Raoult's law. Give the sign of $Δ_{mi x} H$ and $Δ_{mi x} V$ in each case and describe how the total vapour pressure curve differs from the ideal straight line.
Hint
Compare A–B forces with A–A and B–B forces.
Answer
Positive (ethanol + acetone): A–B attraction weaker, molecules escape easily, $Δ_{mi x} H > 0$ , $Δ_{mi x} V > 0$ ; curve lies above the ideal line. Negative (chloroform + acetone): A–B attraction stronger, molecules held back, $Δ_{mi x} H < 0$ , $Δ_{mi x} V < 0$ ; curve lies below the ideal line.
Q7 · 3 marks · Short answerHOTS / competency
Two liquids A and B form an ideal solution, and pure A is three times as volatile as pure B ( $p_{A}^{0} = 3 p_{B}^{0}$ ). For a liquid with $x_{A} = 0.25$ , find the mole fraction of A in the vapour. What does your answer show about why repeated distillation separates A and B?
Hint
Let $p_{B}^{0} = p$ ; then $p_{A} = 0.75 p$ and $p_{B} = 0.75 p$ .
Answer
$p_{A} = 0.25 \times 3 p = 0.75 p$ and $p_{B} = 0.75 p$ , so $y_{A} = 0.5$ . The vapour (50% A) is twice as rich in A as the liquid (25% A). Condensing and re-evaporating this vapour enriches A again at each step, which is what a fractionating column does.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a chemistry teacher shows two bottles. One holds an ethanol–water mixture that the label says is about 95% ethanol by volume and was obtained by careful fractional distillation. The other holds nitric acid labelled about 68% $H N O_{3}$ by mass, which boils at about 393.5 K, higher than both water (373 K) and pure nitric acid.
(a) What is an azeotrope? (1 mark)
(b) Which bottle contains a minimum boiling azeotrope and which a maximum boiling one? Link each to the type of deviation. (2 marks)
(c) A student says 'more careful distillation of the 95% ethanol will give 100% ethanol'. Is the student right? Explain. (1 mark)
Hint
At the azeotropic composition, liquid and vapour have the same composition.
Answer
(a) A binary mixture that boils at a constant temperature with the same composition in liquid and vapour. (b) Ethanol–water: minimum boiling azeotrope, from large positive deviation. Nitric acid–water: maximum boiling azeotrope (boils above both components), from large negative deviation. (c) No. At about 95% the vapour has the same composition as the liquid, so further distillation cannot change it; other methods are needed.

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.State Raoult's law for a non-volatile solute.Show answer
The relative lowering of vapour pressure of the solvent equals the mole fraction of the solute: $\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = x_{2}$ .
2.Two conditions for an ideal solution?Show answer
$Δ_{mi x} H = 0$ and $Δ_{mi x} V = 0$ (it obeys Raoult's law at all compositions).
3.Which deviation does acetone + carbon disulphide show?Show answer
Positive deviation.
4.What kind of azeotrope does large negative deviation give?Show answer
A maximum boiling azeotrope.
5.Why is the vapour richer in the more volatile component?Show answer
Its partial pressure $x_{i} p_{i}^{0}$ is larger relative to its mole fraction, so $y_{i} > x_{i}$ for it.
6.Approximate composition of the ethanol–water azeotrope?Show answer
About 95% ethanol by volume (minimum boiling).

Raoult's Law, Ideal and Non-ideal Solutions: frequently asked questions

What is Raoult's law in Class 12 Chemistry?

For a solution of volatile liquids, the partial vapour pressure of each component equals its mole fraction in the solution times its vapour pressure when pure, p_i = x_i p_i^0.

What is the difference between ideal and non-ideal solutions?

An ideal solution obeys Raoult's law at every composition and has \Delta_{mix}H = 0 and \Delta_{mix}V = 0. A non-ideal solution does not obey it and shows positive or negative deviation, with non-zero enthalpy and volume of mixing.

Why does a mixture of chloroform and acetone show negative deviation?

Chloroform's C–H hydrogen bonds to the oxygen of acetone. This A–B attraction is stronger than in the pure liquids, so fewer molecules escape and the vapour pressure is lower than ideal.

Can an azeotrope be separated by fractional distillation?

No. At the azeotropic composition the liquid and vapour have the same composition, so distillation cannot change it.

Is relative lowering of vapour pressure a colligative property?

Yes. It equals the mole fraction of the solute, so it depends only on the number of solute particles, not on their nature.