JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Vapour Pressure Raoults Law
TOPIC 2 OF 23
Vapour Pressure Raoults Law
🎓 Class 12
Chemistry
CBSE
Theory
Ch 1 – Solutions
⏱ ~14 min
🌐 Language:
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Vapour Pressure Raoults Law
Upload images, PDFs, or Word documents to include their content in assessment generation.
Vapour Pressure Raoults Law
Recap and Roadmap
In Part 1 we saw how to express "how much solute" in a solution. But the story of solutions does not end with concentration. When we pour ethanol into water, the vapour above the mixture contains both species; when we boil the mix, the resulting distillate is richer in one than the other. To handle such behaviour we need the idea of vapour pressure — the pressure exerted by the vapours of a liquid in equilibrium with itself. This second part develops that idea into Raoult's law, classifies solutions as ideal or non-ideal, and ends with the fascinating world of azeotropes.
1.4 Vapour Pressure of Liquid Solutions
1.4.1 Vapour Pressure of Liquid–Liquid Solutions — Raoult's Law
Consider a binary solution made from two volatile liquids 1 and 2. Each liquid contributes its own vapour above the solution. In 1886 François-Marie Raoult proposed:
Raoult's Law: For a solution of volatile liquids, the partial pressure of each component is equal to the product of its pure-liquid vapour pressure and its mole fraction in the solution.
\(p_{1}=p_{1}^{\circ}\,x_{1} \qquad\text{and}\qquad p_{2}=p_{2}^{\circ}\,x_{2}\)
By Dalton's law, the total pressure is the sum:
\(p_{total}=p_{1}+p_{2}=p_{1}^{\circ}\,x_{1}+p_{2}^{\circ}\,x_{2} = p_{2}^{\circ}+(p_{1}^{\circ}-p_{2}^{\circ})\,x_{1}\)
The last rearrangement shows that ptotal is a linear function of x1 — the plot is a straight line joining p2° at x1 = 0 to p1° at x1 = 1.An everyday example is benzene + toluene, which behaves almost ideally because the two molecules have very similar shapes, sizes and intermolecular forces.
1.4.2 Composition of the Vapour — Dalton's Law
If y1 is the mole fraction of component 1 in the vapour, Dalton's law gives:
\(y_{1}=\dfrac{p_{1}}{p_{total}}=\dfrac{p_{1}^{\circ}\,x_{1}}{p_{1}^{\circ}\,x_{1}+p_{2}^{\circ}\,x_{2}}\)
Because p1° / p2° ≠ 1 in general, the vapour is richer in the more volatile component. This single fact makes fractional distillation possible.
Raoult's Law as a Special Case of Henry's Law
If component 2 is a non-volatile solute (x2 stays small), then the dissolved molecules behave like a gas in a liquid. Writing Henry's law in the form p = KH x and comparing with Raoult's law p = p° x, we see that Raoult's law is simply Henry's law with KH = p°. This unification is one of the most elegant ideas in physical chemistry.
Worked Examples — Raoult's Law
Example 1.8 — Total vapour pressure of benzene–toluene
At 353 K, the vapour pressures of pure benzene and pure toluene are 98.4 kPa and 36.4 kPa respectively. Calculate the total pressure and vapour composition over a solution containing 3.0 mol of benzene and 2.0 mol of toluene.
xbenzene = 3/(3+2) = 0.60; xtoluene = 0.40.
\(p_{total}=98.4(0.60)+36.4(0.40)=59.04+14.56=73.60\,\text{kPa}\)
Vapour-phase mole fractions:
\(y_{benzene}=\dfrac{59.04}{73.60}=0.802;\quad y_{toluene}=0.198\)
Answer: ptotal = 73.60 kPa; the vapour is 80.2 % benzene although the liquid is only 60 % benzene — the vapour is enriched in the more volatile component.
Example 1.9 — Mole fraction of liquid from given vapour composition
At 300 K, vapour pressures of pure A and pure B are 500 mm Hg and 200 mm Hg. A solution has a total vapour pressure of 350 mm Hg. Find xA in the liquid.
\(p_{total}=p_{B}^{\circ}+(p_{A}^{\circ}-p_{B}^{\circ})x_{A}\)
\(350=200+(500-200)x_{A}\;\Rightarrow\;x_{A}=\dfrac{150}{300}=0.50\)
Answer: xA = 0.50.
Example 1.10 — Vapour composition from liquid composition
Using the same A–B data above with the 0.50–0.50 liquid, find yA.
\(y_{A}=\dfrac{p_{A}^{\circ}x_{A}}{p_{total}}=\dfrac{500(0.5)}{350}=\dfrac{250}{350}=0.714\)
Although the liquid is 50 % A, the vapour is 71.4 % A. Repeated condensation–revaporisation cycles in a distillation column enrich A further — this is how refineries separate mixtures.
1.5 Ideal and Non-Ideal Solutions
Ideal Solutions
A solution is called ideal if it obeys Raoult's law over the entire composition range. Equivalently, when two liquids mix to form an ideal solution:
- ΔHmix = 0 — no heat is absorbed or evolved.
- ΔVmix = 0 — the total volume equals the sum of the component volumes.
- A–B intermolecular interactions are essentially identical to A–A and B–B interactions.
Examples: benzene + toluene; n-hexane + n-heptane; chlorobenzene + bromobenzene; ethyl bromide + ethyl iodide.
Non-Ideal Solutions
Most real solutions deviate from Raoult's law. Two patterns emerge:
Positive deviation
Observed ptotal > Raoult's-law prediction. The A–B attraction is weaker than A–A and B–B attractions, so molecules escape the liquid more easily. Mixing is endothermic (ΔHmix > 0) and expansive (ΔVmix > 0). Examples: ethanol + acetone, acetone + CS₂, ethanol + water, carbon disulphide + acetone.
Negative deviation
Observed ptotal < Raoult's-law prediction. A–B attractions are stronger than A–A, B–B (often because of hydrogen bonding or ion–dipole attractions). Mixing is exothermic (ΔHmix < 0) and contractive (ΔVmix < 0). Examples: chloroform + acetone, HNO₃ + water, HCl + water, phenol + aniline.
Azeotropes
Azeotropes are binary mixtures with a constant boiling point, at which the vapour has the same composition as the liquid. Two types exist:
- Minimum-boiling azeotrope: formed by solutions that show strong positive deviation. The enhanced escaping tendency of the vapour gives a ptotal maximum at some intermediate composition — and a corresponding minimum in the boiling-point curve. Example: ethanol + water at about 95.6 % ethanol, b.p. 351.15 K.
- Maximum-boiling azeotrope: formed by solutions showing strong negative deviation. A pressure minimum → boiling-point maximum. Example: HNO₃ + water at about 68 % HNO₃, b.p. 393.5 K.
Why it matters: At an azeotropic composition, the vapour and liquid have the same composition, so distillation produces no enrichment — the mixture is a "distillation dead-end". Industrial alcohol manufacturers solve this by adding benzene or CaO to break the azeotrope.
Example 1.11 — Classify the mixture
When acetone and chloroform are mixed at room temperature, the beaker feels warm. Also, the observed vapour pressure is lower than what Raoult's law predicts. What type of solution is this?
The mixing is exothermic (warm beaker) and the vapour pressure is below Raoult's law → this is a negative deviation. The driving force is the strong C–H⋯O=C hydrogen bond between chloroform's H and acetone's carbonyl.
Example 1.12 — Ethanol–acetone mixture
Ethanol self-associates by O–H⋯O hydrogen bonds. When acetone is added, some of these bonds are disrupted. Predict the type of deviation.
Breaking ethanol's own hydrogen bonds raises the escaping tendency of ethanol molecules; the new O–H⋯O=C bonds between ethanol and acetone are weaker than the ones they replaced. Result: observed ptotal > Raoult prediction — a positive deviation, which gives the well-known ethanol–acetone azeotrope.
Example 1.13 — Azeotrope identification
A 68 % w/w HNO₃–water mixture has a boiling point of 393.5 K, higher than either pure component. Explain and classify.
HNO₃ and water are strongly attracted by H-bonding and proton transfer, giving a negative deviation. The total pressure shows a minimum, so the boiling point shows a maximum — a maximum-boiling azeotrope.
Activity 1.2 — "Warm" Mixing Demonstration
L3 Apply
Predict: If you mix equal volumes of ethanol and water, will the total volume (a) stay the same, (b) be a little less than the sum, or (c) be a little more? Why?
- Measure 50.0 mL of ethanol and 50.0 mL of distilled water in two separate measuring cylinders.
- Combine them in a 100-mL graduated cylinder. Note the final volume.
- Touch the outer wall of the cylinder — is it warm or cool?
- Compare with the Raoult's-law prediction (ΔVmix = 0 for ideal solutions).
Observation: the final volume is about 96 mL (not 100 mL) — a volume contraction of ~4 %. The cylinder feels slightly warm (exothermic). Interpretation: ethanol's H-bond network is broken, but the new ethanol-water hydrogen bonds are denser and more tightly packed — a mild negative deviation from Raoult's law. (Interestingly, at some compositions ethanol–water also shows positive features — the full curve is complex, which is why an azeotrope forms at 95.6 % ethanol.)
Interactive: Raoult's Law Plotter L3 Apply
Enter p₁°, p₂° and the current mole fraction x₁; the tool computes partial pressures, total pressure, and the vapour composition y₁.
p₁° (kPa):
p₂° (kPa):
x₁ (liquid):
Computed values will appear here.
Competency-Based Questions
At 300 K the vapour pressures of pure cyclohexane and pure n-hexane are 98 kPa and 60 kPa, respectively. A student prepares a mixture with xcyclo = 0.40. She also studies three "real" mixtures: (i) benzene–toluene, (ii) ethanol–water, and (iii) chloroform–acetone.
Q1. L2 Understand Which pair is expected to behave most nearly as an ideal solution?
Answer: A. Benzene and toluene are similar in size, shape and intermolecular forces. The other pairs have specific H-bonding or strong A–B interactions that cause deviation.
Q2. L3 Apply Compute ptotal for the cyclohexane–hexane mixture. (2 marks)
ptotal = 98(0.40) + 60(0.60) = 39.2 + 36.0 = 75.2 kPa.
Q3. L3 Apply Find the mole fraction of cyclohexane in the vapour. (2 marks)
ycyclo = 39.2/75.2 = 0.521. The vapour is 52 % cyclohexane while the liquid is only 40 % — the more volatile cyclohexane is enriched.
Q4. L4 Analyse Chloroform–acetone mixing is strongly exothermic. Sketch the ptotal vs xchloroform curve and mark the azeotropic composition. (3 marks)
Because A–B attraction (C–H⋯O=C hydrogen bond) is stronger than A–A or B–B, ptotal lies below the Raoult's-law straight line — a negative deviation with a pressure minimum at about 65 % chloroform. Since pressure is a minimum there, the boiling point is a maximum ≈ 337.8 K — a maximum-boiling azeotrope.
Q5. L5 Evaluate Why is it impossible to obtain 100 % pure ethanol by simple distillation of a dilute ethanol–water mixture? Suggest a practical solution. (3 marks)
Ethanol–water shows positive deviation and forms a minimum-boiling azeotrope at 95.6 % ethanol (b.p. 351.1 K). At this composition the vapour has the same composition as the liquid; further distillation yields no enrichment. To cross the azeotrope, one adds a third component (benzene — "azeotropic distillation") or a chemical drying agent such as CaO, which ties up the last water and lets one distil over the remaining pure ethanol.
Assertion-Reason Questions
Assertion (A): A solution of chloroform and acetone shows a negative deviation from Raoult's law.
Reason (R): Hydrogen bonds form between the H of chloroform and the C=O of acetone, strengthening A–B interactions.
Answer: A. Strong A–B attraction reduces escaping tendency → lower ptotal than Raoult predicts. The stated H-bond is precisely why.
Assertion (A): For an ideal solution, ΔVmix = 0 and ΔHmix = 0.
Reason (R): In an ideal solution A–B intermolecular forces are identical to A–A and B–B forces, so neither heat nor volume changes on mixing.
Answer: A. Ideality is defined by equality of molecular interactions, and this equality makes both ΔH and ΔV zero.
Assertion (A): An azeotropic mixture cannot be separated by simple fractional distillation.
Reason (R): At the azeotropic composition, the vapour and the liquid have the same composition.
Answer: A. Distillation works only when the vapour differs from the liquid. At an azeotrope they are identical, so no enrichment is possible.
Frequently Asked Questions - Vapour Pressure Raoults Law
What is the main concept covered in Vapour Pressure Raoults Law?
In NCERT Class 12 Chemistry Chapter 1 (Solutions), "Vapour Pressure Raoults Law" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Vapour Pressure Raoults Law useful in real-life or applied chemistry?
Real-life applications of "Vapour Pressure Raoults Law" from NCERT Class 12 Chemistry Chapter 1 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Vapour Pressure Raoults Law?
Key reactions in "Vapour Pressure Raoults Law" (NCERT Class 12 Chemistry Chapter 1 Solutions) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 1?
NCERT Class 12 Chemistry Chapter 1 (Solutions) is structured so each part builds chemical understanding sequentially. "Vapour Pressure Raoults Law" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Vapour Pressure Raoults Law?
CBSE board questions from "Vapour Pressure Raoults Law" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Vapour Pressure Raoults Law" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
AI Tutor
Chemistry Class 12 Part I – NCERT (2025-26)
Ready
🎯 Practise Chemistry
Sit a full paper on what you have been studying, marked question by question.
Mock exams
JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
JEE Main Mock Test — Full Paper (75 Questions, 300 Marks)
Part A - Physics (Section A) · Part A - Physics (Section B) · Part B - Chemistry (Section A) · Part B - Chemistry (Section B) · Part C - Mathematics (Section A) · Part C - Mathematics (Section B)
🎁 Join our community and get free AI credits!