JEE (Advanced) · Chemistry, Class 12
Solutions: JEE Advanced PYQ, 2007–2026
Solutions (Chemistry, Class 12) has been asked 30 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.8 questions a paper, 4% of all Chemistry questions. It is the 9th most-asked of 31 Chemistry chapters. The two 2026 papers had 3 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 1in 2 papers |
| 2008 | 3in 2 papers |
| 2009 | 1in 2 papers |
| 2010 | 0in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 1in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 2in 2 papers |
| 2015 | 1in 2 papers |
| 2016 | 2in 2 papers |
| 2017 | 2in 2 papers |
| 2018 | 2in 2 papers |
| 2019 | 2in 2 papers |
| 2020 | 1in 2 papers |
| 2021 | 2in 2 papers |
| 2022 | 1in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 1in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 3in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- van't Hoff factor9
- Colligative properties8
- Raoult's law8
- Concentration terms5
Try it
10 Solutions questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Solutions
Four aqueous solutions of equal molality (0.1 m) are prepared: (I) glucose, (II) KNO₃, (III) Na₃PO₄, (IV) CaCl₂. Assuming complete dissociation of the electrolytes, which of the following gives the correct order of increasing freezing point?
Answer: (C)
ΔTf = i·Kf·m, and at equal molality ΔTf is proportional to i. The solute with the largest i lowers the freezing point the most. Na₃PO₄ → 3Na⁺ + PO₄³⁻ gives i = 4. CaCl₂ → Ca²⁺ + 2Cl⁻ gives i = 3. KNO₃ → K⁺ + NO₃⁻ gives i = 2. Glucose does not ionise, so i = 1. The largest depression, and so the lowest freezing point, belongs to Na₃PO₄, and the smallest depression, so the highest freezing point, to glucose. The order is III < IV < II < I.
Q2 · Chemistry · Solutions
200 mg of a protein is dissolved in water to make 50 mL of solution. At 300 K the osmotic pressure of this solution is 2.5 mm Hg. Assuming the protein is a non-electrolyte and the solution behaves ideally, what is the approximate molar mass of the protein? (R = 0.0821 L atm K⁻¹ mol⁻¹; 1 atm = 760 mm Hg)
Answer: (A)
For a dilute solution of a non-electrolyte, π = CRT = (w/M)(RT/V), so M = wRT/(πV). Convert units first. w = 200 mg = 0.200 g. V = 50 mL = 0.050 L. π = 2.5/760 = 3.29 × 10⁻³ atm. M = (0.200 × 0.0821 × 300)/(3.29 × 10⁻³ × 0.050) = 4.926/1.645 × 10⁻⁴ ≈ 3.0 × 10⁴ g mol⁻¹. Missing the mL to L conversion gives about 30, missing mm Hg to atm gives about 39, and missing mg to g gives about 3 × 10⁷.
Q3 · Chemistry · Solutions
At 300 K, the total vapour pressure P (in torr) of an ideal solution of volatile liquids A and B varies with the mole fraction x_A of A in the liquid as P = 180 + 120 x_A. What is the partial vapour pressure of B above the solution when x_A = 0.4?
Answer: (D)
For an ideal solution, P = x_A p°_A + (1 − x_A) p°_B = p°_B + (p°_A − p°_B) x_A. Comparing with P = 180 + 120 x_A gives p°_B = 180 torr and p°_A = 300 torr. At x_A = 0.4, x_B = 0.6, so p_B = 0.6 × 180 = 108 torr. As a check, p_A = 0.4 × 300 = 120 torr and the sum is 228 torr, which matches the given equation.
Q4 · Chemistry · Solutions
Liquids A and B form an ideal solution. A solution of 2 mol of A and 3 mol of B has a vapour pressure of 320 torr. When 5 mol more of A is added at the same temperature, the vapour pressure becomes 260 torr. What is the mole fraction of A in the vapour in equilibrium with the first solution?
Answer: (B)
First solution, x(A) = 0.4: 0.4p°(A) + 0.6p°(B) = 320. Second, x(A) = 7/10: 0.7p°(A) + 0.3p°(B) = 260. Doubling the second and subtracting the first: p°(A) = 200 torr, and then p°(B) = 400 torr. Over the first solution p(A) = 0.4 × 200 = 80 torr, so the mole fraction of A in the vapour is 80/320 = 0.25.
Q5 · Chemistry · Solutions
An aqueous solution of a non-volatile, non-electrolyte solute boils at a higher temperature than pure water at the same external pressure. Which statement gives the correct reason?
Answer: (D)
A liquid boils when its vapour pressure equals the external pressure. A non-volatile solute lowers the vapour pressure of the solvent at every temperature, because it reduces the solvent's mole fraction at the surface. At the boiling point of pure solvent the solution's vapour pressure is therefore below the external pressure. The solution must be heated further to ΔTb above the solvent's boiling point before its vapour pressure matches the external pressure. Changes in external pressure, in intermolecular forces, or in bubble nucleation are not the cause of the colligative effect.
Q6 · Chemistry · Solutions
Liquids A and B form an ideal solution. At a certain temperature the vapour pressures of the pure liquids are 600 torr (A) and 200 torr (B). The solution is in equilibrium with its vapour, and the vapour contains equal numbers of moles of A and B. What is the total vapour pressure of the solution at this temperature?
Answer: (A)
Equal moles in the vapour means y(A) = y(B) = 0.5. By Raoult's law and Dalton's law, p(A) = x(A)p°(A) = y(A)P and p(B) = x(B)p°(B) = y(B)P. Therefore x(A) = y(A)P/p°(A) and x(B) = y(B)P/p°(B). Since x(A) + x(B) = 1, P(y(A)/p°(A) + y(B)/p°(B)) = 1. So 1/P = 0.5/600 + 0.5/200 = 1/1200 + 3/1200 = 4/1200, and P = 300 torr. As a check, x(A) = 0.5 × 300/600 = 0.25, and P = 0.25 × 600 + 0.75 × 200 = 300 torr. The liquid is richer in the less volatile B, so the 400 torr from simple averaging is wrong.
Q7 · Chemistry · Solutions
0.1 mol of CaCl₂ is dissolved in 90 g of water (molar mass 18 g/mol). Assuming CaCl₂ dissociates completely and the solution is ideal, what is the relative lowering of vapour pressure of the solution?
Answer: (A)
Relative lowering of vapour pressure equals the mole fraction of solute particles: (p° − p)/p° = i·n₂/(n₁ + i·n₂). Moles of water n₁ = 90/18 = 5. For complete dissociation of CaCl₂ → Ca²⁺ + 2Cl⁻, i = 3, so the effective solute moles are 3 × 0.1 = 0.3. The relative lowering is 0.3/(5 + 0.3) = 0.3/5.3 ≈ 0.057.
Q8 · Chemistry · Solutions
100 mL of 0.2 M aqueous glucose solution is mixed with 100 mL of 0.1 M aqueous Na2SO4 solution. Assume volumes are additive and Na2SO4 is 85% dissociated in the mixture. What is the osmotic pressure of the mixture at 300 K? (R = 0.083 L bar mol⁻¹ K⁻¹)
Answer: (A)
Principle: osmotic pressure depends on the total molar concentration of particles, π = (Σ iC)RT. Step 1, dilution: the volume doubles, so glucose becomes 0.10 M and Na2SO4 becomes 0.05 M. Step 2, the van't Hoff factor: Na2SO4 → 2Na⁺ + SO4²⁻ has n = 3, so i = 1 + (3 − 1)(0.85) = 2.7. Its particle concentration is 0.05 × 2.7 = 0.135 M, and glucose (i = 1) contributes 0.10 M. The total is 0.235 M. Step 3: π = 0.235 × 0.083 × 300 = 5.85 bar. Complete dissociation would give 6.23 bar, skipping dilution gives 11.70 bar, and using i = 1 + α gives 4.79 bar.
Q9 · Chemistry · Solutions
11.1 g of anhydrous CaCl2 (molar mass 111 g mol⁻¹) is dissolved in 250 g of water. Assume CaCl2 dissociates completely. Kb for water = 0.52 K kg mol⁻¹, and pure water boils at 100.000 °C at the given pressure. What is the boiling point of the solution?
Answer: (C)
Boiling point elevation for an electrolyte is ΔTb = i·Kb·m. Moles of CaCl2 = 11.1/111 = 0.100 mol. Molality = 0.100 mol / 0.250 kg = 0.400 mol kg⁻¹. Complete dissociation CaCl2 → Ca²⁺ + 2Cl⁻ gives i = 3. ΔTb = 3 × 0.52 × 0.400 = 0.624 K, so the boiling point is 100.624 °C. Dropping i gives 100.208 °C, using i = 2 gives 100.416 °C, and using the wrong solvent mass gives 100.156 °C.
Q10 · Chemistry · Solutions
Liquids A and B form an ideal solution. At a certain temperature the vapour pressures of the pure liquids are 240 torr (A) and 60 torr (B). A liquid mixture with mole fraction of A equal to 0.5 is placed in a closed vessel and equilibrated. The vapour is then completely condensed, and the condensate is allowed to come to equilibrium with its own vapour at the same temperature. What is the mole fraction of A in the vapour in equilibrium with the condensate?
Answer: (A)
For an ideal solution, P_i = x_i P_i° and the vapour mole fraction is y_i = P_i / P_total. Stage 1: with x_A = 0.5, P_A = 0.5 × 240 = 120 torr and P_B = 0.5 × 60 = 30 torr. Total = 150 torr, so y_A = 120/150 = 0.80. Condensing the vapour gives a liquid with x_A = 0.80 and x_B = 0.20. Stage 2: P_A = 0.8 × 240 = 192 torr and P_B = 0.2 × 60 = 12 torr. Total = 204 torr, so y_A = 192/204 = 0.941 ≈ 0.94. Each stage of evaporation and condensation enriches the more volatile component, which is the principle behind fractional distillation.
Free with an account
All 30 real JEE Advanced questions from Solutions.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
Solve them freeQuestions
About Solutions in JEE Advanced.
How many questions come from Solutions in JEE Advanced?
Solutions (Chemistry, Class 12) has been asked 30 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.8 questions a paper, 4% of all Chemistry questions. It is the 9th most-asked of 31 Chemistry chapters. The two 2026 papers had 3 from it.
Which topics of Solutions are asked most in JEE Advanced?
van't Hoff factor (9), Colligative properties (8), Raoult's law (8), Concentration terms (5), counted over the papers from 2007–2026.
Is Solutions important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 9 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Solutions JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Solutions with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
Your JEE Advanced 2027 plan, chapter by chapter.
A week-wise plan that gives the heavy chapters their time. Free, from a non-profit trust.