JEE (Main) · Chemistry, Class 12
Solutions: JEE Main PYQ, 2020–2025
Solutions (Chemistry, Class 12) has been asked 135 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.2 questions a paper, 5% of all Chemistry questions. It is the 8th most-asked of 21 Chemistry chapters. The 2025 shifts had 23 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 14in 14 papers |
| 2021 | 32in 26 papers |
| 2022 | 27in 24 papers |
| 2023 | 14in 12 papers |
| 2024 | 25in 18 papers |
| 2025 | 23in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Colligative properties47
- Concentration terms32
- van't Hoff factor32
- Raoult's law24
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10 Solutions questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Solutions
The Henry's law constant for nitrogen in water at 298 K is 8 × 10⁴ atm. What is the mole fraction of nitrogen dissolved in water in contact with air in which the partial pressure of nitrogen is 0.8 atm?
Answer: (B)
x = p/K_H = 0.8/(8 × 10⁴) = 1.0 × 10⁻⁵.
Q2 · 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.
Q3 · 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.
Q4 · Chemistry · Solutions
When Al₂(SO₄)₃ dissolves completely in water, how many ions are produced per formula unit, which is the van't Hoff factor i if the solute dissociates completely?
Answer: (C)
The van't Hoff factor i gives the ratio of actual particles to formula units for a solute. For complete dissociation, i equals the total number of ions formed per formula unit. Writing the dissociation equation Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻ shows 5 ions, so i = 5 for ideal complete dissociation. Balance the equation first, then add the ions.
Q5 · 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.
Q6 · Chemistry · Solutions
When liquids A and B are mixed, the mixture becomes warm and its volume is slightly less than the sum of the two volumes taken. Such a solution is expected to:
Answer: (B)
Heat evolved (Δ_mixH < 0) and a contraction in volume (Δ_mixV < 0) show that A–B attractions are stronger than A–A and B–B attractions. The molecules escape less readily, so the vapour pressure is lower than Raoult’s law predicts: negative deviation. A solution with a large negative deviation forms a maximum boiling azeotrope, as nitric acid and water do.
Q7 · Chemistry · Solutions
A non-volatile solute is dissolved in a volatile solvent. Which statement correctly describes the vapour above the solution?
Answer: (A)
A non-volatile solute has zero vapour pressure, so it cannot enter the gas phase. The vapour is therefore pure solvent vapour, and its partial pressure is reduced according to Raoult's law as p_solvent = x_solvent p°_solvent.
Q8 · Chemistry · Solutions
A sample of sulphuric acid is 49 % H₂SO₄ by mass and has a density of 1.4 g mL⁻¹. What is its molarity? (M of H₂SO₄ = 98 g mol⁻¹)
Answer: (B)
One litre of the solution weighs 1.4 × 1000 = 1400 g, of which 49 % is acid: 0.49 × 1400 = 686 g. Moles of H₂SO₄ = 686 / 98 = 7.0 mol in one litre, so the molarity is 7.0 M.
Q9 · 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.
Q10 · Chemistry · Solutions
A solution is prepared by mixing 46 g of ethanol (C₂H₅OH, molar mass 46 g mol⁻¹) with 54 g of water (molar mass 18 g mol⁻¹). What is the mole fraction of ethanol in the solution?
Answer: (C)
Mole fraction of a component is its moles divided by the total moles of all components in the solution. First convert each mass to moles using n = mass / molar mass: ethanol gives 1 mol and water gives 3 mol. The total is 4 mol, so the mole fraction of ethanol is 1/4 = 0.25. The mole fractions of all components in a solution must add up to 1, which gives a check: 0.25 + 0.75 = 1.
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All 135 real JEE Main 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.
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About Solutions in JEE Main.
How many questions come from Solutions in JEE Main?
Solutions (Chemistry, Class 12) has been asked 135 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.2 questions a paper, 5% of all Chemistry questions. It is the 8th most-asked of 21 Chemistry chapters. The 2025 shifts had 23 from it in all.
Which topics of Solutions are asked most in JEE Main?
Colligative properties (47), Concentration terms (32), van't Hoff factor (32), Raoult's law (24), counted over the papers from 2020–2025.
Is Solutions important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 8 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Solutions JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main 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.
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