Solutions Class 12 Chemistry Notes for JEE and NEET

ChemistryClass 12 Chemistry, Unit 1For JEE and NEET

The whole chapter on one page for JEE and NEET: concentration terms, Henry's and Raoult's laws, ideal and non-ideal solutions, the four colligative properties and the van't Hoff factor, each with the condition under which it holds. A table shows how often each idea was needed in past JEE and NEET papers, and ten past-paper questions from both exams are solved at the end. For the full explanation of a topic, follow the links to our concept pages below.

1. Types of solutions and concentration

A solution is a homogeneous mixture. The component in the larger amount is usually the solvent. Any of the three states can dissolve in any other: oxygen and nitrogen (gas in gas), chloroform vapour in nitrogen (liquid in gas), camphor in nitrogen (solid in gas), oxygen in water (gas in liquid), ethanol in water (liquid in liquid), glucose in water (solid in liquid), hydrogen in palladium (gas in solid), an amalgam of mercury with sodium (liquid in solid) and copper in gold (solid in solid).

Term	Definition	Note
Mass percentage	mass of the component ÷ mass of the solution × 100	w/w; volume percentage (v/v) and mass by volume (w/v) are defined the same way
Parts per million (ppm)	parts of the component in $1 0^{6}$ parts of the solution	for traces, such as gases in water
Mole fraction	$x_{i}$ $= n_{i} / (n_{1} + n_{2} + \dots)$	all mole fractions add up to 1
Molarity (M)	moles of solute per litre of solution	changes with temperature (the volume changes)
Molality (m)	moles of solute per kg of solvent	does not change with temperature

Molality from molarity and density

m = \frac{1000 M}{1000 d - M M _{2}}

$d$ is the density in g/mL and $M_{2}$ the molar mass of the solute; work with 1 L of solution.

Mole fraction from molality

x_{2} = \frac{m M _{1}}{1000 + m M _{1}}

$M_{1}$ is the molar mass of the solvent (g/mol).

Syllabus: JEE Main and NEET list molality, molarity, mole fraction and percentages in this unit; JEE Advanced lists mole fraction, molarity, molality and normality under General topics. No 2026 text names ppm or the types of solutions, but JEE Main 2025 and NEET 2025 asked them.

2. Solubility of gases: Henry's law

Henry's law

p = K_{H} x

$p$ is the partial pressure of the gas above the solution and $x$ its mole fraction in the solution. A larger $K_{H}$ means a less soluble gas. Holds at constant temperature for a gas at low pressure that does not react with the solvent or ionise in it (not $N H_{3}$ or $HCl$ in water).

A gas dissolves less as the temperature rises ( $K_{H}$ rises), so aquatic life is more comfortable in cold water.
Soft drinks are sealed under a high pressure of $C O_{2}$ ; the gas comes out when the bottle is opened.
Divers breathing compressed air dissolve more $N_{2}$ in their blood; rising too fast releases bubbles (the bends).
At high altitude the partial pressure of oxygen is low, so less oxygen dissolves in the blood (anoxia).

Syllabus: JEE Advanced lists Henry's law under Solutions; JEE Main and NEET list it under Equilibrium (physical processes). All three exams have it.

3. Vapour pressure and Raoult's law

Two volatile liquids

p = x_{A} p_{A}^{\circ} + x_{B} p_{B}^{\circ}, y_{A} = \frac{x _{A} p _{A}^{\circ}}{p}

For an ideal solution (one that obeys Raoult's law): each liquid's partial pressure is its mole fraction times its pure vapour pressure; $y_{A}$ is its mole fraction in the vapour. The vapour is richer in the more volatile liquid: $y_{A} / y_{B}$ $= (p_{A}^{\circ} / p_{B}^{\circ}) (x_{A} / x_{B})$ .

A non-volatile solute

\frac{p _{1}^{\circ} - p}{p _{1}^{\circ}} = x_{2}

The solvent's vapour pressure falls in proportion to the mole fraction of solute particles.

Solution	Mixing	Examples
Ideal: obeys Raoult's law at every composition	$Δ_{mix} H = 0$ , $Δ_{mix} V = 0$	n-hexane + n-heptane, benzene + toluene, bromoethane + chloroethane
Positive deviation: A-B attractions weaker than A-A and B-B	$Δ_{mix} H > 0$ , $Δ_{mix} V > 0$	ethanol + acetone, carbon disulphide + acetone
Negative deviation: A-B attractions stronger	$Δ_{mix} H < 0$ , $Δ_{mix} V < 0$	chloroform + acetone (a hydrogen bond forms), phenol + aniline, nitric acid + water

Figure 1: Total vapour pressure against composition, computed for

p_{B}^{\circ} = 60

and

p_{A}^{\circ} = 100

(any units) with a simple model of the deviations. An ideal mixture follows Raoult's law, a straight line. A large positive deviation gives a maximum (a minimum-boiling azeotrope); a large negative deviation gives a minimum (a maximum-boiling azeotrope).

An azeotrope is a mixture whose vapour has the same composition as the liquid, so it boils at a constant temperature and fractional distillation cannot separate it. A large positive deviation gives a minimum-boiling azeotrope (ethanol and water, about 95% ethanol by volume); a large negative deviation gives a maximum-boiling azeotrope (nitric acid and water, about 68% nitric acid by mass).

Syllabus: non-ideal solutions and vapour pressure-composition plots (where deviations and azeotropes come from) are named for JEE Main and NEET; no 2026 text names azeotropes. The JEE Advanced 2026 text names only ideal solutions.

4. Colligative properties

In a dilute solution of a non-volatile solute these depend only on the number of solute particles, not on what they are. Each one gives the molar mass $M_{2}$ of the solute ( $w_{2}$ g in $w_{1}$ g of solvent):

Property	Formula	Molar mass
Relative lowering of vapour pressure	$\frac{p _{1}^{\circ} - p}{p _{1}^{\circ}}$ $= x_{2} \approx \frac{n _{2}}{n _{1}}$	$M_{2}$ $= \frac{w _{2} M _{1} p _{1}^{\circ}}{w _{1} ( p _{1}^{\circ} - p )}$ (dilute)
Elevation of boiling point	$Δ T_{b} = K_{b} m$	$M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$
Depression of freezing point	$Δ T_{f} = K_{f} m$	$M_{2} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}}$
Osmotic pressure	$π = C R T$	$M_{2} = \frac{w _{2} R T}{π V}$

For water $K_{b} = 0.52$ and $K_{f} = 1.86$ K kg mol $^{- 1}$ , so the freezing point falls more than the boiling point rises for the same solution.

Kb and Kf from the solvent's data

K_{b} = \frac{R T _{b}^{2} M _{1}}{1000 Δ _{v a p} H}, K_{f} = \frac{R T _{f}^{2} M _{1}}{1000 Δ _{f u s} H}

$M_{1}$ in g/mol and the enthalpy in J/mol. For water they give 0.51 and 1.86 K kg mol $^{- 1}$ , close to the measured 0.52 and 1.86. Not named in any 2026 syllabus, but JEE Advanced 2026 and JEE Main 2025 asked it.

Osmosis. Solvent passes through a semipermeable membrane from the dilute side to the concentrated side. The osmotic pressure $π$ is the extra pressure that must be applied on the solution, when it is separated from pure solvent, to stop this flow; when it is shown by a column of solution of height $h$ , $π = ρ g h$ . Solutions with the same osmotic pressure are isotonic: 0.9% (w/v) sodium chloride (normal saline) is isotonic with blood. In a more concentrated (hypertonic) solution a cell shrinks; in a hypotonic one it swells. A pressure larger than $π$ on the solution side reverses the flow (reverse osmosis), which purifies sea water. Osmotic pressure is used for the molar masses of proteins and polymers: it is measured near room temperature (biomolecules break down when heated) and it is large even for very dilute solutions.

5. Abnormal molar mass and the van't Hoff factor

The van't Hoff factor

i = \frac{normal molar mass}{observed molar mass} = \frac{observed colligative property}{calculated colligative property}

Every colligative formula takes $i$ : $Δ T_{b} = i K_{b} m$ , $Δ T_{f} = i K_{f} m$ , $π = i C R T$ , and the relative lowering equals the mole fraction of all solute particles.

Case	Degree $α$	van't Hoff factor
Dissociation into $n$ particles	fraction dissociated	$i = 1 + (n - 1) α$
Association of $n$ molecules into one	fraction associated	$i = 1 - (1 - \frac{1}{n}) α$

Typical values for full dissociation: $NaCl$ 2, $MgS O_{4}$ 2, $K_{2} S O_{4}$ 3, $AlC l_{3}$ 4; urea and glucose 1. Ethanoic acid dimerises in benzene through hydrogen bonds, so $i$ falls towards $\frac{1}{2}$ and the observed molar mass is about twice the true one.

6. Which ideas JEE and NEET use most

We read all 70 questions on solutions in our 55 past papers (JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026) and noted the ideas each solution needs; one question can need several. The percentages are shares of each exam's questions on this chapter. The ideas used most:

Idea or formula	JEE (48 papers)	NEET (7 papers)
van't Hoff factor $i$ and abnormal molar mass	1830%	111%
Concentration: molality, molarity, mole fraction, percentages, ppm	1220%	111%
Elevation of boiling point $Δ T_{b} = i K_{b} m$	1220%	111%
Raoult's law for two liquids, $p = x_{A} p_{A}^{\circ} + x_{B} p_{B}^{\circ}$ , and the vapour's make-up	915%	222%
Osmotic pressure $π = i C R T$ , isotonic solutions, reverse osmosis	1118%	00%
Ideal and non-ideal solutions, deviations, azeotropes	58%	444%
Depression of freezing point $Δ T_{f} = i K_{f} m$	813%	00%
Relative lowering of vapour pressure equals the mole fraction of the solute	610%	00%

Show the other 3

Idea or formula	JEE (48 papers)	NEET (7 papers)
Henry's law $p = K_{H} x$ and the solubility of gases	47%	222%
$K_{b}$ and $K_{f}$ from the solvent's boiling or freezing point	23%	00%
Types of solutions	00%	111%

Questions per paper: JEE 61 in 48 papers (about 1.3 a paper), NEET 9 in 7 papers (about 1.3 a paper). The NEET column is smaller only because we have fewer NEET papers. Our bank has no NEET 2020 or 2021 paper.

7. Exam traps

Watch out

Molality uses the mass of the solvent in kg, not of the solution; molarity changes with temperature, molality does not.
Put the van't Hoff factor $i$ in every colligative formula; association (acid dimers in benzene) gives $i$ below 1.
$K_{f}$ of water (1.86) is larger than $K_{b}$ (0.52), so the freezing point falls more than the boiling point rises.
The relative lowering of vapour pressure equals the mole fraction of the solute, not of the solvent.
For an ideal solution the vapour is richer in the more volatile liquid: $y_{A} / y_{B}$ $= (p_{A}^{\circ} / p_{B}^{\circ}) (x_{A} / x_{B})$ .
A large positive deviation gives a minimum-boiling azeotrope; a large negative deviation (chloroform + acetone, nitric acid + water) a maximum-boiling one.
A larger Henry's constant $K_{H}$ means a less soluble gas.
Isotonic means equal $i \times C$ , not equal $C$ ; reverse osmosis needs a pressure larger than $π$ on the more concentrated side.
Osmotic pressure, not the boiling point, is the colligative property used for proteins and polymers.

8. What the 2026 syllabi say

JEE Main 2026 (NTA syllabus, Unit 5) and NEET (UG) 2026 (NMC syllabus, Unit 5) use the same words (quoted from JEE Main): "Different methods for expressing the concentration of solution - molality, molarity, mole fraction, percentage (by volume and mass both), the vapour pressure of solutions and Raoult's Law - Ideal and non-ideal solutions, vapour pressure - composition, plots for ideal and non- ideal solutions, Colligative properties of dilute solutions - a relative lowering of vapour pressure, depression of freezing point, the elevation of boiling point and osmotic pressure, determination of molecular mass using colligative properties, abnormal value of molar mass, van't Hoff factor and its significance." Both list Henry's law under Equilibrium (Unit 6).

JEE Advanced 2026 (Information Brochure, Annexure-I, Chemistry): "Henry's law; Raoult's law; Ideal solutions; Colligative properties: lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure; van't Hoff factor." Concentration is under General topics: "Concentration in terms of mole fraction, molarity, molality and normality."

So non-ideal solutions are in the JEE Main and NEET texts only, and deviations and azeotropes come with them (no text names azeotropes). Percentages are not in the JEE Advanced text. Parts per million (ppm), the types of solutions and the $K_{b}$ , $K_{f}$ formulas are in none, though they have been asked.

9. Solved past-paper questions

Ten questions from JEE Main, JEE Advanced and NEET, chosen to cover the ideas asked most. The question and the answer key are as set in the paper; the working is ours.

PYQ 1: Molarity to molality

JEE Main 2025 Jan 22 Shift 2, Chemistry Q11

Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is :

(A) 1.79 m
(B) 2 m
(C) 3 m
(D) 2.79 m

Solution:

Take 1 L of solution: mass $= 1000 \times 1.25$ $= 1250$ g. It holds 3 mol $NaCl$ $= 3 \times 58.5 = 175.5$ g.
Mass of water $= 1250 - 175.5$ $= 1074.5$ g $= 1.0745$ kg, so molality $= 3/1.0745 = 2.79$ m (D).

Answer key (D) 2.79 m

PYQ 2: Raoult's law and the vapour

JEE Advanced 2026 Paper 2, Chemistry Section 4 Q2

Question Stem (Q1-Q2). Two volatile liquids $A$ and $B$ form an ideal solution. Consider a $5$ molal solution of $B$ in $A$ inside a closed container having a total vapour pressure of $100$ mm Hg at $300$ K. The vapour pressure of pure $A$ at $300$ K is $105$ mm Hg. Assume that $A$ and $B$ behave as ideal gases in the vapour phase.

Given: Gas constant $R = 0.08$ L atm K $^{- 1}$ mol $^{- 1}$ ; molar mass of $A$ is $50$ g mol $^{- 1}$ ; molar mass of $B$ is $57$ g mol $^{- 1}$ ; density of liquid $B$ at $300$ K is $0.5$ g/mL; $1$ atm $= 760$ mm Hg.

Q2. The mole fraction of $B$ in vapour phase which is in equilibrium with this solution is _____.

Solution:

5 molal: 5 mol B in 1000 g A, which is $1000/50 = 20$ mol A. So $x_{A} = 20/25 = 0.8$ and $x_{B} = 0.2$ .
Raoult's law: $100$ $= 0.8 \times 105 + 0.2 p_{B}^{\circ}$ , so $p_{B}^{\circ} = 80$ mm Hg.
Partial pressure of B $= 0.2 \times 80 = 16$ mm Hg, so $y_{B} = 16/100 = 0.16$ .

Answer key 0.16

PYQ 3: Which deviation?

NEET 2025, Chemistry Q36

$5$ moles of liquid X and $10$ moles of liquid Y make a solution having a vapour pressure of $70$ torr. The vapour pressures of pure X and Y are $63$ torr and $78$ torr respectively. Which of the following is true regarding the described solution?

(A) The solution shows positive deviation.
(B) The solution shows negative deviation.
(C) The solution is ideal.
(D) The solution has volume greater than the sum of individual volumes.

Solution:

$x_{X} = 5/15 = 1/3$ and $x_{Y} = 2/3$ . Raoult's law predicts $63/3 + 78 \times 2/3$ $= 21 + 52$ $= 73$ torr.
The observed 70 torr is lower, so the solution shows negative deviation (B): X-Y attractions are stronger than X-X and Y-Y attractions, and the volume shrinks on mixing (so D is wrong).

Answer key (B) The solution shows negative deviation.

PYQ 4: Ions per formula unit from the lowering

JEE Advanced 2022 Paper 2, Chemistry Section 1 Q2

An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35°C. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is 59.724 mm of Hg. Vapor pressure of water at 35°C is 60.000 mm of Hg. The number of ions present per formula unit of the ionic salt is _______.

Solution:

The relative lowering $(60.000 - 59.724) /60.000$ is 0.0046. Water: $1800/18 = 100$ mol; salt particles: $0.1 i$ mol.
$0.0046$ $= \frac{0.1 i}{100 + 0.1 i}$ gives $i = 4.62$ .
With 90% dissociation into $n$ ions, $i = 1 + 0.9 (n - 1)$ , so $n$ $= 1 + 3.62/0.9$ $= 5.0$ : 5 ions per formula unit.

Answer key 5

PYQ 5: Highest boiling point

NEET 2025, Chemistry Q34

Which of the following aqueous solution will exhibit highest boiling point?

(A) 0.01 M Urea
(B) 0.01 M KNO $_{3}$
(C) 0.01 M Na $_{2}$ SO $_{4}$
(D) 0.015 M C $_{6}$ H $_{12}$ O $_{6}$

Solution:

In dilute water solutions molality is about equal to molarity, and the salts are taken as fully dissociated, so the elevation is proportional to $i \times m$ . Urea: $1 \times 0.01$ ; $KN O_{3}$ : $2 \times 0.01$ ; $N a_{2} S O_{4}$ : $3 \times 0.01$ ; glucose: $1 \times 0.015$ .
$N a_{2} S O_{4}$ has the largest $i m$ (0.03), so it boils highest (C).

Answer key (C) 0.01 M Na $_{2}$ SO $_{4}$

PYQ 6: Ka from the freezing point

JEE Main 2025 Apr 8 Shift 2, Chemistry Q9

$H A (a q) ⇌ H^{+} (a q) + A^{-} (a q)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 $^{\circ}$ C. The dissociation constant for the acid is

Given: $K_{f} (H_{2} O) = 1.8$ K kg mol $^{- 1}$ , molality $\equiv$ molarity

(A) $1.38 \times 1 0^{- 3}$
(B) $1.1 \times 1 0^{- 2}$
(C) $1.90 \times 1 0^{- 3}$
(D) $1.89 \times 1 0^{- 1}$

Solution:

$Δ T_{f} = i K_{f} m$ : $0.20 = i \times 1.8 \times 0.1$ , so $i = 10/9$ .
For $HA ⇌ H^{+} + A^{-}$ , $i = 1 + α$ , so $α = 1/9$ .
$K_{a}$ $= \frac{C α ^{2}}{1 - α}$ $= \frac{0.1 \times ( 1/81 )}{8/9}$ $= \frac{1}{720}$ $= 1.39 \times 1 0^{- 3}$ , closest to option (A), $1.38 \times 1 0^{- 3}$ .

Answer key (A) $1.38 \times 1 0^{- 3}$

PYQ 7: Molar mass of a protein from osmosis

JEE Main 2026 Apr 5 Shift 2, Chemistry Q23

20 g of hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by a semi-permeable membrane. At equilibrium the height of solution in a tube dipped in solution (A) is found to be 80.0 mm higher than the tube dipped in water.

The molar mass of hemoglobin is _____ kg mol $^{- 1}$ . (Nearest integer)

Given: $g = 10$ ms $^{- 2}$ , $R = 8.3$ kPa·dm $^{3}$ ·K $^{- 1}$ ·mol $^{- 1}$ , density of solution $= 1000$ kg m $^{- 3}$ .

Solution:

$π$ $= ρ g h$ $= 1000 \times 10 \times 0.080$ $= 800$ Pa $= 0.800$ kPa.
Units: kPa $\times$ L $=$ J, so $R = 8.3$ J/(mol K) works with $V$ in litres.
$π = \frac{w}{M} \cdot \frac{R T}{V}$ , so $M$ $= \frac{20 \times 8.3 \times 300}{0.800 \times 1}$ $= 62250$ g mol $^{- 1}$ , about 62 kg mol $^{- 1}$ .

Answer key 62

PYQ 8: Degree of ionisation from the molar mass

JEE Main 2025 Jan 24 Shift 2, Chemistry Q21

The observed and normal masses of compound $M X_{2}$ are 65.6 and 164 respectively. The percent degree of ionisation of $M X_{2}$ is ________%. (Nearest integer)

Solution:

$i = 164/65.6 = 2.5$ .
$M X_{2}$ gives three ions: $i = 1 + 2 α$ , so $α = 0.75$ , that is 75%.

Answer key 75

PYQ 9: Kb from the solvent, then the degree of dissociation

JEE Advanced 2026 Paper 2, Chemistry Section 3 Q3

In a solvent $S$ , a compound $B$ is partially dissociated into $C$ and $D$ as given below:

$B ⇌ 2 C + 2 D$

$B$ , $C$ and $D$ are non-volatile in nature. The molar mass of $B$ is $10$ times the molar mass of $S$ . The standard boiling point and the standard enthalpy of vaporization of $S$ are $400$ K and $10 R$ J mol $^{- 1}$ , respectively ( $R$ is the gas constant in J K $^{- 1}$ mol $^{- 1}$ ). A solution of $B$ in $S$ with an initial concentration of $B$ as $0.25%$ (mass/mass) has a boiling point of $408$ K at $1$ bar pressure. In this solution, the mole percent of $B$ that has been dissociated is _____.

Solution:

$K_{b}$ $= \frac{R T _{b}^{2} M _{S}}{1000 Δ _{v a p} H}$ $= \frac{R \times 40 0 ^{2} \times M _{S}}{1000 \times 10 R}$ $= 16 M_{S}$ K kg mol $^{- 1}$ .
100 g of solution holds 0.25 g B in 99.75 g S, with $M_{B} = 10 M_{S}$ : $m$ $= \frac{0.25/ ( 10 M _{S} )}{0.09975}$ $= \frac{0.2506}{M _{S}}$ .
$Δ T_{b}$ $= 8$ $= i \times 16 M_{S} \times \frac{0.2506}{M _{S}}$ $= 4.010 i$ and so $i = 1.995$ .
B gives four particles ( $2 C + 2 D$ ): $i = 1 + 3 α$ , so $α = 0.3317$ : 33.17 mole percent.

Answer key 33.17

PYQ 10: Henry's law and solubility

NEET 2024, Chemistry Q25

The Henry's law constant ( $K_{H}$ ) values of three gases (A, B, C) in water are 145, $2 \times 1 0^{- 5}$ and 35 kbar, respectively. The solubility of these gases in water follow the order:

(A) A $>$ C $>$ B
(B) A $>$ B $>$ C
(C) B $>$ A $>$ C
(D) B $>$ C $>$ A

Solution:

$x = p / K_{H}$ : at the same pressure, the smaller $K_{H}$ , the more soluble the gas.
$K_{H}$ : B ( $2 \times 1 0^{- 5}$ ) < C (35) < A (145) kbar, so the solubility is B > C > A (D).

Answer key (D) B $>$ C $>$ A

Every number on this page is checked by a script, and every worked answer is recomputed and matched with the answer key. Past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026; the questions were read by hand (lists checked 2026-10-19).

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