How much of JEE Main comes from Solutions?

JEEGenie estimates about 3% of JEE Main marks come from Solutions, based on recent papers. The share varies from year to year.

Which formulas matter most in Solutions?

The Solutions sheet on this page has 15 formula entries, starting with Concentration Terms, Molarity & Molality Conversion, Henry's Law, Raoult's Law for Volatile Liquids, Ideal & Non-Ideal Solutions. Each one comes with when to use it and the trap to avoid.

What topics does Solutions cover for JEE?

Solutions covers Concentration Units, Raoult's Law, Colligative Properties, Osmotic Pressure, Van't Hoff Factor, Ideal and Non-ideal Solutions.

How should I revise Solutions before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Solutions questions on JEEGenie.

How many previous-year questions are there on Solutions?

JEEGenie has 323 previous-year questions on Solutions, and this page shows a few with worked explanations.

Is practice on Solutions free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Solutions for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 3% of JEE Main marks come from Solutions. This page covers 15 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Molarity depends on temperature (volume changes), molality does not.

Key points to know

Solution & Its Components. A solution is a homogeneous mixture of two or more substances; the component present in larger amount is the solvent, and the smaller amount is the solute.
Concentration Terms. Concentration expresses the amount of solute per amount of solution or solvent; common terms include mass percentage, mole fraction, molarity, molality, and ppm.
Solubility & Factors. Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature; it depends on nature of solute/solvent, temperature, and pressure (for gases).
Ideal & Non-Ideal Solutions. Ideal solutions obey Raoult's law at all concentrations and have no enthalpy or volume change on mixing; non-ideal solutions show positive or negative deviations.
Van't Hoff Factor (i). The van't Hoff factor accounts for the number of particles a solute dissociates into; for non-electrolytes $i = 1$ , for electrolytes $i > 1$ (or <1 if association).
Key Constants for Solvents. Memorize $K_{f}$ and $K_{b}$ for common solvents like water, benzene, and camphor.
Common Exam Problems: Mixing Solutions. When mixing two solutions of same solute, use the formula for final concentration: $M_{1} V_{1} + M_{2} V_{2} = M_{f} (V_{1} + V_{2})$ for molarity; for molality, use mass of solvent.
Henry's Law Constant Trends. Higher $K_{H}$ means lower solubility; $K_{H}$ increases with temperature for gases.
Key Points for JEE Numericals. Always check units: mass in grams, volume in liters, temperature in Kelvin; use $R = 0.0821$ L atm/mol K for osmotic pressure in atm.

Formula sheet

Name	Formula	When to use
Concentration Terms	Mass % = $\frac{mass of solute}{mass of solution} \times 100$ ; Mole fraction $x_{A} = \frac{n_{A}}{n_{A} + n_{B}}$ ; Molarity $M = \frac{n_{solute}}{V_{solution (L)}}$ ; Molality $m = \frac{n_{solute}}{mass of solvent (kg)}$ ; ppm = $\frac{mass of solute}{mass of solution} \times 10^{6}$	Use to calculate or convert between concentration units in problems.
Molarity & Molality Conversion	$m = \frac{1000 M}{1000 ρ - M \cdot M_{solute}}$ where $ρ$ is density in g/mL, $M_{solute}$ is molar mass of solute.	When a problem gives molarity and density and asks for molality or vice versa.
Henry's Law	$p = K_{H} \cdot x$ where $p$ is partial pressure, $x$ is mole fraction of gas in solution, $K_{H}$ is Henry's constant.	Use to calculate gas solubility at different pressures or to find Henry's constant.
Raoult's Law for Volatile Liquids	$p_{A} = x_{A} p_{A}^{0}$ , $p_{B} = x_{B} p_{B}^{0}$ ; Total pressure $P = p_{A} + p_{B} = x_{A} p_{A}^{0} + x_{B} p_{B}^{0}$	Use to calculate vapor pressure of a mixture of volatile liquids or to find composition of vapor.
Ideal & Non-Ideal Solutions	For ideal: $Δ H_{mix} = 0$ , $Δ V_{mix} = 0$ ; Positive deviation: $P > x_{A} p_{A}^{0} + x_{B} p_{B}^{0}$ ; Negative deviation: $P < x_{A} p_{A}^{0} + x_{B} p_{B}^{0}$	Use to classify solutions and predict deviations based on intermolecular forces.
Relative Lowering of Vapor Pressure	$\frac{p^{0} - p}{p^{0}} = x_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$	Use to calculate molar mass of non-volatile solute from vapor pressure data.
Boiling Point Elevation	$Δ T_{b} = K_{b} \cdot m$ where $K_{b}$ is ebullioscopic constant, $m$ is molality.	Use to find molar mass of solute or to calculate boiling point of a solution.
Freezing Point Depression	$Δ T_{f} = K_{f} \cdot m$ where $K_{f}$ is cryoscopic constant.	Use to find molar mass of solute or to calculate freezing point of a solution.
Osmotic Pressure	$Π = i M R T$ where $Π$ is osmotic pressure, $M$ is molarity, $R$ is gas constant, $T$ is temperature in Kelvin.	Use to determine molar mass of macromolecules or to compare osmotic pressures.
Van't Hoff Factor (i)	$i = \frac{observed colligative property}{expected for non-electrolyte}$ ; For dissociation: $i = 1 + α (n - 1)$ where $α$ is degree of dissociation, $n$ is number of ions.	Use when dealing with electrolytes or solutes that associate/dissociate.
Abnormal Molar Mass & van't Hoff Factor	$i = \frac{normal molar mass}{observed molar mass}$	Use to determine if a solute dissociates or associates from experimental molar mass data.
Determination of Molar Mass Using Colligative Properties	From freezing point depression: $M = \frac{i K_{f} \cdot w_{solute} \times 1000}{Δ T_{f} \cdot w_{solvent}}$ ; From osmotic pressure: $M = \frac{i w_{solute} R T}{Π V}$	Use when experimental data for colligative property is given to find molar mass.
Common Exam Problems: Mixing Solutions	$M_{f} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$	Use when diluting or mixing solutions of same solute.
Raoult's Law for Non-Volatile Solute	$p = x_{solvent} p^{0}$ ; $\frac{p^{0} - p}{p^{0}} = x_{solute}$	Use to calculate vapor pressure of solution or molar mass of non-volatile solute.
Colligative Properties of Electrolytes	$Δ T_{f} = i K_{f} m$ ; $Π = i M R T$	Use when solute is an electrolyte or when degree of dissociation is given.

Open the Solutions formula page

What JEE asks

Solution & Its Components. JEE often asks to identify solute/solvent in non-aqueous solutions like iodine in CCl4.
Concentration Terms. JEE frequently tests conversions between molarity and molality, especially when density is given.
Molarity & Molality Conversion. JEE often gives density and molarity, then asks for molality or mole fraction.
Solubility & Factors. JEE asks about effect of temperature on solubility of gases and solids, often with Henry's law.
Henry's Law. JEE often asks to compare solubilities of gases using $K_{H}$ values or to calculate pressure needed for a given concentration.
Raoult's Law for Volatile Liquids. JEE often asks to find vapor composition using Dalton's law: $y_{A} = p_{A} / P$ .
Ideal & Non-Ideal Solutions. JEE asks to identify examples of positive (e.g., acetone + CS2) and negative (e.g., acetone + chloroform) deviations.
Colligative Properties: Overview. JEE often asks to compare colligative properties of different solutions or calculate molar mass using one of these properties.
Relative Lowering of Vapor Pressure. JEE often gives vapor pressure of pure solvent and solution, asks for molar mass.
Boiling Point Elevation. JEE often gives $K_{b}$ and asks for molar mass or compares boiling points of different solutions.
Freezing Point Depression. JEE often uses freezing point depression to determine molar mass, especially for non-electrolytes.
Osmotic Pressure. JEE often asks to calculate molar mass from osmotic pressure data, especially for polymers.
Van't Hoff Factor (i). JEE often gives degree of dissociation and asks for $i$ or uses $i$ to find molar mass.
Abnormal Molar Mass & van't Hoff Factor. JEE often asks to calculate degree of dissociation from abnormal molar mass.
Determination of Molar Mass Using Colligative Properties. JEE frequently gives numerical problems to calculate molar mass using $Δ T_{f}$ or $Π$ .
Key Constants for Solvents. JEE often provides these constants, but knowing them saves time.
Azeotropes. JEE asks to identify examples of azeotropes and their deviation type.
Common Exam Problems: Mixing Solutions. JEE often gives two solutions and asks for final concentration after mixing.
Raoult's Law for Non-Volatile Solute. JEE often asks to find molar mass from vapor pressure lowering data.
Henry's Law Constant Trends. JEE asks to arrange gases in order of solubility given $K_{H}$ values.
Colligative Properties of Electrolytes. JEE often gives degree of dissociation and asks for $Δ T_{f}$ or $Π$ .
Key Points for JEE Numericals. JEE often includes unit conversion traps; careful with density given in g/mL.

Common traps

Do not confuse solvent with the component that dissolves; the solvent is the component present in larger amount and whose physical state is retained. Solution & Its Components
Molarity depends on temperature (volume changes), molality does not. Concentration Terms
Density must be in g/mL; if given in kg/L, convert to g/mL. Molarity & Molality Conversion
For solids in liquids, solubility generally increases with temperature (exothermic dissolution may decrease). Solubility & Factors
Henry's constant $K_{H}$ is temperature-dependent; higher $K_{H}$ means lower solubility. Henry's Law
Raoult's law applies only to ideal solutions; deviations occur for non-ideal solutions. Raoult's Law for Volatile Liquids
Positive deviation occurs when A-B interactions are weaker than A-A and B-B (e.g., ethanol + water, acetone + CS2); negative deviation when A-B interactions are stronger (e.g., acetone + chloroform, HNO3 + water). Ideal & Non-Ideal Solutions
Colligative properties are valid only for dilute solutions. Colligative Properties: Overview
For dilute solutions, $x_{solute} \approx \frac{n_{solute}}{n_{solvent}}$ . Relative Lowering of Vapor Pressure
$K_{b}$ is solvent-specific (e.g., water $K_{b} = 0.512 K kg/mol$ ). Boiling Point Elevation
$K_{f}$ is solvent-specific (e.g., water $K_{f} = 1.86 K kg/mol$ ). Freezing Point Depression
Use molarity, not molality, in osmotic pressure formula. Osmotic Pressure
For association (e.g., dimerization), $i < 1$ . Van't Hoff Factor (i)
For dissociation, observed molar mass < normal molar mass, so $i > 1$ . Abnormal Molar Mass & van't Hoff Factor

Topics in Solutions

Concentration Units
Raoult's Law
Colligative Properties
Osmotic Pressure
Van't Hoff Factor
Ideal and Non-ideal Solutions

Previous-year questions on Solutions

JEEGenie has 323 previous-year questions on Solutions. A few worked examples follow.

Year	Questions
2026	23
2025	23
2024	17
2023	27
2022	23
2021	29
2020	13
2019	38
2018	8
2017	11
2016	9
2015	5
2014	8
2013	7
2012	6
2011	5
2010	2
2009	4
2008	6
2007	4
2006	4
2005	4
2004	8
2003	5
2002	3
2001	1
2000	1
1999	2
1998	1
1996	2
1994	3
1993	2
1992	1
1991	2
1990	2
1989	2
1988	1
1987	1
1986	2
1985	1
1983	2
1982	1
1981	2
1978	2

Question 1 medium

19.5 g of fluoro acetic acid (molar mass = 78 g mol−1) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1

\circ

C. What is

K_{a}

of fluoro acetic acid? (For water,

K_{f} = 1.86

K kg mol−1). Assume molarity and molality to have same values.

$10^{- 6}$
$4 \times 10^{- 4}$
$3 \times 10^{- 5}$
$3 \times 10^{- 3}$

Answer: D

Explanation. 1. Moles of acid

= \frac{19.5}{78} = 0.25

mol. Molality

m = \frac{0.25}{0.500} = 0.5

mol kg

^{- 1}

. 2. Observed freezing point depression:

Δ T_{f} = 1.0 = i \cdot K_{f} \cdot m = i \times 1.86 \times 0.5

. Hence

i = \frac{1.0}{1.86 \times 0.5} = \frac{1.0}{0.93} \approx 1.075

. 3. For weak acid dissociation

H A ⇌ H^{+} + A^{-}

i = 1 + α

, so

α = i - 1 = 0.075

. 4. Initial concentration

C = 0.5

M (given molarity = molality). At equilibrium:

[H^{+}] = [A^{-}] = C α

[H A] = C (1 - α)

. 5.

K_{a} = \frac{C α^{2}}{1 - α} = \frac{0.5 \times (0.075)^{2}}{1 - 0.075}

. 6. Compute:

(0.075)^{2} = 5.625 \times 10^{- 3}

0.5 \times 5.625 \times 10^{- 3} = 2.8125 \times 10^{- 3}

, divide by

0.925

K_{a} \approx 3.04 \times 10^{- 3}

. 7. Therefore,

K_{a} \approx 3 \times 10^{- 3}

. **Answer: D)

3 \times 10^{- 3}

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 easy

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water

= 0.52 K k g {m o l}^{- 1}

)

377.3 K
379.2 K
375.3 K
277.3 K

Answer: A

Explanation. Ethylene glycol is non-electrolyte, glucose is non-electrolyte. Total moles of solute

= 2 + 2 = 4

moles. Mass of water

= 500 g = 0.5 kg

. Molality

m = \frac{4}{0.5} = 8 {mol kg}^{- 1}

. Boiling point elevation:

Δ T_{b} = K_{b} \cdot m = 0.52 \times 8 = 4.16 K

. Boiling point of pure water

= 373.15 K

. Boiling point of solution

= 373.15 + 4.16 = 377.31 K \approx 377.3 K

. Thus answer is option A.

From NTA/IIT papers: JEE Main 2025 · 3 Apr Shift 1

Question 3 hard

Vessel-1 contains

w_{2}

g of a non-volatile solute

X

dissolved in

w_{1}

g of water. Vessel-2 contains

w_{2}

g of another non-volatile solute

Y

dissolved in

w_{1}

g of water. Both the vessels are at the same temperature and pressure. The molar mass of

X

is 80% of that of

Y

. The van't Hoff factor for

X

is 1.2 times of that of

Y

for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is ______ % of the solution in Vessel-2.

Answer: 150

Explanation. For solute

X

Δ T_{b, X} = i_{X} K_{b} m_{X} = i_{X} K_{b} \frac{w_{2} / M_{X}}{w_{1} / 1000}

For solute

Y

Δ T_{b, Y} = i_{Y} K_{b} m_{Y} = i_{Y} K_{b} \frac{w_{2} / M_{Y}}{w_{1} / 1000}

Given

M_{X} = 0.8 M_{Y}

and

i_{X} = 1.2 i_{Y}

. Ratio:

\frac{Δ T_{b, X}}{Δ T_{b, Y}} = \frac{i_{X}}{i_{Y}} \cdot \frac{M_{Y}}{M_{X}} = 1.2 \cdot \frac{1}{0.8} = 1.5

Thus,

Δ T_{b, X} = 1.5 \times Δ T_{b, Y}

, which means it is

150 %

of the elevation in Vessel-2.

150

From NTA/IIT papers: JEE Advanced 2024

Question 4 medium

Mass of Urea

({N H}_{2} {C O N H}_{2})

required to be dissolved in

1000 g

of water in order to reduce the vapour pressure of water by

25 %

is _________ g. (Nearest integer) Given: Molar mass of N, C, O and H are

14, 12, 16

and

1 g {m o l}^{- 1}

respectively

Answer: 1111

Explanation. **Principle:** Relative lowering of vapour pressure (Raoult's law for non-volatile solute):

\frac{p^{\circ} - p}{p^{\circ}} = \frac{n_{solute}}{n_{solute} + n_{solvent}}

**Given:** Vapour pressure reduced by

25 %

, so

\frac{p^{\circ} - p}{p^{\circ}} = 0.25

. Water:

1000 g

M = 18

, so

n_{water} = \frac{1000}{18} = 55.56

mol. Urea

M = 14 + 2 (1) + 12 + 16 + 14 + 2 (1) = 60 {g mol}^{- 1}

. Let

w

g urea be dissolved, so

n_{urea} = \frac{w}{60}

0.25 = \frac{w / 60}{w / 60 + 55.56}

0.25 (\frac{w}{60} + 55.56) = \frac{w}{60}

0.25 \cdot \frac{w}{60} + 13.89 = \frac{w}{60}

13.89 = \frac{w}{60} (1 - 0.25) = 0.75 \cdot \frac{w}{60}

w = \frac{13.89 \times 60}{0.75} = 1111.2 g

Nearest integer:

1111

. Answer: 1111

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

An aqueous solution is prepared by dissolving

0.1 m o l

of an ionic salt in

1.8 k g

of water at

35^{\circ} C

. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is

59.724 m m

of Hg. Vapour pressure of water at

35^{\circ} C

60.000 m m

of Hg. The number of ions present per formula unit of the ionic salt is _____.

Answer: 5

Explanation. Given, - Moles of salt

= 0.1

mol - Mass of water

= 1.8

\Rightarrow

moles of water

= \frac{1800}{18} = 100

mol -

p^{0} = 60.000

mm Hg,

p = 59.724

mm Hg - Degree of dissociation

α = 0.90

Using Raoult’s law for non-volatile solute:

\frac{p^{0} - p}{p^{0}} = \frac{i n_{solute}}{n_{solute} + n_{solvent}}

Substitute values:

\frac{60.000 - 59.724}{60.000} = \frac{i (0.1)}{0.1 + 100}

\frac{0.276}{60.000} = \frac{0.1 i}{100.1}

0.0046 = \frac{0.1 i}{100.1}

i = \frac{0.0046 \times 100.1}{0.1} = 4.6046 \approx 4.6

For dissociation:

i = 1 + (n - 1) α

4.6 = 1 + (n - 1) (0.9)

3.6 = 0.9 (n - 1)

n - 1 = 4 \Rightarrow n = 5

5

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Solutions?: JEEGenie estimates about 3% of JEE Main marks come from Solutions, based on recent papers. The share varies from year to year.
Which formulas matter most in Solutions?: The Solutions sheet on this page has 15 formula entries, starting with Concentration Terms, Molarity & Molality Conversion, Henry's Law, Raoult's Law for Volatile Liquids, Ideal & Non-Ideal Solutions. Each one comes with when to use it and the trap to avoid.
What topics does Solutions cover for JEE?: Solutions covers Concentration Units, Raoult's Law, Colligative Properties, Osmotic Pressure, Van't Hoff Factor, Ideal and Non-ideal Solutions.
How should I revise Solutions before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Solutions questions on JEEGenie.
How many previous-year questions are there on Solutions?: JEEGenie has 323 previous-year questions on Solutions, and this page shows a few with worked explanations.
Is practice on Solutions free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.