Solutions Formula Sheet — JEE Main Chemistry

Q: What are the most important Solutions formulas for JEE Main?

This Solutions formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Methods of expressing concentration, Molality, molarity, mole fraction, percentage, Vapour pressure of solutions, Raoult's Law, Ideal and non-ideal solutions — each shown with the key result and, where useful, a worked example.

Q: Is this Solutions formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Solutions formulas?

Blurt the Solutions formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Solutions formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 14 sub-topics

Methods of expressing concentration
Molality, molarity, mole fraction, percentage
Vapour pressure of solutions
Raoult's Law
Ideal and non-ideal solutions
Vapour pressure-composition plots
Colligative properties of dilute solutions
Relative lowering of vapour pressure
Depression of freezing point
Elevation of boiling point
Osmotic pressure
Determination of molecular mass using colligative properties
Abnormal value of molar mass
van't Hoff factor and significance

Expressing Concentration

A solution is a homogeneous mixture

▸ $Solvent$ = component in larger amount; $solute$ = in smaller amount.
▸ $Binary$ solution = two components; can be solid/liquid/gas in solid/liquid/gas (9 types).
▸Concentration = how much solute per fixed amount of solution or solvent.

Mass %

\frac{w _{solute}}{w _{solution}} \times 100

ppm

\frac{w _{solute}}{w _{solution}} \times 1 0^{6}

Mole fraction

x_{A} = \frac{n _{A}}{n _{A} + n _{B}}, \sum x = 1

Molarity

M = \frac{n _{solute}}{V _{solution} ( L )}

Molality

m = \frac{n _{solute}}{w _{solvent} ( kg )}

Volume % ; mass/volume %

\frac{V _{solute}}{V _{solution}} \times 100; \frac{w _{solute} ( g )}{V _{solution} ( mL )} \times 100

🎯 Exam · Temperature dependence

Molarity

changes with temperature (volume expands).

Molality

mole fraction

and

mass %

are temperature-

independent

(based on mass). Prefer molality for colligative-property work.

💡 Tip · Quick link

For dilute aqueous solutions, molarity

\approx

molality (since density

\approx 1

L^{- 1}

). 1 litre of water

= 55.5

mol.

🚫 Examiner Trap · Examiner traps

(1)

Molarity

is temperature-dependent (volume expands);

molality

, mole fraction, mass% are NOT — use molality for colligative work. (2) Molality uses kg of

solvent

, molarity uses L of

solution

. (3) Dilute aqueous:

M \approx m

(density

\approx 1

). (4) ppm

=

mass-ratio

\times 1 0^{6}

, for very dilute solutions.

Solubility & Henry's Law

★ Remember · Syllabus note for this page

Solubility of solids/gases ('like dissolves like', exo/endothermic dissolution) is NOT in the 2026 syllabus — kept as background.

Henry’s law IS in the syllabus

, listed under Unit 6 (Equilibrium) as a physical equilibrium.

Solubility

▸ $Solid in liquid$ : a saturated solution is in dynamic equilibrium with undissolved solute; mostly increases with temperature for endothermic dissolution.
▸ $Gas in liquid$ : solubility rises with $pressure$ and falls with $temperature$ (dissolution is exothermic — Le Chatelier).
▸Polar solutes dissolve in polar solvents ('like dissolves like').

Henry's law: At constant temperature, the partial pressure of a gas over a solution is proportional to its mole fraction in the solution:

p = K_{H} x

. A larger Henry's constant

K_{H}

means a less soluble gas.

Straight-line plot of gas mole fraction versus partial pressure with slope one over K_H, and the consequences that pressure raises and temperature lowers gas solubility — $p = K_{H} x$ : dissolved gas $\propto$ its partial pressure.

Henry's law in action

▸ $Soda water / soft drinks$ : bottled under high $C O_{2}$ pressure to force gas into solution.
▸ $Scuba diving (bends)$ : high pressure dissolves more $N_{2}$ in blood; rapid ascent releases bubbles — tanks use diluted air (He).
▸ $Anoxia$ : low $O_{2}$ partial pressure at high altitude lowers blood $O_{2}$ .

★ Remember · Raoult is a special case of Henry

Both say partial pressure

\propto

mole fraction. When the solute is the solvent itself,

K_{H} \to p^{0}

(pure VP), and Henry's law becomes Raoult's law.

✎ Example · Why warm soda goes flat

Explain using Henry's law.

Opening the bottle drops the $C O_{2}$ partial pressure above the liquid.
Lower $p \Rightarrow$ lower $x_{C O_{2}}$ , so gas escapes.
Warming further lowers gas solubility.

∴ Lower p and higher T both reduce dissolved

C O_{2}

🚫 Examiner Trap · Examiner traps

(1) Gas solubility

↑

with PRESSURE,

↓

with TEMPERATURE (dissolution exothermic). (2) Henry

p = K_{H} x

: a

larger

K_{H}

means a

less

soluble gas. (3) Raoult is the special case of Henry when

K_{H} \to p^{0}

. (4) Soda fizzes on opening (lower

p_{C O_{2}}

); divers get bends (more

N_{2}

dissolves at depth).

Raoult's Law & Vapour Pressure

Raoult's law: For a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction:

p_{1} = p_{1}^{0} x_{1}

p_{2} = p_{2}^{0} x_{2}

, where

p^{0}

is the pure-component vapour pressure.

Total vapour pressure (Dalton + Raoult)

p_{t o t a l} = p_{1}^{0} x_{1} + p_{2}^{0} x_{2} = p_{1}^{0} + (p_{2}^{0} - p_{1}^{0}) x_{2}

Varies linearly with

x_{2}

between

p_{1}^{0}

and

p_{2}^{0}

Vapour pressure versus mole fraction for an ideal solution: two dashed partial-pressure lines and a solid total-pressure line varying linearly between the pure vapour pressures — Ideal solution: straight-line $p_{t o t a l}$ between $p_{1}^{0}$ and $p_{2}^{0}$ .

Vapour-phase composition

y_{i} = \frac{p _{i}}{p _{t o t a l}}

y_{i}

= mole fraction of component i in the vapour. The vapour is always

richer

in the more volatile component (higher

p^{0}

💡 Tip · Non-volatile solute

If the solute is non-volatile, only the solvent contributes to vapour pressure:

p_{solution} = p_{1}^{0} x_{1} < p_{1}^{0}

— the basis of all colligative properties.

✎ Example · Vapour-phase enrichment

CHC l_{3}

(

p^{0} = 200

) and

C H_{2} C l_{2}

(

p^{0} = 415

mm Hg), with

x_{C H_{2} C l_{2}} = 0.688

. Find

y_{C H_{2} C l_{2}}

$p_{C H_{2} C l_{2}} = 0.688 \times 415 = 285.5$ ; $p_{CHC l_{3}} = 0.312 \times 200 = 62.4$ .
$p_{t o t a l} = 347.9$ mm Hg.
$y = 285.5/347.9$ .

∴

y_{C H_{2} C l_{2}} = 0.82

(vapour richer in the more volatile component).

🚫 Examiner Trap · Examiner traps

(1) The vapour is always

richer in the more volatile

component (higher

p^{0}

). (2) Non-volatile solute: only solvent contributes,

p = p_{1}^{0} x_{1} < p_{1}^{0}

(basis of colligative properties). (3)

p_{t o t a l}

varies

linearly

with

x_{2}

only for an ideal solution. (4) Use mole fractions in the LIQUID for Raoult, in the VAPOUR (

y_{i}

) for composition.

Ideal & Non-ideal Solutions

Ideal solution: Obeys Raoult's law over the entire composition range, with

Δ_{mi x} H = 0

and

Δ_{mi x} V = 0

. Occurs when solute-solvent (A–B) interactions equal A–A and B–B (e.g. benzene + toluene, n-hexane + n-heptane).

Two vapour-pressure plots: positive deviation curving above the Raoult line, and negative deviation curving below it — Real solutions deviate when A–B interactions differ from A–A / B–B.

	Positive dev.	Negative dev.
A–B force	weaker	stronger
Vapour pressure	$>$ Raoult	$<$ Raoult
$Δ_{mi x} H$	$+$ (endo)	$-$ (exo)
$Δ_{mi x} V$	$+$	$-$
Azeotrope	min. boiling	max. boiling

Recognising the deviations

▸ $Positive$ : A–B attraction weaker $\Rightarrow$ molecules escape easily $\Rightarrow$ VP higher. Ex: ethanol + acetone, $C S_{2}$ + acetone.
▸ $Negative$ : A–B attraction stronger (often new H-bonds) $\Rightarrow$ VP lower. Ex: phenol + aniline, chloroform + acetone, $HN O_{3}$ + water.

Azeotrope: A binary mixture that boils at constant temperature with the same composition in liquid and vapour, so it

cannot

be separated by fractional distillation. Large positive deviation → minimum-boiling (e.g. 95% ethanol-water); large negative deviation → maximum-boiling (e.g. 68%

HN O_{3}

-water).

★ Remember · Sign logic

Positive deviation

\Rightarrow

weaker forces

\Rightarrow

Δ_{mi x} H > 0

Δ_{mi x} V > 0

. Negative deviation

\Rightarrow

stronger forces

\Rightarrow

Δ_{mi x} H < 0

Δ_{mi x} V < 0

. Ideal

\Rightarrow

both zero.

🚫 Examiner Trap · Examiner traps

(1)

Positive

deviation: A–B weaker

\Rightarrow

>

Raoult,

Δ_{mi x} H > 0

, min-boiling azeotrope (ethanol+water).

Negative

: A–B stronger, VP

<

Raoult,

Δ_{mi x} H < 0

, max-boiling azeotrope (HN

O_{3}

+water). (2) Ideal

\Rightarrow Δ_{mi x} H = Δ_{mi x} V = 0

. (3) An azeotrope CANNOT be separated by fractional distillation. (4) Don't confuse min-boiling (positive) with max-boiling (negative).

Colligative Properties — VP, Boiling & Freezing

Colligative property: A property that depends only on the number of solute particles, not their nature. The four are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Relative lowering of VP

\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = x_{2}

Boiling-point elevation

Δ T_{b} = K_{b} m

Freezing-point depression

Δ T_{f} = K_{f} m

Two plots: solution vapour-pressure curve below the solvent curve, raising the boiling point at 1 atm and lowering the freezing point where it meets the solid curve — A non-volatile solute lowers VP, so $T_{b}$ rises and $T_{f}$ falls.

Molar mass from colligative data

M_{2} = \frac{1000 w _{2} K _{b}}{Δ T _{b} w _{1}} = \frac{1000 w _{2} K _{f}}{Δ T _{f} w _{1}}

w_{2}

= solute mass,

w_{1}

= solvent mass (g),

m = \frac{1000 w _{2}}{M _{2} w _{1}}

. Also

M_{2} = \frac{w _{2} M _{1}}{( Δ p / p _{1}^{0} ) w _{1}}

from VP lowering.

Solvent	$K_{b}$	$K_{f}$ (K kg mo $l^{- 1}$ )
Water	0.52	1.86
Benzene	2.53	5.12
Chloroform	3.63	4.79
Camphor	5.95	40.0

🎯 Exam · Why salt/glycol are used

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

K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}

are properties of the

solvent

only. Salt melts ice (fp depression); ethylene glycol is antifreeze + coolant (raises

T_{b}

, lowers

T_{f}

🚫 Examiner Trap · Examiner traps

(1) Colligative properties depend on the

number

of solute particles, NOT their nature. (2)

K_{b}, K_{f}

are properties of the

solvent

only (camphor has a huge

K_{f} = 40

). (3) Use

molality

Δ T = K m

(not molarity). (4) A non-volatile solute raises

T_{b}

AND lowers

T_{f}

— opposite directions.

Osmotic Pressure

Osmosis & osmotic pressure (

Π

): Across a semipermeable membrane, solvent flows from pure solvent (or dilute) to a more concentrated solution — osmosis. The excess pressure that must be applied on the solution to just stop this flow is the osmotic pressure

Π

U-tube with a semipermeable membrane: solvent flows into the solution raising its level by height h, with osmotic-pressure equations and tonicity definitions — $Π = h ρ g$ ; it is the best colligative property for large molecules.

van't Hoff equation for osmotic pressure

Π = CRT = \frac{n _{2}}{V} RT \Rightarrow M_{2} = \frac{w _{2} RT}{Π V}

C = molarity, T in K,

R = 0.0821

L atm

K^{- 1}

l^{- 1}

. Measured at room temperature with large values even for dilute solutions.

Tonicity (same

T

)

▸ $Isotonic$ : equal $Π$ — no net osmosis (0.9% $NaCl$ saline is isotonic with blood).
▸ $Hypertonic$ : higher $Π$ — cell loses water and $shrinks$ (crenation).
▸ $Hypotonic$ : lower $Π$ — cell takes in water and $swells$ (may burst).
▸Two solutions are isotonic when $i_{1} C_{1} = i_{2} C_{2}$ (particle concentrations equal), e.g. $0.1$ M glucose $\equiv 0.05$ M NaCl.

🎯 Exam · Reverse osmosis & uses

Applying

p > Π

on the solution pushes pure solvent

out

—

reverse osmosis

, used for sea-water desalination. Osmotic pressure is preferred for

macromolecules/proteins

(large, easily measured, room-T).

✎ Example · Molar mass of a protein

200 cm^{3}

solution has

1.26

g protein,

Π = 2.57 \times 1 0^{- 3}

bar at 300 K.

$M_{2} = \frac{w _{2} RT}{Π V}$ with $R = 0.083$ L bar $K^{- 1}$ mo $l^{- 1}$ , $V = 0.200$ L.
$M_{2} = \frac{1.26 \times 0.083 \times 300}{2.57 \times 1 0 ^{- 3} \times 0.200}$ .

∴

M_{2} \approx 61000 g mol^{- 1}

🚫 Examiner Trap · Examiner traps

(1)

Π = CRT

uses

molarity

C and T in kelvin. (2) Osmotic pressure is the BEST colligative property for macromolecules/proteins (large, measurable at room T). (3) Hypertonic

\Rightarrow

cell shrinks; hypotonic

\Rightarrow

swells/bursts; isotonic

\Rightarrow

no net flow. (4) Reverse osmosis needs

p > Π

(desalination).

van't Hoff Factor & Abnormal Molar Mass

Abnormal molar mass: When a solute dissociates or associates in solution, the number of particles changes, so the measured colligative property — and hence the calculated molar mass — differs from the expected value.

KCl dissociating into two ions giving i greater than 1, versus two acetic acid molecules associating into a dimer giving i less than 1 — Dissociation raises the particle count ( $i > 1$ ); association lowers it ( $i < 1$ ).

van't Hoff factor

i = \frac{normal molar mass}{abnormal molar mass} = \frac{observed colligative property}{calculated value} = \frac{particles after}{particles before}

i > 1

: dissociation (e.g.

KCl \to 2

K_{2} S O_{4} \to 3

i < 1

: association (e.g. acetic acid dimer in benzene,

i \to 0.5

i = 1

: no change.

Relative lowering of VP

\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = i \frac{n _{2}}{n _{1}}

Boiling-point elevation

Δ T_{b} = i K_{b} m

Freezing-point depression

Δ T_{f} = i K_{f} m

Osmotic pressure

Π = i CRT

★ Remember · Degree of dissociation / association

Dissociation into n ions:

i = 1 + (n - 1) α

. Association of n monomers:

i = 1 - (1 - \frac{1}{n}) α

, where

α

is the degree of dissociation/association.

🎯 Exam · Particle count drives everything

For the same molality,

Δ T_{f}

order:

K_{2} S O_{4} (i \approx 3) > NaCl (i \approx 2) > glucose (i = 1)

. Electrolytes depress freezing point more than non-electrolytes.

🚫 Examiner Trap · Examiner traps

(1)

i > 1

dissociation (KCl

\to 2

K_{2} S O_{4} \to 3

);

i < 1

association (acetic acid dimer in benzene

i \to 0.5

). (2) Dissociation:

i = 1 + (n - 1) α

; association:

i = 1 - (1 - \frac{1}{n}) α

. (3) Multiply EVERY colligative formula by i for electrolytes. (4) Same molality:

Δ T_{f}

order

K_{2} S O_{4} >

NaCl

>

glucose (particle count).

More JEE Main Chemistry formula sheets

Frequently Asked Questions

What are the most important Solutions formulas for JEE Main?

This Solutions formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Methods of expressing concentration, Molality, molarity, mole fraction, percentage, Vapour pressure of solutions, Raoult's Law, Ideal and non-ideal solutions — each shown with the key result and, where useful, a worked example.

Is this Solutions formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Solutions formulas?

Blurt the Solutions formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheetsJEE Main previous-year papersmost important chapters.