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Solutions: one-page cheat sheet

NEET Chemistry · Unit 5 · Physical Chemistry

ScoreChemby CSJ Sir
Weightage: 2–3 questions · 8–12 marks

1 · Concentration recap

Mass-basedmass %, ppm, mole fraction, molality: do not change with

T

Volume-basedmolarity falls on heating

Mole fraction

x_{1} + x_{2} = 1

Solid solutionsbrass (Cu–Zn), bronze (Cu–Sn), H

_{2}

in Pd, Na–Hg amalgam

All formulas: see the Some Basic Concepts cheat sheet.

2 · Henry's law

p = K_{H} x

Solubilityat the same

p

, higher $K_{H}$ means lower solubility

Effect of

T

K_{H}

rises with

T

, so gas solubility falls

Usessoda bottled under pressure; scuba tanks use helium (very low solubility, avoids the bends); anoxia at altitude

3 · Raoult's law

p_{i} = p_{i}^{\circ} x_{i}, p_{total} = p_{1}^{\circ} + (p_{2}^{\circ} - p_{1}^{\circ}) x_{2}

Vapour phase

y_{i} = p_{i} / p_{total}

; richer in the more volatile component

Non-vol. solute

p_{soln} = p_{1}^{\circ} x_{1}

4 · Ideal and non-ideal solutions

	Ideal	Positive dev.	Negative dev.
A–B forces	$\approx$ A–A, B–B	weaker	stronger
$p_{obs}$	$=$ Raoult	$>$ Raoult	$<$ Raoult
$Δ_{mix} H$	0	$> 0$	$< 0$
$Δ_{mix} V$	0	$> 0$	$< 0$
Azeotrope	none	minimum boiling	maximum boiling
Example	benzene + toluene	ethanol + acetone	chloroform + acetone

Azeotropes95% ethanol (351.15 K, minimum); 68% HNO

_{3}

(393.5 K, maximum). Cannot be separated by fractional distillation.

Chloroform + acetone: H-bond C–H···O=C makes A–B stronger, so negative deviation.

5 · Colligative properties

\frac{p _{1}^{\circ} - p _{1}}{p _{1}^{\circ}} = x_{2}, Δ T_{b} = K_{b} m, Δ T_{f} = K_{f} m

Water

K_{b} = 0.52

K_{f} = 1.86 K kg mol^{- 1}

Osmotic pressure

Π = C R T = \frac{w _{2} R T}{M _{2} V}

(uses molarity, works at room

T

)

Molar mass

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

(

K = K_{b}

K_{f}

)

Equal massat the same

w

and

V

Π \propto 1/ M_{2}

6 · Osmosis

Isotonicequal

Π

; no net flow

Hypertonichigher

Π

; cell shrinks (plasmolysis)

Hypotoniclower

Π

; cell swells

Reverse osmosisapplied pressure

> Π

pushes solvent out; desalination

7 · van 't Hoff factor

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

Usemultiply each colligative property by

i

Dissociation

i = 1 + (n - 1) α

Association

i = 1 - (1 - \frac{1}{n}) α

; dimer in benzene gives

i = 0.5

Full ionisationNaCl, MgSO

_{4}

: 2; K

_{2}

_{4}

, CaCl

_{2}

: 3; K

_{4}

[Fe(CN)

_{6}

], Al

_{2}

(SO

_{4}

)

_{3}

: 5

Seen in NEET 2025

Highest boiling point among 0.015 M glucose, 0.01 M urea, 0.01 M KNO $_{3}$ , 0.01 M Na $_{2}$ SO $_{4}$ ?

i \times m : 0.015, 0.010, 0.020, 0.030

Na $_{2}$ SO $_{4}$ has the most particles.

Seen in NEET 2024

$Π$ vs $C$ is a straight line of slope $25.73$ L bar mol $^{- 1}$ . Find $T$ .

T = \frac{slope}{R} = \frac{25.73}{0.083} = 310 K = 3 7^{\circ} C

Seen in NEET 2020

$K_{f}$ (benzene) $= 5.12$ ; solution is $0.078 m$ in a non-electrolyte.

Δ T_{f} = K_{f} m = 5.12 \times 0.078 = 0.40 K

Six traps to remember

Higher Henry's constant means lower gas solubility.
Only molarity changes with temperature.
Positive deviation: weaker A–B forces, minimum-boiling azeotrope.
Rank colligative properties by $i \times$ concentration, not concentration alone.
Osmotic pressure at equal mass is highest for the smallest molar mass.
Dimerisation gives $i < 1$ ; dissociation gives $i > 1$ .

Solutions: free one-page cheat sheet