Every formula, rule and trap from the chapter on one page. Keep it open while you solve PYQs, or download it and stick it above your desk.

Some Basic Concepts of Chemistry: one-page cheat sheet

NEET Chemistry · Unit 1 · Physical Chemistry

ScoreChemby CSJ Sir
Weightage: 1–2 questions · 4–8 marks

1 · Matter & measurement

Pureelement (one kind of atom) or compound (fixed mass ratio)

Mixturevariable composition, separated physically

SI basem, kg, s, A, K, mol, cd

Temp.

K =^{\circ} C + 273.15,^{\circ} F = \frac{9}{5}^{\circ} C + 32

Precisioncloseness of repeated readings

Accuracycloseness to the true value

Milk and blood are heterogeneous (colloids); dry, dust-free air is homogeneous. Mass is constant, weight is not.

2 · Significant figures

Digits	Counted?	Example
Non-zero, trapped zeros	always	$2.18, 2.008$
Leading zeros	never	$0.0034 \to 2$
Trailing, with point	yes	$4.200 \to 4$
Trailing, no point	no	$2500 \to 2$
Exact counts	$\infty$	20 apples

+ and −keep the least decimal places:

18.0 + 12.11 = 30.1

× and ÷keep the least sig. figures:

2.5 \times 1.25 = 3.1

3 · Laws of chemical combination

Law	By	Idea
Mass conservation	Lavoisier	mass is conserved
Definite proportions	Proust	fixed mass % in a compound
Multiple proportions	Dalton	simple ratio of B masses
Gaseous volumes	Gay-Lussac	simple volume ratios
Avogadro's law	Avogadro	equal $V$ , equal molecules

ExampleCO, CO

_{2}

: for 12 g C, O is 16 g and 32 g (

1 : 2

)

Dalton's theory explains the first three laws but not Gay-Lussac's law.

4 · The mole

n = \frac{w}{M} = \frac{N}{N _{A}} = \frac{V _{STP}}{22.4 L}

Constants

N_{A} = 6.022 \times 1 0^{23} mol^{- 1}

;

1 amu = 1.66 \times 1 0^{- 24} g

Molar vol.22.4 L (273 K, 1 atm); 22.7 L (273 K, 1 bar)

1 particlemass

= M / N_{A}

Atoms

n \times N_{A} \times atomicity

Avg. mass

\sum (abundance % \times isotopic mass) /100

“Maximum atoms”: compare

(mass / M) \times atomicity

. 1 g Li = 0.143 mol atoms; 1 g O

_{2}

= 0.0625 mol atoms.

5 · Empirical & molecular formula

MF = (EF)_{x}, x = \frac{molar mass}{EF mass}, M = 2 \times V.D.

Step 1divide each mass % by its atomic mass

Step 2divide all by the smallest value

Step 3fractions:

1.5 \to \times 2

1.33 \to \times 3

Mass %

(element in 1 mol / M) \times 100

6 · Limiting reagent

Methodconvert to moles, then compare

n / coefficient

for each reactant

Limitingthe smaller ratio; it fixes the yield

% yield

(actual / theoretical) \times 100

Puritypure mass

=

sample

\times

purity

/100

Limiting reagent is not the one with fewer grams or moles. Divide by the coefficient first.

7 · Concentration terms

Term	Formula	Changes with $T$
Mass %	$w_{solute} / w_{soln} \times 100$	no
Mole fraction	$χ_{A} = n_{A} / (n_{A} + n_{B})$	no
ppm	$w_{solute} / w_{soln} \times 1 0^{6}$	no
Molality $m$	$n_{solute} / kg solvent$	no
Molarity $M$	$n_{solute} / L solution$	yes

Heating increases volume, so only $M$ falls.

8 · Shortcuts

Dilution

M_{1} V_{1} = M_{2} V_{2}

Mixing

M = (M_{1} V_{1} + M_{2} V_{2}) / (V_{1} + V_{2})

M

from %

M = % w/w \times d \times 10/ M_{B}

m

from

M

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

d

in g mL

^{- 1}

Seen in NEET 2024

X: 32% A, 20% B, rest C (A = 64, B = 40, C = 32).

\frac{32}{64} : \frac{20}{40} : \frac{48}{32} = 0.5 : 0.5 : 1.5 = 1 : 1 : 3

Empirical formula: $ABC_{3}$ .

Seen in NEET 2023

Mass of CO $_{2}$ from heating 20 g of 20% pure limestone (CaCO $_{3}$ ).

\frac{20 \times 20}{100} = 4 g, \frac{4}{100} = 0.04 mol, 0.04 \times 44 = 1.76 g

Use the pure mass (4 g), not 20 g.

Six traps to remember

Milk and blood are heterogeneous; dry air is homogeneous.
Trailing zeros without a decimal point are not significant.
Dalton's theory could not explain the law of gaseous volumes.
For maximum atoms, multiply moles by atomicity.
Limiting reagent: compare $n$ / coefficient, not grams or moles.
Only molarity changes with temperature; molality, mole fraction and mass % do not.

Some Basic Concepts of Chemistry: free one-page cheat sheet