MHT-CET Chemistry · Teaching notes

Some Basic Concepts of Chemistry — MHT-CET Chemistry

The arithmetic backbone of MHT-CET Chemistry — the chapter that turns grams, litres and molecule-counts into each other. It is heavily tested and mostly straightforward: master a handful of bridges and nearly every question falls in one or two steps. It teaches in four movements, foundations first: (1) SI units, physical properties, matter classification and average atomic mass — the measurement groundwork; (2) the laws of chemical combination, Dalton's atomic theory and percentage composition; (3) the mole concept and its interconversions — the engine room: n = m/M, n = V/22.4, N = n·NA, and vapour density; (4) stoichiometry and concentration — reading mole ratios off a balanced equation, limiting reagent, combining gas volumes and H2O2 volume strength. Mostly formula concepts with worked numbers; the named laws live in reference tables. Gas laws, Dalton's law of partial pressures and the kinetic theory are taught in the separate States of Matter chapter, following the Maharashtra State Board syllabus. Every PYQ tagged.

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Subtopic notes

PYQ weightage by concept

20 concepts · 53 PYQs — where the marks actually sit, so you know what to drill first

SI Units, Physical Properties and Atomic Abundance6 PYQs · 11%

Concept	PYQs	Share
Common SI derived units	2	4%
The seven SI base units	1	2%
Intensive vs extensive properties	1	2%
Average atomic mass from isotopic abundance	1	2%
Abundance of elements on Earth	1	2%
Classification of matterfoundation	—	—

Laws of Chemical Combination and Percentage Composition7 PYQs · 13%

Concept	PYQs	Share
The five laws of chemical combination	5	9%
Percentage composition by mass	1	2%
Percent atom economy	1	2%
Dalton's atomic theoryfoundation	—	—

The Mole and Its Interconversions31 PYQs · 58%

Concept	PYQs	Share
Moles from mass and molar mass	11	21%
Molar volume at STP	8	15%
Counting molecules and atoms from moles	4	8%
Counting ions and electrons	4	8%
The mole, Avogadro's number, and one single particle	2	4%
Ratio of molecules from a mass ratio	1	2%
Vapour density to molar mass	1	2%

Stoichiometry and Concentration9 PYQs · 17%

Concept	PYQs	Share
Mole ratios from a balanced equation	6	11%
Combining gaseous volumes and the limiting reagent	2	4%
Concentration: percent by mass and H2O2 volume strength	1	2%

Formula & revision sheet

13 formulas · 7 reference tables · 30 gotchas across all subtopics — the exam-eve cheat-sheet

SI Units, Physical Properties and Atomic Abundance

Formulas (1)

Average atomic mass from isotopic abundance · Average atomic mass $\bar{m} = \sum_i m_i\,f_i \quad\text{where}\quad \sum_i f_i = 1$

Reference tables (5)

Classification of matter7 rows

Category	Definition	Example
Element	A pure substance that cannot be broken into simpler substances by a chemical change.	Gold, oxygen, iron
Compound	A pure substance of two or more elements combined in a fixed proportion by mass.	Water, table salt, mercuric oxide
Homogeneous mixture	A mixture with uniform composition throughout (a solution).	Salt water, air
Heterogeneous mixture	A mixture whose composition is not uniform throughout.	Sand in water, oil and water
Metal	Lustrous, malleable and ductile; a good conductor of heat and electricity.	Copper, silver, iron
Non-metal	Dull and brittle; a poor conductor of heat and electricity.	Nitrogen, iodine, carbon Exceptions the paper likes: graphite (a non-metal) conducts electricity; diamond and iodine have lustre.
Metalloid	An element with properties intermediate between metals and non-metals.	Silicon, germanium, arsenic

Sort by composition first (pure vs mixture), then refine (element/compound, metal/non-metal/metalloid).

The seven SI base units7 rows

Base quantity	Unit	Symbol
Mass	kilogram	kg
Length	metre	m
Time	second	s
Temperature	kelvin	K Note kelvin has no degree sign: write $300\ \text{K}$ , not $300\ ^\circ\text{K}$ .
Amount of substance	mole	mol
Electric current	ampere	A
Luminous intensity	candela	cdQ The odd one out that PYQs love — candela measures luminous intensity, not energy, force or work.

Learn the pairing in both directions — quantity to unit and unit to quantity.

Common SI derived units6 rows

Quantity	Defining relation	SI derived unit
Volume	length cubed	$\text{m}^3$
Density	mass / volume	$\text{kg m}^{-3}$
Force	mass $\times$ acceleration	newton $\text{N} = \text{kg m s}^{-2}$
Pressure	force / area	pascal $\text{Pa} = \text{N m}^{-2}$
Rate of diffusion	volume / time	$\text{dm}^3\,\text{s}^{-1}$ Q
Coefficient of viscosity	stress / velocity gradient	$\text{N s m}^{-2} = \text{Pa s}$ Q Watch the exponents on $\text{s}$ and $\text{m}$ : the correct form is $\text{N s m}^{-2}$ , not $\text{N s}^{-1}\text{m}^{-2}$ .

Every derived unit is the base units of its defining formula, combined.

Intensive vs extensive properties10 rows

Property	Type	Why
Mass	Extensive	Doubles when the sample doubles.
Volume	Extensive	Scales directly with amount.
Internal energy	Extensive	Total energy grows with amount.
Heat capacity	Extensive	Whole-sample quantity; scales with mass.
Temperature	Intensive	A drop and a bucket of the same liquid share it.
Density	Intensive	Ratio mass/volume — the amounts cancel.
Boiling point	Intensive	Fixed for a pure substance, any amount.
Surface tension	Intensive	A material property, independent of quantity.Q
Viscosity	Intensive	Same for a drop or a barrel of the liquid. Surface tension and viscosity are the intensive pair the bank tests — both material properties, unchanged by sample size.
Specific heat	Intensive	Heat capacity per unit mass — a ratio, so amounts cancel.

Change the sample size in your head: if the value moves, it is extensive.

Abundance of elements on Earth5 rows

Domain	Most abundant element	Approx. share
Earth's crust (by mass)	Oxygen	about 46%Q This is the default 'most abundant element on Earth' answer the bank wants — oxygen.
Earth's crust (2nd)	Silicon	about 28%
Earth's crust (3rd)	Aluminium	about 8%
Whole Earth (by mass)	Iron	about 32%
Universe (by mass)	Hydrogen	about 74%

The winner changes with the domain — match the answer to what the question asks.

Watch out for (6)

A compound is a pure substance, not a mixture
→ Classification of matter
Candela measures luminous intensity, not energy
→ The seven SI base units
The exponents on the viscosity unit matter
→ Common SI derived units
Heat capacity is extensive; specific heat is intensive
→ Intensive vs extensive properties
Weight by abundance, not a plain average
→ Average atomic mass from isotopic abundance
Crust versus universe versus whole Earth
→ Abundance of elements on Earth

Laws of Chemical Combination and Percentage Composition

Formulas (2)

Percentage composition by mass · Mass percentage of an element $\%\,\text{element} = \dfrac{n \times A}{M} \times 100$
Percent atom economy · Percent atom economy $\%\,\text{atom economy} = \dfrac{\text{FW of desired product}}{\sum \text{FW of reactants}} \times 100$

Reference tables (2)

The five laws of chemical combination5 rows

Law	Statement	Stock example
Law of conservation of mass	Matter can neither be created nor destroyed in a chemical reaction; total mass of reactants = total mass of products.	$1.7$ g $\text{AgNO}_3 + 0.585$ g $\text{NaCl}$ give $1.435$ g $\text{AgCl} + 0.85$ g $\text{NaNO}_3$ ; both sides total $2.285$ g.
Law of definite (constant) proportions	A given pure compound always contains the same elements in the same fixed proportion by mass, whatever its source.	Water is always $1 : 8$ hydrogen to oxygen by mass.Q Also called Proust's law. The tell-tale phrase in an MCQ is 'a given compound always contains the same proportion of elements'.
Law of multiple proportions	When the same two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a ratio of small whole numbers.	$\text{CO}$ and $\text{CO}_2$ : oxygen masses per fixed carbon are in a $1 : 2$ ratio.Q The most-asked law here. It ONLY applies when both compounds contain the SAME two elements — this is the whole basis of the 'which pair cannot demonstrate it' questions.
Gay-Lussac's law of combining volumes	Gases combine (and form gaseous products) in volume ratios that are simple whole numbers, at the same temperature and pressure.	$1$ volume $\text{N}_2 + 3$ volumes $\text{H}_2 \to 2$ volumes $\text{NH}_3$ (a $1 : 3 : 2$ ratio). Sometimes phrased as the law of reciprocal proportions in older texts — both express fixed combining relationships; for gases the paper uses the combining-volumes form.
Avogadro's law	Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.	$22.4$ L of any gas at STP contains $1$ mole $(6.022\times10^{23}$ molecules $)$ .

Recognise the law from either its definition or a worked example.

Dalton's atomic theory4 rows

Postulate	What it states	Explains / modern status
Atoms exist	Matter is made of extremely small, indivisible particles called atoms.	Modern caveat: the atom IS divisible into protons, neutrons and electrons.
Atoms of an element are identical	All atoms of a given element have the same mass and chemical properties.	Modern caveat: isotopes are atoms of one element with different masses.
Small whole-number combining ratio	Atoms of different elements combine in simple whole-number ratios to form compounds.	Explains the laws of definite and multiple proportions.
Atoms are conserved	Atoms are neither created nor destroyed in a chemical reaction; they are only rearranged.	Explains the law of conservation of mass.

The postulates that underlie the laws of chemical combination; two are now known to have exceptions.

Watch out for (5)

Multiple proportions needs the SAME two elements
→ The five laws of chemical combination
Definite vs multiple proportions
→ The five laws of chemical combination
Two postulates have modern exceptions
→ Dalton's atomic theory
Multiply by the number of atoms, not just the atomic mass
→ Percentage composition by mass
Product over reactants, not the other way round
→ Percent atom economy

The Mole and Its Interconversions

Formulas (7)

The mole, Avogadro's number, and one single particle · Mass and volume of one particle $m_{1} = \dfrac{M}{N_A}, \qquad V_{1} = \dfrac{M}{N_A \, \rho}$
Moles from mass and molar mass · Moles from mass $n = \dfrac{m}{M}$
Molar volume at STP · Moles from gas volume at STP $n = \dfrac{V}{22.4}$
Counting molecules and atoms from moles · Particles from moles $N = n \, N_A = \dfrac{m}{M}\, N_A$
Counting ions and electrons · Ions / electrons from moles $N_{\text{ion/e}} = n \, N_A \times k$
Ratio of molecules from a mass ratio · Molecule ratio of two gases $\dfrac{N_A}{N_B} = \dfrac{m_A / M_A}{m_B / M_B}$
Vapour density to molar mass · Molar mass from vapour density $M = 2 \times \text{V.D.}$

Watch out for (13)

Divide by $N_A$ , not multiply, for one particle
→ The mole, Avogadro's number, and one single particle
Volume of one molecule needs the density
→ The mole, Avogadro's number, and one single particle
Convert kg to grams before dividing
→ Moles from mass and molar mass
Use the molar mass of the WHOLE molecule
→ Moles from mass and molar mass
Atoms of an element vs formula units
→ Moles from mass and molar mass
22.4 dm^3 only at STP, and only for gases
→ Molar volume at STP
$1\,\text{m}^3 = 1000\,\text{dm}^3$
→ Molar volume at STP
Atoms need an extra multiplier
→ Counting molecules and atoms from moles
Convert the given volume to dm^3 first
→ Counting molecules and atoms from moles
Ca2+ and Cl- counts differ for the same salt
→ Counting ions and electrons
Electrons per molecule = sum of atomic numbers
→ Counting ions and electrons
Lighter molecule wins for the same mass
→ Ratio of molecules from a mass ratio
Vapour density is HALF the molar mass
→ Vapour density to molar mass

Stoichiometry and Concentration

Formulas (3)

Mole ratios from a balanced equation · Mass of product from a known reactant quantity $m_{\text{target}} = n_{\text{known}} \times \dfrac{\text{coeff}_{\text{target}}}{\text{coeff}_{\text{known}}} \times M_{\text{target}}$
Combining gaseous volumes and the limiting reagent · Product volume from a limiting gaseous reactant $V_{\text{product}} = V_{\text{limiting}} \times \dfrac{\text{coeff}_{\text{product}}}{\text{coeff}_{\text{limiting}}}$
Concentration: percent by mass and H2O2 volume strength · H2O2 molarity from volume strength, and percent by mass $M_{\text{H}_2\text{O}_2} = \dfrac{\text{volume strength}}{11.2} \qquad \%\text{ by mass} = \dfrac{\text{mass of solute}}{\text{mass of solution}}\times 100$

Watch out for (6)

Coefficients are moles, not grams
→ Mole ratios from a balanced equation
Do not skip the fractional ratio
→ Mole ratios from a balanced equation
Identify the limiting reagent before scaling
→ Combining gaseous volumes and the limiting reagent
Volumes need the same T and P
→ Combining gaseous volumes and the limiting reagent
Divide volume strength by 11.2, not 22.4
→ Concentration: percent by mass and H2O2 volume strength
Percent by mass uses the mass of solution, not solvent
→ Concentration: percent by mass and H2O2 volume strength