Some Basic Concepts of Chemistry

Physical Chemistry Weightage: 1–2 Questions (4–8 Marks) NMC Unit 1

“Chemistry is quantitative. A single mole of any molecular species contains an incomprehensible 6.022 × 10²³ particles, yet fits into a modest beaker. Mastering the conversion pathway between mass, moles, volume, and atomicity forms the bedrock of every numerical on the NEET paper.”

— SCORECHEM ACADEMIC TEAM

1. Classification of Matter & States

Pure Substances: Fixed chemical composition throughout.
- Elements: Consist of a single species of atom ( $\text{Na}, \text{Cu}$ ) or homonuclear molecules ( $\text{H}_2, \text{P}_4, \text{S}_8$ ).
- Compounds: Definite ratio of different elements by mass. Characteristic properties differ entirely from constituents (combustible $\text{H}_2$ + supporter $\text{O}_2 \rightarrow$ fire-retardant $\text{H}_2\text{O}$ ).
Mixtures: Variable composition; separated by physical techniques.
- Homogeneous: Single phase throughout (e.g., air, true aqueous salt solutions).
- Heterogeneous: Distinct phase boundaries visible (e.g., colloids like milk and blood, suspensions).

⚠️ NEET Trap: Homogeneous vs Heterogeneous Dry, dust-free air is a homogeneous mixture of gases. Milk and blood are heterogeneous systems (colloids) despite appearing uniform without a microscope.

2. Significant Figures & Rounding Off

Critical Counting Rules

All non-zero digits are significant ( $2.18\text{ g} \rightarrow 3\text{ SF}$ ).
Leading zeros preceding the first non-zero digit are never significant ( $0.0034 \rightarrow 2\text{ SF}$ ).
Trapped zeros between non-zero digits are always significant ( $2.008 \rightarrow 4\text{ SF}$ ).
Terminal zeros:
- With decimal point: Significant ( $4.200 \rightarrow 4\text{ SF}$ ; $100. \rightarrow 3\text{ SF}$ ).
- Without decimal point: Non-significant ( $100 \rightarrow 1\text{ SF}$ ; $2500 \rightarrow 2\text{ SF}$ ).
Counting numbers of discrete objects have infinite significant figures ( $20\text{ apples} \rightarrow \infty\text{ SF}$ ).

Calculation Operations

Addition / Subtraction: Result must not exceed the least number of decimal places: $18.0\text{ (1 dec place)} + 12.11 + 1.012 = 31.122 \longrightarrow \mathbf{31.1}$
Multiplication / Division: Result must match the smallest count of significant figures: $2.5\text{ (2 SF)} \times 1.25\text{ (3 SF)} = 3.125 \longrightarrow \mathbf{3.1}$

3. Laws of Chemical Combination

Law	Formulated By	Core Exam Rule	Example
Conservation of Mass	Lavoisier (1789)	Mass is neither created nor destroyed in chemical changes.	Hydrocarbon combustion
Definite Proportions	Proust (1799)	Compound always contains identical mass percentages of elements.	Natural vs synthetic $\text{CuCO}_3$
Multiple Proportions	Dalton (1803)	Masses of $B$ combining with a fixed mass of $A$ are in simple integer ratios.	$\text{CO}$ vs $\text{CO}_2$ ( $16:32 = 1:2$ )
Gaseous Volumes	Gay-Lussac (1808)	Gases react and form in simple volume ratios at constant $T, P$ .	$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ ( $2:1:2$ )
Avogadro's Law	Avogadro (1811)	Equal volumes of gases at same $T, P$ have equal molecules ( $V \propto n$ ).	Standard molar volume

⚠️ NEET Trap: Failure of Dalton's Atomic Theory Dalton's hypothesis successfully explained mass conservation, definite proportions, and multiple proportions, but **failed to explain Gay-Lussac's Law of Gaseous Volumes** because Dalton did not accept that atoms of the same element could unite into diatomic molecules.

4. The Master Mole Engine

\text{Moles } (n) = \frac{w\text{ (g)}}{M\text{ (molar mass)}} = \frac{N\text{ (particles)}}{N_A} = \frac{V_{\text{gas at STP (L)}}}{22.4\text{ L}}

Atomicity Factor: $\text{Total Atoms} = n \times N_A \times \text{Atomicity}$ .
Mass of 1 single atom: $\frac{\text{Atomic weight (g)}}{N_A}$ .
Mass of 1 single molecule: $\frac{\text{Molecular weight (g)}}{N_A}$ .

⚠️ NEET Trap: Maximum Number of Atoms To quickly evaluate "Which has the maximum number of atoms?", compute: $$\text{Atoms} \propto \frac{\text{Given Mass}}{\text{Molar Mass}} \times \text{Atomicity}$$ * $1\text{ g }\text{Li} \implies \frac{1}{7} \times 1 = 0.143\text{ mol atoms}$ * $1\text{ g }\text{O}_2 \implies \frac{1}{32} \times 2 = 0.0625\text{ mol atoms}$ Smallest molar mass with highest atomicity always yields the maximum atoms.

5. Empirical & Molecular Formulae

Empirical Formula (EF): Lowest whole-number ratio of atoms present in a molecule.
Molecular Formula (MF): Actual count of each constituent atom. $\text{MF} = (\text{EF})_n \quad \text{where} \quad n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}$
Vapour Density (V.D.): $\text{Molar Mass} = 2 \times \text{Vapour Density}$

6. Stoichiometry & Limiting Reagent (LR)

For any balanced equation: $aA + bB \longrightarrow cC + dD$

Convert all reactant amounts into moles ( $n_A, n_B$ ).
Compute the mole-to-coefficient ratio: $\text{Compare } \frac{n_A}{a} \quad \text{vs.} \quad \frac{n_B}{b}$
The species giving the lower ratio is the Limiting Reagent (LR).
The LR is completely consumed and determines the exact theoretical yield of products.

7. Expression of Concentration of Solutions

Concentration Term	Formula	Temperature Dependence
Mass Percent ( $\% \text{ w/w}$ )	$\frac{w_{\text{solute}}}{w_{\text{solute}} + w_{\text{solvent}}} \times 100$	Independent
Mole Fraction ( $\chi_A$ )	$\chi_A = \frac{n_A}{n_A + n_B} \quad (\sum \chi_i = 1)$	Independent
Molarity ( $M$ )	$\frac{n_{\text{solute}}}{V_{\text{solution (L)}}} = \frac{w_B \times 1000}{M_B \times V\text{ (mL)}}$	Dependent ( $T \uparrow \implies V \uparrow \implies M \downarrow$ )
Molality ( $m$ )	$\frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}} = \frac{w_B \times 1000}{M_B \times W_{\text{solvent (g)}}}$	Independent

Vital Examination Shortcuts

Dilution: $M_1 V_1 = M_2 V_2$
Mixing Solutions: $M_{\text{final}} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$
Molality from Molarity ( $M$ ) & Density ( $d\text{ in g mL}^{-1}$ ): $m = \frac{1000 \times M}{(1000 \times d) - (M \times M_B)}$