Some Basic Concepts in Chemistry

Physical Chemistry Weightage: 1-2 Questions (4-8 Marks) JEE Unit 1

“Welcome to the JEE Main track! These notes assume you have already read the NCERT chapter and know the basic definitions, so they skip the textbook narration and focus only on what JEE actually tests. This chapter is small but it is pure marks: about one or two questions per paper, almost always a calculation (limiting reagent, concentration conversion, formula finding or gas-volume analysis), and often a numerical-answer question with no options to guess from. The trick is speed and a clean method, not theory. Learn the five tools below, watch the red trap boxes, and this chapter becomes the easiest 4 to 8 marks in your paper. Oxidation numbers and redox balancing live in Unit 7, and colligative properties in Unit 5, so they are not repeated here.”

— SCORECHEM ACADEMIC TEAM

1. The Mole Toolkit

Every question in this chapter routes through moles (n). Convert to n first, work in n, convert back last.

Fig. 1: Every mole-concept question routes through n.

1 amu = 1/N_A g = 1.66 × 10⁻²⁴ g (because 1 mol of C-12 is 12 g). Useful for "mass of one atom" questions: mass of one atom = molar mass / N_A.
Average atomic mass of an isotope mix = Σ(mass × abundance %) / 100.
Mean molar mass of a gas mixture = total mass / total moles = Σ(n_iM_i) / Σn_i. Vapour density = M / 2.
Counting what is asked: always write "per molecule" numbers first. H₂SO₄: 7 atoms, 50 electrons, 98 g/mol. In 1 mol H₂O: 10 N_A electrons, 10 N_A protons, 8 N_A neutrons (for ¹⁶O with common H).

⚠️ JEE Trap: Which molar volume? The default in JEE questions is 22.4 L at 273 K and 1 atm. If the question says "1 bar" the molar volume is 22.7 L. If the conditions are not STP, use PV = nRT. Also, "33.6 L H₂ per mole of Al" is true only at STP, never "regardless of temperature and pressure".

2. Finding Formulae

Empirical formula (from % composition): percentage → moles (divide by atomic mass) → divide by the smallest → make whole numbers.

Ratio after dividing by smallest	Multiply all by
1.5 (x.5)	2
1.33 or 1.67 (x.33, x.67)	3
1.25 or 1.75	4

Molecular formula = (empirical formula)_n, where n = molar mass / empirical formula mass.

Option-check shortcut (fastest in JEE): when the molar mass and one percentage are given with four candidate formulae, just test each option's molar mass and %C. Example: 42.1% C, 6.4% H, molar mass 342. Only C₁₂H₂₂O₁₁ gives 144/342 = 42.1%, so the answer is found in about ten seconds.

⚠️ JEE Trap: Rounding to the wrong whole number. A ratio of 1.5 is NOT 1 or 2; it means the true ratio is 3:2. Never round 1.33 to 1. Multiply first, then round only tiny errors (like 1.98 or 2.02).

3. Stoichiometry and Limiting Reagent

Fig. 2: Limiting reagent in five steps, with a worked check.

Balance first. Moles reacted ÷ coefficient is the same for every species in a balanced equation.
% yield = (actual / theoretical) × 100. Theoretical always comes from the limiting reagent. Example: 9.3 g aniline (0.1 mol) with excess Br₂ water gives at most 0.1 mol of 2,4,6-tribromoaniline (330 g/mol) = 33 g; if 26.4 g is obtained, yield = 80%.
POAC (atom conservation) saves balancing: moles of an atom before = after. For KClO₃ → KCl + O₂: 3 n(KClO₃) = 2 n(O₂) and n(KClO₃) = n(KCl).
Gas reactions at the same T and P: volume ratio = mole ratio, so 6 mol HCl : 3 mol H₂ means 12 L HCl : 6 L H₂ directly.
Sequential reactions: the moles of the intermediate that is carried over are the "given" for the next step; use POAC on the key atom to skip the middle.

⚠️ JEE Trap: The "correct statements" item. Questions such as "A + 2B → AB₂, which statements are correct?" need all four checks: LR, moles of product, mass of product using the product's molar mass (not A or B), and leftover of the excess reagent. A single wrong molar mass sinks the option.

4. Gas Volume Analysis (Eudiometry)

Fig. 3: Gas-volume (eudiometry) bookkeeping for a hydrocarbon.

KOH (or alkali) absorbs CO₂. Volume drop on KOH = V(CO₂). Other absorbers: alkaline pyrogallol takes up O₂, turpentine oil takes up O₃. The absorbing agent tells you which gas is removed.
Volumes are proportional to moles, so you never need to convert to grams.
Two equations solve most problems: V(CO₂) / V(hc) = x, and O₂ used / V(hc) = x + y/4.
Worked check: 10 mL C₂H₆ + 50 mL O₂. O₂ used 35, CO₂ formed 20, O₂ left 15. After cooling the volume is 35 mL; after KOH it is 15 mL. Contraction on cooling = 25 mL = 10 × (1 + 6/4).

⚠️ JEE Trap: Is water a gas? "After cooling to room temperature" means H₂O is liquid and adds nothing to the volume. If the products are measured hot (above 100 °C), water vapour must be counted, and the contraction formula changes.

5. Concentration Conversions

Fig. 4: Concentration conversions. When in doubt, assume 1 L of solution or 100 g of solution.

Temperature dependence: molarity and % w/v depend on temperature (volume changes). Molality, mole fraction, % w/w and ppm do not.
Method that never fails: pick a convenient basis (1 L of solution, or 100 g of solution), find the mass of solute, mass of solvent and moles, then compute what is asked. Use density only to move between mass and volume of the solution.
Dilution: M₁V₁ = M₂V₂. Mixing the same solute: M = (M₁V₁ + M₂V₂) / (V₁ + V₂).
Check: a 3 molal NaOH solution of density 1.12 g/mL is 3.0 M (1000 × 1.12 × 3 / (1000 + 3 × 40)). If molality and molarity of a dilute solution are given, they are nearly equal only when the density is close to 1.

⚠️ JEE Trap: Solvent mass vs solution mass. Molality uses the mass of SOLVENT in kg, and molarity uses the volume of SOLUTION in litres. Students who use "1000 g" or "1 L" for the wrong one lose every conversion question. For an aqueous solution, solvent mass = solution mass − solute mass; it is never simply "1000 g".

6. H₂O₂, Oleum and Hardness

Topic	Key relation	Remember
H₂O₂ volume strength	V = 11.2 × M; % w/v = 0.3036 × V	2H₂O₂ → 2H₂O + O₂; 1 mol gives 11.2 L O₂ at STP; n-factor is 2 in every reaction
Oleum labelled y%	% free SO₃ = (40/9)(y − 100)	109% oleum means 100 g oleum absorbs 9 g water, i.e. contains 40 g free SO₃
Hardness (ppm)	ppm of CaCO₃ = mass of CaCO₃ equivalent per 10⁶ g water	Convert every Ca²⁺/Mg²⁺ salt to its CaCO₃ equivalent (molar mass 100)
Temporary vs permanent hardness	Bicarbonates vs chlorides/sulphates of Ca and Mg	Boiling or Clark's process removes temporary; washing soda or ion exchange removes permanent

7. How JEE Frames Questions

If you see	Do this
Two reactant masses, "which statements are correct"	Find LR; check product mass and leftover (Section 3)
Reagent volume from a % and density (e.g. HCl 38.55% w/w, d = 1.13)	Mass of solution = V × d; mass of HCl = × %; convert to moles
Three or four candidate formulae + one %, + molar mass	Test each option's %C; do not do the full derivation
A hydrocarbon burnt with O₂, volumes after cooling and after KOH	Eudiometry: x from CO₂, y from O₂ used
"M molar, density d, find m" or the reverse	Basis method or formula from the table (Section 5)
"Per mole of X, how many litres of gas"	Mole ratio from the equation, then 22.4 L; check the stated T and P
"Nearest integer" answer with % yield or dilution	Keep 4 significant figures until the last step; round once at the end

8. Quick Sheet and Checklist

Point	Formula
Moles	n = w/M = N/N_A = V/22.4 (STP)
Empirical to molecular	MF = (EF)_n, n = M / M(EF)
LR	smallest n / coefficient
% yield	actual / theoretical × 100
M to m	m = 1000 M / (1000 d − M M_w)
m to M	M = 1000 d m / (1000 + m M_w)
% w/w to M	M = 10 d (%) / M_w
Hydrocarbon burning	O₂ needed = x + y/4 per mole C_xH_y
H₂O₂	V = 11.2 M
Oleum	free SO₃ % = (40/9)(y − 100)

Before the exam, check you can:

Convert between M, m, x and % w/w in under a minute using a basis.
Identify the limiting reagent and compute leftover and product mass without balancing errors.
Solve a gas-volume problem where a hydrocarbon burns and KOH absorbs CO₂.
Decide which molar volume (22.4 or 22.7 L) a question intends.
Find a molecular formula from % composition and molar mass, using the option check.