JEE Main Chemistry · Mole Concept PYQ

JEE Main Some Basic Concepts (Mole Concept) PYQ (2002–2026)

Mole Concept (Some Basic Concepts of Chemistry) is a P2 Physical Chemistry foundation chapter — ~4% JEE Main weightage with 1 question per session. Prerequisite for Stoichiometry, Equilibrium and Electrochemistry. Deceptively simple — small arithmetic errors break the answer.

Mole Concept PYQs from 2002 to 2026 covering moles, Avogadro's number, stoichiometry, percentage composition, empirical formula and concentration units.

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Some Basic Concepts (Mole Concept) at a Glance

Weightage

~4%

approx · 10-yr avg

Priority

Strong secondary

Year range

2002–2026

PYQ coverage

Typical in paper

per session

Try 5 Real Some Basic Concepts (Mole Concept) PYQs

5 real JEE Main questions on Some Basic Concepts (Mole Concept), straight from past papers. Try them, then check the answer.

Q1JEE Main 2026 (Online) 21st January Evening Shift
Aqueous HCl reacts with MnO₂(s) to form MnCl₂(aq), Cl₂(g), and H₂O(l). What is the weight (in g) of Cl₂ liberated when 8.7 g of MnO₂(s) is reacted with excess aqueous HCl solution? (Given Molar mass in g mol⁻¹ Mn = 55, Cl = 35.5, O = 16, H = 1)
1. $14.2$
2. $21.3$
3. $7.1$
4. $71$
Step-by-step solution — free with sign-up
Q2JEE Main 2025 (Online) 22nd January Evening Shift
Density of 3 M NaCl solution is $1.25 \mathrm{~g} / \mathrm{mL}$. The molality of the solution is :
1. $2.79 m$
2. $2 m$
3. $1.79 m$
4. $3 m$
Step-by-step solution — free with sign-up
Q3JEE Main 2024 (Online) 27th January Morning Shift
$Mass of methane required to produce 22 g of $\mathrm{CO}_2$ after complete combustion is _______ g. (Given Molar mass in g mol-1 $\mathrm{C}$${}=12.0$, $\mathrm{H}$${}=1.0$, $\mathrm{O}$${}=16.0)$$
Step-by-step solution — free with sign-up
Q4JEE Main 2023 (Online) 24th January Morning Shift
$When $\mathrm{Fe_{0.93}O}$ is heated in presence of oxygen, it converts to Fe_2O_3. The number of correct statement/s from the following is ________ A. The equivalent weight of $\mathrm{Fe_{0.93}O}$ is ${{\mathrm{Molecular\,weight}} \over {0.79}}$ B. The number of moles of Fe$^{2+}$ and Fe$^{3+}$ in 1 mole of $\mathrm{Fe_{0.93}O}$ is 0.79 and 0.14 respectively C. $\mathrm{Fe_{0.93}O}$ is metal deficient with lattice comprising of cubic closed packed arrangement of O$^{2-}$ ions D. The % composition of Fe$^{2+}$ and Fe$^{3+}$ in $\mathrm{Fe_{0.93}O}$ is 85% and 15% respectively$
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Q5JEE Main 2022 (Online) 29th June Evening Shift
Using the rules for significant figures, the correct answer for the expression ${{0.02858 \times 0.112} \over {0.5702}}$ will be
1. $0.005613$
2. $0.00561$
3. $0.0056$
4. $0.006$
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Practise the full Some Basic Concepts (Mole Concept) set

Every past-paper question of this chapter, with solutions, timing and your accuracy tracked.

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Key Sub-Topics & What's Tested

Mole Concept & Avogadro's Number

1 mole = 6.022 × 10²³ particles (Avogadro's number, N_A). Molar mass, molar volume at STP (22.4 L for ideal gas).

Mole-Mass-Volume Interconversion

Moles = mass/molar mass = volume at STP/22.4 = particles/N_A. Master the four-way conversion.

Percentage Composition

% of element = (mass of element in formula/molar mass) × 100%. Useful in finding empirical formula from combustion analysis.

Empirical & Molecular Formula

Empirical = simplest whole-number ratio of atoms. Molecular = actual formula, integer multiple of empirical (ratio = molar mass/empirical mass).

Stoichiometry

Balanced chemical equations, mole ratios for reactants/products, limiting reagent identification, theoretical vs actual yield.

Limiting Reagent Problems

In a reaction with multiple reactants, the one that runs out first is the limiting reagent — determines product amount.

Concentration Units

Molarity (mol/L), molality (mol/kg), normality (equivalents/L), mole fraction (moles of component/total moles).

Equivalent Weight & Normality

Equivalent weight = molar mass/n-factor (n = valence, number of H⁺/OH⁻, electrons transferred). N = eq/L. N × V = N × V for mixing.

Question Type Distribution

Question Type	Share (approx)	Example Pattern
Mole Calculation	25%	Number of atoms in 9 g of water (H₂O, M = 18 g/mol).
Stoichiometry Problem	25%	2H₂ + O₂ → 2H₂O. If 4g H₂ and 16g O₂ react, what mass of water forms?
Empirical Formula	15%	Compound has 40% C, 6.67% H, 53.33% O. Find empirical formula.
Concentration / Dilution	15%	Dilute 100 mL of 2M HCl to 500 mL — find new molarity.
Limiting Reagent	10%	N₂ + 3H₂ → 2NH₃. Given 14g N₂ and 3g H₂, identify limiting reagent.
Equivalent Weight / Normality	10%	Equivalent weight of KMnO₄ in acidic medium? (M/5 = 158/5 = 31.6)

How to Solve Some Basic Concepts (Mole Concept) PYQs

1
Always start with balanced equation. Un-balanced equations give wrong mole ratios. Always balance before calculating.
2
Master the mole conversions. Mass ↔ moles (÷M), moles ↔ particles (×N_A), moles ↔ volume at STP (×22.4L). These three ratios handle 90% of problems.
3
For limiting reagent, compute moles of each reactant. Divide by stoichiometric coefficient. Smallest ratio = limiting reagent. Everything else is excess.
4
Equivalent weight depends on reaction context. KMnO₄: M/5 in acidic, M/3 in neutral, M/1 in basic medium. Context determines n-factor.
5
Dilution: C₁V₁ = C₂V₂ (for molarity, if no chemical change). For normality mixing (acid-base, redox): N₁V₁ = N₂V₂. Same logic, different unit.

Common Mistakes That Cost Marks

Wrong molar volume value. Molar volume of ideal gas at STP is 22.4 L (not 22.7 or 24 or 25.7). Temperature of 0°C = 273 K.
Forgetting stoichiometric coefficients. 2H₂ + O₂ → 2H₂O: 2 moles H₂ gives 2 moles H₂O (not 1 mole). Coefficient is crucial.
Using molarity in mixing without volume conservation. Mixing V₁ of C₁ with V₂ of C₂ → new concentration C = (C₁V₁ + C₂V₂)/(V₁+V₂). Don't average.
Wrong n-factor for KMnO₄/K₂Cr₂O₇. KMnO₄: n=5 (acidic). K₂Cr₂O₇: n=6 (acidic only). Using wrong n gives wrong equivalent weight.
Treating empirical formula = molecular formula. They're the same only if ratio = 1. For ratio > 1, molecular = empirical × ratio. Ethene empirical = CH₂, molecular = C₂H₄.

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JEE Main Chemical Thermodynamics PYQ

Uses moles for ΔH, ΔS calculations.

JEE Main Equilibrium PYQ

Kp/Kc depend on mole concentrations — direct application.

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Frequently asked questions

How many Mole Concept PYQs should I solve?

Target 40–60 PYQs across 2010–2026. Foundation chapter — strong practice here accelerates every subsequent Physical Chemistry chapter.

Is Mole Concept really tested in JEE Main?

Yes — 1 question per paper consistently. Short, straightforward calculations that reward careful arithmetic. Don't skip.

What's the most-tested pattern?

Limiting reagent + theoretical yield calculation. Appears 2-3 times per 10-year window. Master the stepwise approach.

Do I need to memorise atomic masses?

Common ones yes — H=1, C=12, N=14, O=16, Na=23, Mg=24, Al=27, Cl=35.5, K=39, Ca=40, Fe=56, Cu=63.5. These appear in PYQs without atomic weights provided.

How important is equivalent weight concept?

Moderately — appears in 1 PYQ per 2-3 papers. Know KMnO₄, K₂Cr₂O₇ n-factors and how normality-molarity relates.

Does Mole Concept connect with Thermodynamics?

Yes — mole-based reaction progress tracking feeds into Thermodynamics ΔH calculations and Equilibrium ICE tables.

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