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JEE Mole Concept PYQs with Solutions PDF, Download Now

Dakshita Bhatia

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Mar 27, 2026

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JEE Mole Concept PYQs with Solutions PDF, Download Now

JEE Mole Concept PYQs

JEE Mole Concept PYQs are an important part of the JEE Chemistry syllabus. They help you understand the kind of questions asked from this chapter and check how well you know the main ideas, such as mole, molar mass, Avogadro number, empirical formula, molecular formula, percentage composition, and stoichiometry.

In the exam, mole concept questions usually come as direct numerical problems or simple concept-based questions. The good thing is that this chapter becomes much easier when your basics are clear. Once you understand the concepts properly and know which formula or method to use, solving questions feels much more manageable. You do not need to think of mole concept as a very difficult chapter. With regular revision and smart practice, it can become one of the more scoring parts of JEE Chemistry.

In this blog, you will find a simple formula PDF, a section for important JEE Mole Concept PYQs in download format, a few practice questions with answers, and some extra questions to solve on your own. You will also learn about common mistakes students often make and a few simple tips to save time in the exam.

JEE Mole Concept Important PYQs PDF

This PDF can include the most important previous year questions from mole concept. It may cover topics like mole calculation, molar mass, number of particles, Avogadro number, percentage composition, empirical formula, molecular formula, limiting reagent, and stoichiometric calculations.

Practicing these questions will help you understand the exam pattern better. It will also improve your speed, accuracy, and confidence before the exam.

Important Formulas for JEE Mole Concept PYQs

You only need a few important formulas and ideas to solve most mole concept questions in JEE. These formulas help you understand mole calculations, mass relations, particle count, and reaction-based numerical problems more clearly.

You can download the full formula PDF from the link above. Here is a quick look at some of the main formulas:

ConceptFormula
Number of MolesMoles = Given Mass / Molar Mass
Number of ParticlesParticles = Moles Γ— Avogadro Number
Avogadro Number6.022 Γ— 10Β²Β³ particles/mol
Number of MoleculesMolecules = Moles Γ— 6.022 Γ— 10Β²Β³
Number of AtomsAtoms = Moles Γ— Avogadro Number Γ— Atomicity
Molar MassSum of atomic masses of all atoms in a compound
Percentage Composition(Mass of Element / Molar Mass of Compound) Γ— 100
Empirical FormulaSimplest whole number ratio of atoms
Molecular Formula(Empirical Formula) Γ— n
Value of nMolecular Mass / Empirical Formula Mass
MolarityM = Moles of Solute / Volume of Solution in L

These formulas are commonly used in questions based on moles, particles, composition, formulas, and stoichiometry. If you revise them properly, many JEE questions will start to feel much easier.

Top 5 Common Mistakes to Avoid in JEE Mole Concept PYQs

Many students find mole concept confusing at first because it includes a lot of calculations. But most mistakes happen because small details are missed while solving. Here are some common mistakes you should avoid:

Mixing up mass, moles, and number of particles
These three are related, but they are not the same. Many students rush into solving and forget to check what the question is actually asking.

Using molar mass incorrectly
To convert mass into moles, you need the correct molar mass. Even a small mistake here can change the whole answer.

Making mistakes in empirical and molecular formula questions
Students often get the ratio correctly but forget to convert it into the simplest whole number ratio.

Ignoring the limiting reagent in reaction-based questions
In stoichiometry problems, the final result depends on the limiting reagent. If you do not identify it correctly, the answer can go wrong.

Not checking units carefully
Always see whether the question gives mass in grams, volume in litres, or particles in numbers. Even when the method is right, a unit mistake can lead to the wrong answer.

List of JEE Mole Concept PYQs

Here is a short set of JEE-style mole concept questions for practice. These include common question types from mole calculation, particles, molar mass, formulas, and stoichiometry. Solving them regularly can help you become faster and more confident.

Question 1

By usual analysis, 1.00 g of compow1d (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer)
(Given, molar mass in $$gmol^{-1}$$ : 0 = 16, Mg = 24, P = 31 )

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Question 2

Given below are four compounds :
(a) n-propyl choride
(b) iso-propyl chloride
(c) sec-butyl chloride
(d) neo-pentyl chloride
Percentage of carbon in the one which exhibits optical isomerism is:

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Question 3

$$A + 2B \rightarrow AB_{2}$$
36.0 g of ' A' (Molar mass : 60 g $$mol^{-1}$$) and 56.0 g of 'B' (Molar mass: 80 g $$mol^{-1}$$) are allowed to react. Which of the following statements are correct?
A. 'A' is the limiting reagent.
B. 77.0 g of $$AB_{2}$$ is formed.
C. Molar mass of $$AB_{2} is 140 g $$mol^{-1}$$
D. 15.0 g of A is left unreacted after the completion of reaction.
Choose the correct answer from the options given below :

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Question 4

Aqueous HCI reacts with $$MnO_{2} \left(s\right)$$ to form $$MnCl_{2}\left(aq\right)$$, $$Cl_{2}\left(g\right)$$ and $$H_{2}O\left(l\right)$$. What is the weight (in g) of Cl_{2} liberated when 8.7 g of $$MnO_{2} \left(s\right)$$ is reacted with excess aqueous HCI solution ?
(Given Molar mass in g $$mol^{-1}$$ Mn = 55, Cl = 35.5, 0 = 16, H = l )

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Question 5

x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M $$Ba(OH)_{2}$$ solution is used when the HCl solution was titrated against it. The numerical value of x is ______$$\times 10^{-1}$$. (nearest integer)
Given : Molar mass of HCl and $$Ba(OH)_{2}$$ are 36.5 and 171.0 gΒ $$mol^{-1}$$ respectively.

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Question 6

0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced $$0.9_{g}H_{2}O$$. Molar mass of (X) is_____$$mol^{-1}$$.

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Question 7

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is_______ - (Nearest integer) Given : Molar mass of Al is 27.0 g $$mol^{-1}$$ Molar mass of O is 16.0 g $$mol^{-1}$$.

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Question 8

The observed and normal molar masses of compoundΒ  $$MX_2$$ areΒ  $$65.6$$Β  andΒ  $$164$$Β  respectively. The percent degree of ionisation ofΒ  $$MX_2$$ Β isΒ  $$\underline{\hspace{2cm}}\%.$$ Β (Nearest integer)

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Question 9

The number of moles of methane required to produce $$11$$ g $$CO_2(g)$$ after complete combustion is : (Given molar mass of methane in $$g mol^{-1} : 16$$)

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Question 10

5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$$^{-1}$$ respectively

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Question 11

Combustion of glucose $$(C_6H_{12}O_6)$$ produces $$CO_2$$ and water. The amount of oxygen (in g) required for the complete combustion of $$900 \text{ g}$$ of glucose is : [Molar mass of glucose in $$\text{gmol}^{-1} = 180$$]

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Question 12

A sample of CaCO$$_3$$ and MgCO$$_3$$ weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of the mixture is: (Given molar mass in g mol$$^{-1}$$, CaCO$$_3$$: 100, MgCO$$_3$$: 84)

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Question 13

Consider the following reaction: $$3PbCl_2 + 2(NH_4)_3PO_4 \rightarrow Pb_3(PO_4)_2 + 6NH_4Cl$$. If $$72$$ mmol $$PbCl_2$$ is mixed with $$50$$ mmol of $$(NH_4)_3PO_4$$, then amount of $$Pb_3(PO_4)_2$$ formed in mmol is (nearest integer):

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Question 14

$$2.8 \times 10^{-3}$$ mol of $$CO_{2}$$ is left after removing $$10^{21}$$ molecules from its 'x' mg sample. The mass of $$CO_{2}$$ taken initially is Given: $$N_{A} = 6.02 \times 10^{23} mol^{-1}$$

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Question 15

The elemental composition of a compound isΒ  $$54.2\%C,\ 9.2\%H$$Β  andΒ  $$36.6\%O.$$ If the molar mass of the compound isΒ  $$132\ \text{g mol}^{-1},$$ the molecular formula of the compound is: [Given: Relative atomic massesΒ  C:H:O = 12:1:16]

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