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:
| Concept | Formula |
|---|---|
| Number of Moles | Moles = Given Mass / Molar Mass |
| Number of Particles | Particles = Moles × Avogadro Number |
| Avogadro Number | 6.022 × 10²³ particles/mol |
| Number of Molecules | Molecules = Moles × 6.022 × 10²³ |
| Number of Atoms | Atoms = Moles × Avogadro Number × Atomicity |
| Molar Mass | Sum of atomic masses of all atoms in a compound |
| Percentage Composition | (Mass of Element / Molar Mass of Compound) × 100 |
| Empirical Formula | Simplest whole number ratio of atoms |
| Molecular Formula | (Empirical Formula) × n |
| Value of n | Molecular Mass / Empirical Formula Mass |
| Molarity | M = 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 )
correct answer:- 4
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:
correct answer:- 2
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}$$
D. 15.0 g of A is left unreacted after the completion of reaction.
Choose the correct answer from the options given below :
correct answer:- 3
Question 4
Aqueous HCI reacts with $$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}$$ Mn = 55, Cl = 35.5, 0 = 16, H = l )
correct answer:- 4
Question 5
x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M $$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}$$ are 36.5 and 171.0 g $$mol^{-1}$$ respectively.
correct answer:- 1825
Question 6
0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced $$0.9_{g}H_{2}O$$. Molar mass of (X) is_____$$mol^{-1}$$.
correct answer:- 100
Question 7
When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is_______ - (Nearest integer) Given : Molar mass of Al is 27.0 g $$mol^{-1}$$ Molar mass of O is 16.0 g $$mol^{-1}$$.
correct answer:- 153
Question 8
The observed and normal molar masses of compound $$MX_2$$ are $$65.6$$ and $$164$$ respectively. The percent degree of ionisation of $$MX_2$$ is $$\underline{\hspace{2cm}}\%.$$ (Nearest integer)
correct answer:- 75
Question 9
The number of moles of methane required to produce $$11$$ g $$CO_2(g)$$ after complete combustion is : (Given molar mass of methane in $$g mol^{-1} : 16$$)
correct answer:- 4
Question 10
5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$$^{-1}$$ respectively
correct answer:- 180
Question 11
Combustion of glucose $$(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$$]
correct answer:- 3
Question 12
A sample of CaCO$$_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$$: 84)
correct answer:- 1
Question 13
Consider the following reaction: $$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$$ formed in mmol is (nearest integer):
correct answer:- 24
Question 14
$$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}$$
correct answer:- 3
Question 15
The elemental composition of a compound is $$54.2\%C,\ 9.2\%H$$ and $$36.6\%O.$$ If the molar mass of the compound is $$132\ \text{g mol}^{-1},$$ the molecular formula of the compound is: [Given: Relative atomic masses C:H:O = 12:1:16]
correct answer:- 4