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JEE Basic Concepts in Chemistry PYQs With Video Solutions PDF

Srikanth Lingamneni

16

Aug 06, 2026

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JEE Basic Concepts in Chemistry PYQs With Video Solutions PDF

JEE Basic Concepts in Chemistry PYQ

The chapter “Some Basic Concepts of Chemistry” is one of the most important chapters for JEE preparation. It includes topics such as the mole concept, molar mass, stoichiometry, empirical formula, molecular formula, concentration terms, and percentage composition.

Solving JEE Basic Concepts in Chemistry PYQ helps you understand how questions are asked in the exam. It also improves your calculation speed and helps you find the topics that need more revision.

This chapter may look simple, but students often lose marks because of small mistakes. These mistakes usually happen while converting units, calculating moles, or using the wrong formula. Regular practice can help you avoid such errors.

On this page, you can revise important formulas, download the PYQ PDF, check common mistakes, and practise questions in a test format.

JEE Basic Concepts in Chemistry PYQ

Previous-year questions are very useful for JEE preparation. They show you the type of questions asked in the exam and the level of difficulty you can expect.

While solving JEE Basic Concepts in Chemistry Questions, do not try to memorise the answers. First, understand what the question is asking. Write down the given values, check the units, and then choose the correct formula.

For questions based on chemical reactions, always balance the equation before starting the calculation. The coefficients in a balanced equation give the correct mole ratio between reactants and products.

In limiting reagent questions, do not simply choose the reactant with the smaller mass. Convert the mass of each reactant into moles and compare the mole ratio with the balanced equation.

A good way to practise is to solve one JEE Main PYQ set within a fixed time. After finishing the set, check your answers and note your mistakes. Try to understand whether the mistake happened because of a weak concept, a wrong formula, or a calculation error.

You can revise these mistakes before attempting your next JEE Mock Test. This simple habit can improve both speed and accuracy.

JEE Basic Concepts in Chemistry PYQs PDF

You can use the JEE Basic Concepts in Chemistry PYQs PDF given below for chapter-wise practice.

First, try to solve all the questions without checking the answers. You can also set a timer to create an exam-like environment.

After completing the PDF, divide the questions into three groups:

  • Easy questions
  • Questions that took more time
  • Questions you could not solve

Revise the concepts used in the difficult questions and attempt them again after a few days.

You can also print the PDF and use it as a practice worksheet. While solving JEE Main questions, write all the calculation steps clearly. This will help you find mistakes during revision.

Do not only read the solutions. Try to solve each question on your own before checking the correct method.

Important Formulas for JEE Basic Concepts in Chemistry PYQs

The following formulas are commonly used in this chapter. You can add them to your JEE Mains Formula Sheet for quick revision.

ConceptFormulaSimple Explanation
Number of molesn = Given mass ÷ Molar massUse this formula when mass is given.
Number of particlesN = n × NₐNₐ is 6.022 × 10²³ particles per mole.
MolarityM = Moles of solute ÷ Volume of solution in litresConvert millilitres into litres before using the formula.
Molalitym = Moles of solute ÷ Mass of solvent in kgUse the mass of the solvent, not the solution.
Mole fractionX = Moles of one component ÷ Total molesThe sum of all mole fractions is equal to 1.
Mass percentageMass of component ÷ Total mass × 100Keep both masses in the same unit.
Percentage compositionMass of element ÷ Molar mass of compound × 100Used to find the percentage of an element in a compound.
Molecular formulaEmpirical formula × nn = Molar mass ÷ Empirical formula mass.
Ideal gas equationPV = nRTUse the correct value of R according to the units.
Percentage yieldActual yield ÷ Theoretical yield × 100Theoretical yield depends on the limiting reagent.

Do not learn these formulas without practising questions. You should also understand when and where each formula is used.

For example, molarity uses the volume of the solution, while molality uses the mass of the solvent.

Common Mistakes to Avoid in JEE Basic Concepts in Chemistry PYQs

Many students make mistakes because they start calculations without balancing the chemical equation. Always balance the equation before using mole ratios.

Another common mistake is mixing up molarity and molality. Molarity depends on the volume of the solution in litres. Molality depends on the mass of the solvent in kilograms.

Students also make errors while finding the limiting reagent. The reactant with the lower mass is not always the limiting reagent. You must first convert both reactants into moles.

Unit conversion is another important area. Remember to:

  • Convert millilitres into litres for molarity.
  • Convert grams into kilograms for molality.
  • Use the correct temperature and pressure units in gas-law questions.
  • Keep all mass values in the same unit.

Avoid rounding numbers too early. Keep one or two extra decimal places during the calculation and round the final answer at the end.

After solving a JEE Main PYQ, also check why the other options are wrong. This will help you understand common traps used in the exam.

List of JEE Electrochemistry PYQs

Use the section below as a practice test. Each question can have four options so that students can solve it like a real exam.

Question 1

The processes of calcination and roasting in metallurgical industries, respectively, can lead to:

Show Answer Explanation

Question 2

A transition metal $$M$$ forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be


Question 3

'$$a$$' and '$$b$$' are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because:


Question 4

The hardness of a water sample containing $$10^{-3}$$ M $$MgSO_4$$ expressed as $$CaCO_3$$ equivalents (in ppm) is ___________.
(molar mass of $$MgSO_4$$ is 120.37 g/mol)


Question 5

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)


Question 6

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Heavy water is used for the study of reaction mechanism.
Reason (R): The rate of reaction for the cleavage of O-H bond is slower than that of O-D bond.
Choose the most appropriate answer from the options given below:

Show Answer Explanation

Question 7

Which of the following is a set of greenhouse gases?

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Question 8

An aqueous solution of a commercially available acid has a density of $$1.25\text{ g mL}^{-1}$$ and is labeled as $$36.5\%$$ by mass of $$\text{HCl}$$. Calculate the molality (m) and molarity (M) of this $$\text{HCl}$$ solution respectively.
(Take molar mass of $$\text{HCl} = 36.5\text{ g mol}^{-1})$$

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Question 9

A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of $$2.0$$ and $$2.5$$ Å respectively. The ratio of densities of lattices fcc to bcc for the metal M is ______ (Nearest integer)

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Question 10

The primary pollutant that leads to photochemical smog is:

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Question 11

Excess of NaOH (aq) was added to 100 mL of FeCl$$_3$$ (aq) resulting into 2.14 g of Fe(OH)$$_3$$. The molarity of FeCl$$_3$$(aq) is: (Given the molar mass of Fe = 56 g mol$$^{-1}$$ and molar mass of Cl = 35.5 g mol$$^{-1}$$)

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Question 12

Identify the pollutant gases largely responsible for the discoloured and lustreless nature of marble of the Taj Mahal.

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Question 13

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N$$_2$$) at 4 bar. The molar mass of the gaseous molecule is

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Question 14

The molecular formula of a commercial resin used for exchanging ions in water softening is $$C_8H_7SO_3Na$$ (molecular weight = 206). What would be the maximum uptake of $$Ca^{2+}$$ ions by the resin if expressed in mol per gm?

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Question 15

The higher concentration of which gas in air can cause stiffness of flower buds?

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Question 16

Taj Mahal is being slowly disfigured and discoloured. This is primarily due to

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Question 17

Among $$10^{-9}$$ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt


Question 18

The molecule that has minimum or no role in the formation of photochemical smog, is:

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Question 19

Match Column I with Column II
Column I      Column II
A. Soda ash      I. CaSO$$_4$$
B. Chlorophyll      II. CaOH$$_2$$
C. Used in Whitewashing      III. Na$$_2$$CO$$_3$$
D. Dentistry, ornamental work      IV. Mg$$^{2+}$$ ions

Show Answer Explanation

Question 20

0.27 g of a long chain fatty acid was dissolved in 100 cm$$^3$$ of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?
[Density of fatty acid = 0.9 g cm$$^{-3}$$; $$\pi = 3$$]


Question 21

Which is wrong with respect to our responsibility as a human being to protect our environment?


Question 22

Biochemical Oxygen Demand (BOD) is the amount of oxygen required (in ppm):

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Question 23

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]

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Question 24

For per gram of reactant, the maximum quantity of N$$_2$$ gas is produced in which of the following thermal decomposition reactions? (Given: Atomic wt. : Cr = 52u, Ba = 137u)

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Question 25

A sample of NaClO$$_3$$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be: (Given: Molar mass of AgCl = 143.5 g mol$$^{-1}$$)

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Question 26

NaClO$$_3$$ is used, even in spacecrafts, to produce O$$_2$$. The daily consumption of pure O$$_2$$ by a person in 492 L at 1 atm, 300K. How much amount of NaClO$$_3$$, in grams, is required to produce O$$_2$$ for the daily consumption of a person at 1 atm, 300K?
NaClO$$_3$$(s) + Fe(s) $$\rightarrow$$ O$$_2$$(g) + NaCl(s) + FeO(s)
R = 0.082 L atm mol$$^{-1}$$ K$$^{-1}$$


Question 27

The correct statements among (a) to (d) regarding $$H_2$$ as a fuel are: (i) It produces less pollutants than petrol. (ii) A cylinder of compressed dihydrogen weighs ~ 30 times more than a petrol tank producing the same amount of energy. (iii) Dihydrogen is stored in tanks of metal alloys like $$NaNi_5$$. (iv) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 kJ, respectively.

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Question 28

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O$$_2$$ for complete combustion, and 40 mL of CO$$_2$$ is formed. The formula of the hydrocarbon is:


Question 29

Water samples with BOD values of 4 ppm and 18 ppm, respectively, are:

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Question 30

A 100 mL solution was made by adding 1.43 g of $$Na_2CO_3 \cdot xH_2O$$. The normality of the solution is 0.1 N. The value of x is __________ (The atomic mass of Na is 23 g/mol)

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