Some Basic Concepts of Chemistry JEE Main previous year questions with solutions

5 solved JEE Main questions on Some Basic Concepts of Chemistry, free to read — no sign-in needed. The full chapter has 210 questions; sign in to attempt the remaining 205 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Mole concept
   Which of the following contain the same number of atoms ? (Given : Molar mass in g mol$^{-1}$ of H, He, O and S are $1, 4, 16$ and $32$ respectively) A. $2$ g of O$_2$ gas B. $4$ g of SO$_2$ gas C. $1400$ mL of O$_2$ at STP D. $0.05$ L of He at STP E. $0.0625$ mol of H$_2$ gas Choose the correct answer from the options given below :
   1. A.A and B only
   2. B.B and C only
   3. C.C and D only
   4. D.A, C and E only
   Show answer & solution

   Answer: (D)

   Number of atoms is calculated as the product of the number of moles, atomicity, and Avogadro's number ($N_A$). For A: Moles of O$_2$ = $\dfrac{2}{32} = \dfrac{1}{16}$ mol. Atomicity of O$_2$ is $2$. Number of atoms = $\dfrac{1}{16} \times 2 \times N_A = \dfrac{1}{8} N_A$ For B: Moles of SO$_2$ = $\dfrac{4}{64} = \dfrac{1}{16}$ mol. Atomicity of SO$_2$ is $3$. Number of atoms = $\dfrac{1}{16} \times 3 \times N_A = \dfrac{3}{16} N_A$ For C: Moles of O$_2$ at STP = $\dfrac{1400}{22400} = \dfrac{1}{16}$ mol. Atomicity of O$_2$ is $2$. Number of atoms = $\dfrac{1}{16} \times 2 \times N_A = \dfrac{1}{8} N_A$ For D: Moles of He at STP = $\dfrac{0.05}{22.4} = \dfrac{1}{448}$ mol. Atomicity of He is $1$. Number of atoms = $\dfrac{1}{448} \times 1 \times N_A = \dfrac{1}{448} N_A$ For E: Moles of H$_2$ = $0.0625 = \dfrac{1}{16}$ mol. Atomicity of H$_2$ is $2$. Number of atoms = $\dfrac{1}{16} \times 2 \times N_A = \dfrac{1}{8} N_A$ Comparing the number of atoms, A, C, and E contain the same number of atoms. Answer: A, C and E only
2. Q2JEE Main 2025 (22 Jan, Shift 2)Concentration terms
   20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is _______ $\times 10^{-2} \mathrm{M}$. (Nearest integer)
   Show answer & solution

   Answer: 57

   $\begin{aligned} & M_F=\frac{M_1 V_1+M_2 V_2}{V_1+V_2} \\ & =\frac{2 \times 20+0.5 \times 400}{420}=0.571 \mathrm{M} \\ & =57.1 \times 10^{-2} \mathrm{M} \\ & =57\end{aligned}$
3. Q3JEE Main 2024 (04 Apr, Shift 2)Atomic Models
   Choose the Incorrect Statement about Dalton's Atomic Theory
   1. A.chemical reactions involve reorganization of atoms
   2. B.Matter consists of indivisible atoms.
   3. C.Compounds are formed when atoms of different elements combine in any ratio.
   4. D.Compounds are formed when atoms of different elements combine in any ratio. All the atoms of a given element have identical properties including identical mass.
   Show answer & solution

   Answer: (C)

   In compound atoms of different elements combine in fixed ratio by mass.
4. Q4JEE Main 2023 (06 Apr, Shift 2)Equivalence concept
   The volume of $0.02M$ aqueous $HBr$ required to neutralize $10.0mL$ of $0.01M$ aqueous $Ba(OH{)}_{2}$ is (Assume complete neutralization)
   1. A.$2.5mL$
   2. B.$5.0mL$
   3. C.$10.0mL$
   4. D.$7.5mL$
   Show answer & solution

   Answer: (C)

   The balanced chemical equation for the reaction between $HBr$ and $Ba(OH{)}_{2}$ is: $2HBr+Ba(OH{)}_{2}\rightarrow {BaBr}_{2}+2{H}_{2}O$ From this equation, we can see that $2$ moles of $HBr$ react with $1$ mole of $Ba(OH{)}_{2}$. Equal equivalents will react. Number of equivalents = Normality $\times$Volume So, ${N}_{1}{V}_{1}={N}_{2}{V}_{2} 0.02\times {V}_{1}=0.02\times 10 {V}_{1}=10ml$
5. Q5JEE Main 2022 (26 Jun, Shift 2)Laws of chemical combination
   The moles of methane required to produce $81g$ of water after complete combustion is____$\times {10}^{-2}mol$. [nearest integer]
   Show answer & solution

   Answer: 225

   ${CH}_{4}+2{O}_{2}⟶{CO}_{2}+2{H}_{2}O$ POAC on H atom ${nCH}_{4}\times 4={nH}_{2}O\times 2$ ${nCH}_{4}=\dfrac{81}{18}\times 2\times \dfrac{1}{4}=\dfrac{81}{36}$ ${nCH}_{4}=2.25$ $=225\times {10}^{-2}$ Hence, the number of moles of methane is $225\times {10}^{-2}$