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 202 questions; sign in to attempt the remaining 197 in the exam simulator.

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
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   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)
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   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 (08 Apr, Shift 1)Quantitative measures in chemical equations
   Combustion of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ produces $\mathrm{CO}_2$ and water. The amount of oxygen (in g) required for the complete combustion of $900 \mathrm{~g}$ of glucose is : [Molar mass of glucose in $\mathrm{g} \mathrm{mol}^{-1}=180$ ]
   1. A.480
   2. B.800
   3. C.960
   4. D.32
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   Answer: (C)

   $\begin{aligned} & \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~s})}+6 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(\theta)} \\ & \frac{900}{180} \\ & =5 \mathrm{~mol} \quad 30 \mathrm{~mol}\end{aligned}$ Mass of $\mathrm{O}_2$ required $=30 \times 32=960 \mathrm{gm}$
4. Q4JEE 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]
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   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 $225\times {10}^{-2}$
5. Q5JEE Main 2026 (05 Apr, Shift 1)Mole concept
   How many grams of residue is obtained by heating $2.76$ g of silver carbonate? (Given: Molar mass of C, O and Ag are $12$, $16$ and $108$ g mol$^{-1}$ respectively)
   1. A.$1.08$ g
   2. B.$2.16$ g
   3. C.$3.24$ g
   4. D.$4.32$ g
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   Answer: (B)

   Molar mass of $Ag_2CO_3$ = $2 \times 108 + 12 + 3 \times 16 = 276$ g mol$^{-1}$ Number of moles of $Ag_2CO_3$ = $\dfrac{2.76}{276} = 0.01$ mol The decomposition reaction of silver carbonate on heating is: $Ag_2CO_3(s) \rightarrow 2Ag(s) + CO_2(g) + \dfrac{1}{2}O_2(g)$ From the stoichiometry of the reaction, $1$ mole of $Ag_2CO_3$ gives $2$ moles of Ag as residue. Number of moles of Ag formed = $2 \times 0.01 = 0.02$ mol Mass of Ag residue = $0.02 \times 108 = 2.16$ g Answer: $2.16$ g