JEE Main 2026 — Some Basic Concepts of Chemistry Question with Solution

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${}_2$ = $\frac{2}{32}=\frac{1}{16}$ mol. Atomicity of O${}_2$ is $2$.
Number of atoms = $\frac{1}{16}\times 2\times N_A=\frac{1}{8}{N}_A$

For B: Moles of SO${}_2$ = $\frac{4}{64}=\frac{1}{16}$ mol. Atomicity of SO${}_2$ is $3$.
Number of atoms = $\frac{1}{16}\times 3\times N_A=\frac{3}{16}{N}_A$

For C: Moles of O${}_2$ at STP = $\frac{1400}{22400}=\frac{1}{16}$ mol. Atomicity of O${}_2$ is $2$.
Number of atoms = $\frac{1}{16}\times 2\times N_A=\frac{1}{8}{N}_A$

For D: Moles of He at STP = $\frac{0.05}{22.4}=\frac{1}{448}$ mol. Atomicity of He is $1$.
Number of atoms = $\frac{1}{448}\times 1\times N_A=\frac{1}{448}{N}_A$

For E: Moles of H${}_2$ = $0.0625=\frac{1}{16}$ mol. Atomicity of H${}_2$ is $2$.
Number of atoms = $\frac{1}{16}\times 2\times N_A=\frac{1}{8}{N}_A$

Comparing the number of atoms, A, C, and E contain the same number of atoms.

Answer: A, C and E only