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ChemMedMain 2026 • 6th Apr - Shift 2

Calculate and Compare Number of Atoms in Given Gas Samples

Which of the following contain the same number of atoms?
(Given : Molar mass in g mol−1\text{g mol}^{-1} of H\text{H}, He\text{He}, O\text{O} and S\text{S} are 11, 44, 1616 and 3232 respectively)

A. 2 g2\text{ g} of O2\text{O}_2 gas
B. 4 g4\text{ g} of SO2\text{SO}_2 gas
C. 1400 mL1400\text{ mL} of O2\text{O}_2 at STP
D. 0.05 L0.05\text{ L} of He\text{He} at STP
E. 0.0625 mol0.0625\text{ mol} of H2\text{H}_2 gas

Choose the correct answer from the options given below :

Options

A

A and B only

B

B and C only

C

C and D only

D

A, C and E only

Correct

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometry

Step-by-Step Solution

To determine which of the given gas samples contain the same number of atoms, we calculate the total number of moles of atoms in each sample using Natoms=nmolecules×atomicity×NAN_{\text{atoms}} = n_{\text{molecules}} \times \text{atomicity} \times N_A, where NAN_A is Avogadro's number:

  1. Sample A: 2 g2\text{ g} of O2\text{O}_2 gas

    • Molar mass of O2=2×16=32 g mol−1\text{O}_2 = 2 \times 16 = 32\text{ g mol}^{-1}
    • Moles of O2\text{O}_2 molecules, nA=2 g32 g mol−1=0.0625 moln_{\text{A}} = \frac{2\text{ g}}{32\text{ g mol}^{-1}} = 0.0625\text{ mol}
    • Atomicity of O2=2\text{O}_2 = 2
    • Total moles of atoms = 0.0625×2=0.125 mol0.0625 \times 2 = 0.125\text{ mol}
    • Number of atoms = 0.125NA0.125 N_A

  2. Sample B: 4 g4\text{ g} of SO2\text{SO}_2 gas

    • Molar mass of SO2=32+(2×16)=64 g mol−1\text{SO}_2 = 32 + (2 \times 16) = 64\text{ g mol}^{-1}
    • Moles of SO2\text{SO}_2 molecules, nB=4 g64 g mol−1=0.0625 moln_{\text{B}} = \frac{4\text{ g}}{64\text{ g mol}^{-1}} = 0.0625\text{ mol}
    • Atomicity of SO2=3\text{SO}_2 = 3
    • Total moles of atoms = 0.0625×3=0.1875 mol0.0625 \times 3 = 0.1875\text{ mol}
    • Number of atoms = 0.1875NA0.1875 N_A

  3. Sample C: 1400 mL1400\text{ mL} of O2\text{O}_2 at STP

    • Molar volume of ideal gas at STP = 22.4 L=22400 mL22.4\text{ L} = 22400\text{ mL}
    • Moles of O2\text{O}_2 molecules, nC=1400 mL22400 mL mol−1=0.0625 moln_{\text{C}} = \frac{1400\text{ mL}}{22400\text{ mL mol}^{-1}} = 0.0625\text{ mol}
    • Atomicity of O2=2\text{O}_2 = 2
    • Total moles of atoms = 0.0625×2=0.125 mol0.0625 \times 2 = 0.125\text{ mol}
    • Number of atoms = 0.125NA0.125 N_A

  4. Sample D: 0.05 L0.05\text{ L} of He\text{He} at STP

    • Moles of He\text{He} atoms, nD=0.05 L22.4 L mol−1≈0.00223 moln_{\text{D}} = \frac{0.05\text{ L}}{22.4\text{ L mol}^{-1}} \approx 0.00223\text{ mol}
    • Atomicity of He=1\text{He} = 1
    • Number of atoms = 0.00223NA0.00223 N_A

  5. Sample E: 0.0625 mol0.0625\text{ mol} of H2\text{H}_2 gas

    • Moles of H2\text{H}_2 molecules, nE=0.0625 moln_{\text{E}} = 0.0625\text{ mol}
    • Atomicity of H2=2\text{H}_2 = 2
    • Total moles of atoms = 0.0625×2=0.125 mol0.0625 \times 2 = 0.125\text{ mol}
    • Number of atoms = 0.125NA0.125 N_A

Comparing the results, samples A, C, and E each contain 0.125NA0.125 N_A atoms.

Hence, the correct option is D (A, C and E only).

Calculate and Compare Number of Atoms in Given Gas Samples | Chemistry PYQ Solution - JEE Challenger