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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 23 January, Morning Shift — Question 33

After removing 102110^{21} molecules from an x mgx\,\mathrm{mg} sample of CO2\mathrm{CO_2}, 2.8×10−3 mol2.8 \times 10^{-3}\,\mathrm{mol} of CO2\mathrm{CO_2} is left. The mass of CO2\mathrm{CO_2} taken initially is:

(NA=6.02×1023 mol−1)(N_A = 6.02 \times 10^{23}\,\mathrm{mol^{-1}})
  1. Option A:

    196.2 mg196.2\,\mathrm{mg}

    Correct
  2. Option B:

    98.3 mg98.3\,\mathrm{mg}

  3. Option C:

    150.4 mg150.4\,\mathrm{mg}

  4. Option D:

    48.2 mg48.2\,\mathrm{mg}

Answer: A

Step-by-step solution

Moles   of   CO2   removed=10216.02×1023=1.66×10−3\text{Moles\; of\; } \mathrm{CO_2} \text{ \;removed} = \frac{10^{21}}{6.02 \times 10^{23}} = 1.66 \times 10^{-3} Moles   of   CO2   left=2.8×10−3\text{Moles \;of \;} \mathrm{CO_2} \text{ \;left} = 2.8 \times 10^{-3} Initial   moles=2.8×10−3+1.66×10−3=4.46×10−3\text{Initial \;moles} = 2.8 \times 10^{-3} + 1.66 \times 10^{-3} = 4.46 \times 10^{-3} Molar   mass   of   CO2=44\text{Molar \;mass\; of \;} \mathrm{CO_2} = 44 Initial   mass=4.46×10−3×44=0.1962 g\text{Initial \;mass} = 4.46 \times 10^{-3} \times 44 = 0.1962\,\mathrm{g} =196.2 mg= 196.2\,\mathrm{mg}

Thus, the correct answer is option A.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Introduction to Mole Concept