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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2025 · 23 January · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

2.8 × 10^ -3 ~mol of CO_2 is left after removing 10^ 21 molecules...

2.8×10−3 mol2.8 \times 10^{-3} \mathrm{~mol} of CO2\mathrm{CO}_2 is left after removing 102110^{21} molecules from its ' xx ' mg sample. The mass of CO2\mathrm{CO}_2 taken initially is Given: NA=6.02×1023 mol−1\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}

Options

  1. A98 .3 mg
  2. B48.2 mg
  3. C196 .2 mg
  4. D150 .4 mg

Correct answer

C. 196 .2 mg

Step-by-step solution

Moles of removed CO2=10216.02×1023 mol=1.66×10−3 mol\mathrm{CO}_2=\frac{10^{21}}{6.02 \times 10^{23}} \mathrm{~mol} =1.66 \times 10^{-3} \mathrm{~mol} mole of CO2\mathrm{CO}_2 left =2.8×10−3moles=2.8 \times 10^{-3} \mathrm{moles} total moles of CO2\mathrm{CO}_2 taken initially =(2.8+1.66)×10−3 mol=(2.8+1.66) \times 10^{-3} \mathrm{~mol} mass of CO2\mathrm{CO}_2 taken initially =4.46×10−3×44=196.24×10−3 g=196.24mg\begin{aligned} & =4.46 \times 10^{-3} \times 44 \\ & =196.24 \times 10^{-3} \mathrm{~g} \\ & =196.24 \mathrm{mg} \end{aligned}
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