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Some Basic Concepts of Chemistry question

2025 · 23 Jan · Shift 1 · Q11
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Some Basic Concepts of Chemistry question

2025 · 23 Jan · Shift 1 · Q11

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
2.8×10−3 mol2.8 \times 10^{-3} \mathrm{~mol}2.8×10−3 mol of CO2\mathrm{CO}_2CO2​ is left after removing 102110^{21}1021 molecules from its 'xxx' mg sample. The mass of CO2\mathrm{CO}_2CO2​ taken initially is Given: NA=6.02×1023 mol−1\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}NA​=6.02×1023 mol−1
  1. A
    98.3 mg
  2. B
    196.2 mg
  3. C
    150.4 mg
  4. D
    48.2 mg
View written solutionFree

Correct answer: B

  1. Let the initial amount of CO2\mathrm{CO_2}CO2​ be nnn mol.

    After removing 102110^{21}1021 molecules, the amount left is given as: 2.8×10−3 mol2.8 \times 10^{-3}\ \text{mol}2.8×10−3 mol

  2. Convert removed molecules into moles using Avogadro's number: nremoved=10216.02×1023 moln_{\text{removed}}=\frac{10^{21}}{6.02\times 10^{23}}\ \text{mol}nremoved​=6.02×10231021​ mol =16.02×10−2=\frac{1}{6.02}\times 10^{-2}=6.021​×10−2 ≈1.66×10−3 mol\approx 1.66\times 10^{-3}\ \text{mol}≈1.66×10−3 mol

  3. Initial moles of CO2\mathrm{CO_2}CO2​: n=nleft+nremovedn = n_{\text{left}} + n_{\text{removed}}n=nleft​+nremoved​ n=2.8×10−3+1.66×10−3n = 2.8\times 10^{-3} + 1.66\times 10^{-3}n=2.8×10−3+1.66×10−3 n≈4.46×10−3 moln \approx 4.46\times 10^{-3}\ \text{mol}n≈4.46×10−3 mol

  4. Molar mass of CO2\mathrm{CO_2}CO2​: M=44 g mol−1M = 44\ \text{g mol}^{-1}M=44 g mol−1

  5. Initial mass: m=nM=4.46×10−3×44 gm = nM = 4.46\times 10^{-3}\times 44\ \text{g}m=nM=4.46×10−3×44 g m≈0.1962 gm \approx 0.1962\ \text{g}m≈0.1962 g

  6. Convert into mg: 0.1962 g=196.2 mg0.1962\ \text{g} = 196.2\ \text{mg}0.1962 g=196.2 mg

  7. Check options:

    • A: 98.398.398.3 mg
    • B: 196.2196.2196.2 mg
    • C: 150.4150.4150.4 mg
    • D: 48.248.248.2 mg

    Hence, the correct option is B.

Final Answer: 196.2 mg196.2\ \text{mg}196.2 mg

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