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Some Basic Concepts of Chemistry question

2023 · 10 Apr · Shift 1 · Q2
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Some Basic Concepts of Chemistry question

2023 · 10 Apr · Shift 1 · Q2

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
The number of molecules and moles in 2.8375 litres of O 2_22​ at STP are respectively
  1. A
    1.505 ×\times× 10 23^{23}23 and 0.250 mol
  2. B
    7.527 ×\times× 10 22^{22}22 and 0.250 mol
  3. C
    7.527 ×\times× 10 23^{23}23 and 0.125 mol
  4. D
    7.527 ×\times× 10 22^{22}22 and 0.125 mol
View written solutionFree

Correct answer: D

  1. Use molar volume at STP

At STP, 111 mole of any ideal gas occupies 22.7 L22.7\ \text{L}22.7 L (approximately).

Given volume of O2\mathrm{O_2}O2​: V=2.8375 LV = 2.8375\ \text{L}V=2.8375 L

So, number of moles is n=V22.7=2.837522.7=0.125 moln = \frac{V}{22.7} = \frac{2.8375}{22.7} = 0.125\ \text{mol}n=22.7V​=22.72.8375​=0.125 mol

  1. Calculate number of molecules

Using Avogadro's number, NA=6.022×1023N_A = 6.022 \times 10^{23}NA​=6.022×1023

Number of molecules: N=nNA=0.125×6.022×1023N = nN_A = 0.125 \times 6.022 \times 10^{23}N=nNA​=0.125×6.022×1023 N=7.5275×1022N = 7.5275 \times 10^{22}N=7.5275×1022

Rounding suitably, N≈7.527×1022N \approx 7.527 \times 10^{22}N≈7.527×1022

  1. Match with the options

We obtained:

  • Moles =0.125 mol= 0.125\ \text{mol}=0.125 mol
  • Molecules =7.527×1022= 7.527 \times 10^{22}=7.527×1022

This matches Option D.

  1. Verification with stored answer

Stored correct answer: D

Our derived answer: D

So, the derived answer agrees with the stored answer.

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