At N.T.P., 11.2 litres of ozone (O3) are contained in a vessel. How many atoms of oxygen are present? (Use Avogadro’s number 6.02 × 10^23 and N.T.P. molar volume 22.4 L/mol.)

Introduction / Context:Counting particles from gas volumes at standard conditions is a staple of stoichiometry. Here we translate a gas volume of ozone into a count of oxygen atoms, reinforcing the link between molar volume, moles of molecules, and constituent atoms.

Given Data / Assumptions:

• N.T.P. molar volume = 22.4 L per mole of ideal gas.
• Avogadro’s number NA = 6.02 × 10^23 mol^−1.
• Species: ozone, O3 (three oxygen atoms per molecule).

Concept / Approach:First convert litres of gas at N.T.P. to moles of O3 using the molar volume. Then multiply by 3 to get moles of oxygen atoms, and finally multiply by Avogadro’s number to obtain the number of atoms.

Step-by-Step Solution:

Verification / Alternative check:Dimensional consistency: litres cancel, leaving moles and then atoms. The factor of 3 accounts for triatomic ozone.

Why Other Options Are Wrong:

• 3.01 × 10^22: too small by a factor of 30.
• 6.02 × 10^23: equals one mole of atoms; our sample has 1.5 moles of atoms.
• 1.20 × 10^24: corresponds to 2 moles of atoms, not 1.5.
• 9.03 × 10^24: an order of magnitude too large.

Common Pitfalls:Accidentally treating ozone as diatomic oxygen or using the wrong molar volume (be consistent about N.T.P. vs S.T.P. definitions).

Final Answer:9.03 × 10^23