The number of moles of oxygen in 1 L of air containing 21% oxygen by volume, at standard conditions, is:
A. 0.186 mol
B. 0.21 mol
C. 2.10 mol
D. 0.0093 mol

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Detailed Solution

The Correct Answer is 0.0093 mol → Option D

Explanation

Given / Assumptions

Volume of air sample, V_air = 1 L
Oxygen content by volume in air = 21%
Standard conditions (STP): molar volume of an ideal gas, V_m = 22.4 L·mol⁻¹

Step-by-step

Volume of O₂ in the sample (by volume %):
V_O2 = 0.21 × V_air = 0.21 × 1 L = 0.21 L
Moles of O₂ at STP (using molar volume):
n_O2 = V_O2 / V_m = 0.21 L / (22.4 L·mol⁻¹) ≈ 0.009375 mol
Rounded to 2–3 significant figures:
n_O2 ≈ 0.0093 mol

Why Other Options Are Incorrect (Quick Check)

0.21 mol (Option B) incorrectly treats 21% of 1 L as moles rather than volume.
0.186 mol and 2.10 mol (Options A & C) are far too large for just 0.21 L of gas at STP.

Option D. 0.0093 mol

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