Air inhaled"=1.80 \\ L"
Oxygen inhaled"=20.9" % of air"=\\frac{20.9}{100}\\times 1.80=0.3762 \\ L"
Number of moles of "O_2" in this amount"=\\frac{0.3762}{22.4}=16.795\\times 10^{-3} \\ mol"
Number of molecules of "O_2=16.795\\times10^{-3}\\times 6.023 \\times 10^{23}=101.156\\times10^{20}"
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