**Volume percent**:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{volume}}{\mathbf{=}}\frac{\mathbf{volume}\mathbf{}\mathbf{component}}{\mathbf{total}\mathbf{}\mathbf{volume}}{\mathbf{\times}}{\mathbf{100}}}$

For 1.00 L sample: (divide %vol by 100)

**V**_{N2}= 0.7808(1.00 L) = 0.**7808 L N**_{2}**V**_{O2}= 0.2094(1.00 L) = 0.2094 L O_{2}**V**_{CO2}= 0.0005(1.00 L) = 0.0005 L CO_{2}**V**_{Ar}=**0.0093(1.00 L) = 0****0.0093****L Ar**

**Ideal gas law**:

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{PV}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}\overline{)\mathbf{RT}}}{\overline{)\mathbf{RT}}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{n}}{\mathbf{=}}\frac{\mathbf{PV}}{\mathbf{RT}}}$

T = 25°C + 273.15 = **298.15 K**

Solving for moles of each gas:

${\mathbf{n}}_{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}\frac{(\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{atm}})(\mathbf{0}\mathbf{.}\mathbf{7808}\mathbf{}\overline{)\mathbf{L}})}{(\mathbf{0}\mathbf{.}\mathbf{08206}\mathbf{}{\displaystyle \frac{\overline{)\mathbf{L}}\mathbf{\xb7}\overline{)\mathbf{atm}}}{\mathbf{mol}\mathbf{\xb7}\overline{)\mathbf{K}}}})(\mathbf{298}\mathbf{.}\mathbf{15}\mathbf{}\overline{)\mathbf{K}})}\mathbf{=}$** 0.0319 mol**

A sample of air contains 78.08% nitrogen, 20.94% oxygen, 0.05% carbon dioxide, and 0.93% argon, by volume. How many molecules of each gas are present in 1.00 L of the sample at 25°C and 1.00 atm?

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