# Problem: 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?

###### FREE Expert Solution

Volume percent:

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

• VN2 = 0.7808(1.00 L) = 0.7808 L N2
• VO2 = 0.2094(1.00 L) = 0.2094 L O2
• VCO2 = 0.0005(1.00 L) = 0.0005 L CO2
• VAr = 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:

0.0319 mol ###### Problem Details

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?