A sample of air contains 78.08% nitrogen, 20.94% oxygen,

Step 1 of 4

Here we have to calculate how many molecules of each gas are present in 1.00 L of the sample at 25°C and 1.00 atm.

Given:

A sample of air contains 78.08% nitrogen, 20.94% oxygen, 0.05% carbon dioxide, and 0.93% argon, by volume.

Volume of sample = 1.00 L

Pressure =  1.00 atm

Temperature = 25oC = 273 + 25 = 298 K

Calculation of number of molecules of \(\mathrm{N}_{2}\) present in the sample:-

1st we have to convert 78.08 % of \(\mathrm{N}_{2}\) into volume, so that by using ideal gas equation we can calculate the number of mole of gas present in the sample and finally by multiplying with Avogadro's number we can find out the number of molecules of \(\mathrm{N}_{2}\) present in the sample.

\(Volume of \mathrm{N}_{2}=\% \text { of Nitrogen } \times 1.00 \mathrm{~L}\)

\(\begin{aligned}=&\frac{78.08\mathrm{~L}\mathrm{~N}_2}{100\mathrm{Lair}}\times1.00\mathrm{~L}\\ =&0.7808\mathrm{~L}\text{ of }\mathrm{N}_2\end{aligned}\)

Now according to ideal gas equation it is known that,

\(\begin{aligned} \mathrm{n} & =\frac{P V}{R T} \\ & =\frac{1 \mathrm{~atm} \times 0.7808 \mathrm{~L}}{0.0821 \mathrm{Latm} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 298 \mathrm{~K}}=0.0319 \simeq 0.032 \mathrm{~mol} \text { of } \mathrm{N}_{2} \end{aligned}\)

Thus the number of molecules of \(\mathrm{N}_{2}\) can be calculated as,

\(0.032 \mathrm{~mol} \text { of } \mathrm{N}_{2} \times\left(6.022 \times 10^{23} \text { molecules of } \mathrm{N}_{2} / 1 \mathrm{~mol} \mathrm{~N}_{2}\right)=1.92 \times 10^{22}\) number of molecules

Thus \(1.92 \times 10^{22}\) number of molecules of Nitrogen is present in 1.00 L of sample.