Question

Air is approximately $79%$ nitrogen and $21%$ oxygen by volume. How many moles of oxygen are present in $22\text{ L}$of ${{24}^{\circ }}C$ and a pressure of $53kPa$?

Answer 1

Hint: It is based on the partial pressure concept of Dalton and first you have to calculate the total number of moles by using the formula as: ${{P}_{t}}V={{n}_{t}}RT$ and after finding the total number of moles, you can easily calculate the individual moles of nitrogen and oxygen by using the given percentage and the total number of moles. Now solve it.

Complete answer:
This numerical is based on the concept of Dalton's law of partial pressures. This law states that the sum of the individual pressures is always equal to the total pressure of the system i.e.
${{P}_{t}}={{p}_{1}}+{{p}_{2}}+{{p}_{3}}----{{p}_{n}}$
In terms of the ideal gas equal ,Dalton’s law of partial pressures can be expressed as;
$\begin{align}
  & {{P}_{t}}V={{n}_{t}}RT \\
 & {{P}_{t}}={{n}_{t}}\dfrac{RT}{V}--------(1) \\
\end{align}$
Here, p represents the total pressure of the system, V represents the volume of the system, T is the temperature of the system and ${{n}_{t}}$ represents the total number of moles of the gases.
Now considering the statement as;
First, we will calculate the total number of moles of the system and using the formula of equation (1)
As we know that-
Volume of the system=$22\text{ L}$(given)
Temperature of the system=${{24}^{\circ }}C=24\times 273K=297K$ (given)
Total pressure of the system=$53kPa=0.52atm\text{ (1Pa=9}\text{.86}\times \text{1}{{\text{0}}^{-6}}atm)$ (given)
Gas constant=$0.08206\text{ L atm mol}{{\text{e}}^{-1}}{{K}^{-1}}$
Now put all these values in equation (1), we get;
 $\begin{align}
  & \Rightarrow {{P}_{t}}={{n}_{t}}\dfrac{RT}{V} \\
 & \Rightarrow {{n}_{t}}=\dfrac{{{P}_{t}}V}{RT} \\
 & \Rightarrow \text{ =}\dfrac{0.52\times 22}{0.08206\times 297}mole \\
 & \Rightarrow \text{ =0}\text{.47 mole} \\
\end{align}$
Now we can easily calculate the number moles of each atom by using the formula as;
$moles\text{ }of\text{ }the\text{ }atom=total\text{ }moles\text{ }\times given\text{ }percentage----(2)$
As we know that,
The percentage of nitrogen in air is=$79%=0.79$ (given)
And the percentage of oxygen in air is=$21%=0.21$ (given)
Then, by using the formula of equation (2), we get;
$moles\text{ }of\text{ }{{N}_{2}}=0.79\text{ }\times 0.47=0.37\text{ }moles$
And $moles\text{ }of\text{ }{{\text{O}}_{2}}=0.21\text{ }\times 0.47=0.099moles$
So, thus when air is approximately $79%$ nitrogen and $21%$ oxygen by volume, then the number of moles of oxygen present in $22\text{ L}$of ${{24}^{\circ }}C$ and a pressure of $53kPa$ is $0.099\text{ }moles$.

Note:
Always remember that while using the values in the numerical first convert them to their standard units and then use them accordingly in the numerical to get the correct answer.