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# Air contains $O_2~$ and $N_2$ ​ in the ratio $1:4~$ calculate the ratio of their solubility in terms of mole fraction at atmospheric pressure and room temperature will be (given Henry’s constant for ${{O}_{2}}=3.3~\times {{10}^{7}}~torr$ for ${{N}_{2}}=6.6~\times {{10}^{7}}~torr$)

Last updated date: 04th Aug 2024
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Hint:The ratio of Henry's law constants of two different gases is equal to the ratio of their solubility.
-Henry’s law develops a relationship between the concentration of the species in aqueous phase and the partial pressure of the gas under consideration.

Henry’s law could be expressed by a simple equation, which is given below,
$H=\dfrac{c}{p}$
Here is $c$ the concentration of a species in the aqueous solution, and $p$ is the partial pressure of that species in the gas phase under equilibrium conditions. And $H$ is the henry’s law constant of proportionality.
We need to calculate the ratio of their solubility in terms of their mole fraction. The values of the henry’s law constants of oxygen and nitrogen are given as
${{O}_{2}}=3.3~\times {{10}^{7}}~torr$ and ${{N}_{2}}=6.6~\times {{10}^{7}}~torr$
We know that the pressure of both the gases are equal as well as the concentration, so in order to find the ratio of their mole fraction,
$\dfrac{{{\chi }_{{{O}_{2}}}}}{{{\chi }_{{{N}_{2}}}}}=\dfrac{{{K}_{H}}({{O}_{2}})\times {p}{{{O}}_{{2}}}}{{{K}_{H}}({{N}_{2}})\times p{{N}_{2}}}$
Where ${{\chi }_{{{O}_{2}}}}$ and ${{\chi }_{{{N}_{2}}}}$ represents the mole fraction of oxygen and nitrogen, ${{K}_{H}}({{O}_{2}})$ represents the henry’s constant of oxygen and the ${{K}_{H}}({{N}_{2}})$ represents the henry’s law constant of nitrogen. And the partial pressures of oxygen and nitrogen are represented by $p{{O}_{2}}$ and $p{{N}_{2}}$ respectively.
The ratio of partial pressure is also given to us as $1:4$.
Now we will substitute all these values in the above equation,
$\dfrac{{{\chi }_{{{O}_{2}}}}}{{{\chi }_{{{N}_{2}}}}}=\dfrac{3.3\times {{10}^{7}}}{6.6\times {{10}^{7}}}\times \dfrac{1}{4}$
$\dfrac{{{\chi }_{{{O}_{2}}}}}{{{\chi }_{{{N}_{2}}}}}=\dfrac{1}{8}$
So the ratio would be $1:8$.