If $\text{N}_2$ gas is bubbled through water at $293\text{ K}$, how many millimoles of $\text{N}_2$ gas would dissolve in $1\text{ litre}$ of water? Assume that $\text{N}_2$ exerts a partial pressure of $0.987\text{ bar}$. Given that Henry's law constant for $\text{N}_2$ at $293\text{ K}$ is $76.48\text{ k bar}$.

$0.0129\text{ m mol/L}$

The correct answer is Option (3) → $0.0129\text{ m mol/L}$ ##

The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry's law.

$x = \frac{\text{P (nitrogen)}}{\text{K}_{\text{H}}}$

$= \frac{0.987\text{ bar}}{76,480\text{ bar}} = 1.29 \times 10^{-5}$

Since, $1\text{ mol} = 1000\text{ m mol}$,

So, $x = 1.29 \times 10^{-5} \times 1000$

$= 0.0129\text{ m mol/L of water}$