Question

If $N_2$ gas is bubbled through water at $293\text{K},$ how many millimoles of $N_2$ would dissolved in $300\text{mole of water.}$ If $N_2$ exerts a partial pressure of $1\text{bar}$. Given that Henry's law constant for $N_{2}293\text{K}$ is $75.00\text{k bar}$.

Answer 1

From Henry's law:
$P_{N_2}=K_H.{\chi }_{N_2}$
Where,
$P$ denotes the partial pressure of the gas in the atmosphere above the liquid
$\chi$ denotes the mole fraction of the dissolved gas
$K_H$ is the Henry’s law constant of the gas
${\chi }_{N_2}=\frac{1}{75000}$
If $n$ represents number of moles of $N_2$ in solution
Total moles $=n+300$
$\therefore {\chi }_{N_2}=\frac{n}{n+300}=\frac{1\text{bar}}{75000\text{bar}}$
$n$ in denominator is neglected as it is $<<300$)
Thus $n=4\times {10}^{-3}\text{mol}=4\text{mmol}$