If `N_(2)` gas is bubbled through water at 293 K, how many millimoles of `N_(2)` gas would dissolve in 1 litre of water ? Assume that `N_(2)` exerts a partial pressure of 0.987 bar. Given that Henry's law constant for `N_(2)` at 293 K is 76.48 k bar.

Explanation

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. Thus :
x (Nitrogen) `= ("p (nitrogen)")/(K_(H))`
`=(0.987" bar")/(76480" bar")=1.29xx10^(-5)`
As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of `N_(2)` in solution,
x(Nitrogen) `= ("n mol")/(n mol +55.5 mol)`
`= (n)/(55.5)=1.29xx10^(-5)`
(n in denominator is neglected as it is `lt lt 55.5`)
Thus, `n = 1.29xx10^(-5)xx55.5 mol = 7.16xx10^(-4)` mol
`= (7.16xx10^(-4)mol xx 1000 m mol)/(1 mol)`
= 0.716 m mol