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")/(76.480" 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)55.5" mol"=7.16xx10^(-4)" mol"`
`=(7.16xx10^(-4)" mol"xx1000" mmol")/(1" mol")=0.716" m mol"`