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

Explanation

The solubility of gas is related to the molar fraction in aqueous solution.
The mole fraction of the gas in the calculated by applying Henry's law. Thus:
`xx ("Nitrogen") =(p("nitrogen"))/(K_(H)) = (0.98"bar")/(76,480 "bar") = 1.29 xx 10^(-5)`
As 1 litre of water contains `55.5 mol` of it, therefore if n represents number of moles of `N_(2)` in solution,
`xx("Nitrogen") =(n "mol")/("n mol"+55.5 mol) = (n)/(55.5) = 1.29 xx 10^(-5)`
(n in denominator is neglected as it is `lt lt 55.5`
Thus `n = 1.29 xx 10^(-5) xx 55.5 mol = 7.16 xx 10^(-4) mol =(7.16 xx 10^(-4) mol xx 1000 mol)/(1 mol) = 0.716 mol`