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") = 1.29 xx10^(-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) = 1.29 xx10^(-5)`
(n in denominator is neglected as it is ` lt lt` 55.5)
Thus n ` = 1.29 xx10^(-5) xx 55.5 mol = 7.16 xx10^(-4) ` mol
` = (7.16 xx10^(-4) mol xx1000 mol )/(1 mol ) = 0.716 ` mmol