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

By appliying Henry's law (Equation, 2.42), we can calculate the mole fraction of the `N_(2)(g)` in the aqueous solution:
`P_(N_(2))=K_(H) chi_(N_(2))`
or `chi_(N_(2))=P_(N_(2))/K_(H)=(0.987 "bar")/(76.48xx10^(3) "bar")
=1.29xx10^(-5)`
By definition
`chi_(N_(L))=n_(N_(2))/(n_(N_(2))+n_(H_(2)O))=1.25xx10^(-5)`
We can find the number of moles of `N_(2)` in solution provided we know the number of moles of `H_(2)O`.
Since the solubility of `N_(2)` in `H_(2)O` is very small, we can take the density of aq. solution to be `1 g mL^(-1)`. Thus, `1 L (1000 mL) of H_(2)O weighs 1000 g`. Now
`n_(H_(2)O)=("mass"_(H_(2)O))/("molar mass"_(H_(2)O))`
`=(1000 g)/(18 g mol^(-1))`
`=55.5 mol`
Substituting this result into mole fraction data, we have
`n_(N_(2))/(n_(N_(2))+55.5)~=n_(N_(2))/55.5=1.29xx10^(-5)`
Note that `n_(N_(2))` in the denominator is neglected as it is `lt lt 55.5`.
Thus
`n_(N_(2))=(55.5 mol)(1.29xx10^(-5))`
`=71.6xx10^(-5) mol`
`=7.16xx10^(-4) mol`
`mmol_(N_(2))=(1000 (mmol)/(mol))(n_(N_(2)))`
`=(1000 mmol//mol)(7.16xx10^(-4) mol)`
`=0.716 mmol`
Thus, there are `0.716 mmol` of dissolved `N_(2)` in 1 litre of `H_(2)O`.