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 its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry's law. Thus,
`X_(N_(2))= P_(N_(2))/K_(H)=(0.987"bar")/(76480"bar")=1.29xx10^(-5)`
AS 1 litre water contains 55.5 mol of it, therefore, if n represents number of moles of `N_(2)` in solution,
`X_(N_(2))=("n "" mol")/("n ""mol"+55.5" mol")~~n/55.5=1.29xx10^(-5)`
Thus , n`=1.29xx10^(-5)xx55.5"mol"=7.16xx10^(-4)"mol"`
`=7.16xx10^(-4)"mol"xx(100m" mol")/(1" mol")=0.716m " mol"`