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 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 `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)xx55.5" mol "=7.16xx10^(-4)`mol
`=(7.16xx10^(-4)" mol "xx1000"m "mol)/(1" mol")=0.716mmol`