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.

Answer

`0.716` mmol.

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. Thud:
`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 number of moles of `N_(2)` in solution.
`x("Nitrogen")=(n "mol")/(n "mol"+55.5"mol")=(n)/(55.5)=1.29xx10^(-5)`
(`n` in denominator is neglected as it is `ltlt55.5`)
Thus `n=1.29xx10^(-5)xx55.5` mol `=7.16xx10^(-4)` mol.
`=(7.16xx10^(-4)"mol"xx1000"mol")/(1"mol")=0.716` mmol.