If `N_(2)` gas is bubbled through water at 293 K, how many of millimoles of `N_(2)` gas would dissolve in 1 litre of water 7 Assume that `N_(1)` exerls 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 its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry's law. Thus,
`chiN_(2) = (pN_(2))/K_(H) =( 0.987 "bar")/(76480 "bar") = 1.29 xx 10^(-5) `
As `1 L` water contains `55.5 mol` of it, therefore, if `n` represents number of moles of `N_(2)` in solution, then
`chiN_(2)=(n mol)/(n mol+55.5 mol )~~ n/55.5 = 1.29 xx 10^(-5)`
Thus , `n=1.29 xx 10^(-5) xx 55.5 mol`
`=7.16 xx 10^(-4) mol`
`=7.16 xx 10^(-4) mol xx( 1000 mmol)/(1 mol)`
`=0.716 mmol`