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 aqueous solution.
The mole fraction of the gas in solution .
`x_(N_(2)) = (P_(N_(2)))/(K_(H)) = (0.987 "bar")/(76480 "bar") = 129 xx 10^(-5)`
If n is the number of moles of `N_(2)` in solution and 1 liter of water contains 55.5 mol , then .
`x_(N_(2)) = (n)/(n+55.5)~~ (n)(55.5) = 1.29 xx 10^(-5)`
(n in the denominator is neglected because it is ` lt lt 55.5`)
`therefore " " n = 55.5 xx 1.29 xx 10^(-5)`
`=7.16 xx 10^(-4) mol`
or `" " = 7.16 xx 10^(-1) mol`
` = 0.716 mol` .