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

`"According to Henry's law, "p_(N_(2))=K_(H)xxx_(N_(2)), thereforex_(N_(2))=(p_(N_(2)))/(K_(H))=("0.987 bar")/("76480 bar")=1.29xx10^(-5)`
If n moles of `N_(2)` are prsent in 1 L of water (i.e., 55.5 moles), `x_(N_(2))=(n)/(n+55.5)~=(n)/(55.5)("as "ltlt 55.5)`
`therefore" "(n)/(55.5)=1.29xx10^(-5)" or "n=1.29xx10^(-5) xx 55.5" moles"=71.595xx10^(-5)" moles = 0.716 millimoles"`