When `N_2` gas is passed through water at 293 K, how many moles of ` 1N_2` would dissolve in one litre water ? Assume that `N_2` exert a partial pressure of 0.987 bar. `K_H` for `N_2` at 293 K is 76.48 kilobar.

Explanation

By Henry.s law, mole fraction of `N_2 = P_(N_2)/K_H` = (0.987 bar)/(76480 bar) =`1.29xx10^(-5)`. One liter water contain 55.5 mols of it. Let .n. be the number of mols of nitrogen in solution , then,
`X_(N_2)= n/(n+55.5)= 1.29xx10^(-5)`
`therefore=1.29xx10^(-5)xx55.5=7.16xx10^(-4)`mol