N₂ gas is bubbled through water at 293 K and the partial pressure of N₂ is 0.987 bar. If the  Henry’s law constant for N₂ at 293 K is 76.48 k.bar, the number of milli moles of N₂ gas  that will dissolve in 1 L of water at 293 K is

${\mathrm{P}}_{{\mathrm{N}}_2}={\mathrm{K}}_{\mathrm{H}}\times {\mathrm{X}}_{{\mathrm{O}}_2}$
According to Henry’s law, 
$\begin{array}{l}{\mathrm{P}}_{{\mathrm{N}}_2}={\mathrm{K}}_{\mathrm{H}}\cdot {\mathrm{X}}_{{\mathrm{N}}_2}\\ \Rightarrow {\mathrm{X}}_{{\mathrm{N}}_2}=\frac{0.987}{76800}=1.29\times {10}^{-5}\end{array}$
Moles of water in$1\mathrm{L}=\frac{1000}{18}=55.56$
Let us take the moles of nitrogen as in Total moles = n + 55.56
$\begin{array}{l}\therefore {\mathrm{X}}_{{\mathrm{N}}_2}=\frac{\mathrm{n}}{\mathrm{n}+55.56}=1.29\times {10}^{-5}\\ \Rightarrow \mathrm{n}\left(1-1.29\times {10}^{-5}\right)-55.56\times 1.29\times {10}^{-5}=0\\ \Rightarrow \mathrm{n}=0.7\times {10}^{-3}=0.7\text{ mi lim oles. }\end{array}$