The solubility of ${\mathrm{N}}_2$ (g) in water exposed to the atmosphere is $5.3\times {10}^{-4}$ M when the partial pressure of Nitrogen is 593 mm. At the same temperature its solubility at 760 mm will be

$\mathrm{s}=\frac{5.3\times {10}^{-4}}{593}\times 760=6.8\times {10}^{-4}\mathrm{M}$

                                                           $\left[\mathrm{OR}\right]$

From Henry's law,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{gas}}={\mathrm{K}}_{\mathrm{H}} {\mathrm{X}}_{\mathrm{gas}} \\ \mathrm{Solubility} \mathrm{\alpha } \mathrm{Pressure} \\ \mathrm{So},\frac{{\mathrm{S}}_1}{{\mathrm{S}}_2} =\frac{ {\mathrm{P}}_1}{{\mathrm{P}}_2} \\ \frac{5.3\times {10}^{-4}}{{\mathrm{S}}_2} = \frac{593}{760} \end{gathered}$$

$$\begin{gathered} {\mathrm{S}}_2=\frac{760}{593}\times 5.3\times {10}^{-4} \\ {\mathrm{S}}_2= 6.8\times {10}^{-4}\mathrm{M} \end{gathered}$$