Henry’s constant of Argon gas at 298K temperature is 40kbar. Calculate the weight of Argon dissolved in 1000 grams of water when packed at a pressure of 7.2 bar and 298 K .

Henry’s law
${\mathrm{P}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{H}}\times {\mathrm{X}}_{\mathrm{A}}$

${\mathrm{X}}_{\mathrm{A}}=\mathrm{\hspace{0.17em}}\frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{solvent}}}$

${\mathrm{n}}_{\mathrm{A}}\mathrm{\hspace{0.17em}}<<<\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{n}}_{\mathrm{solvent}}$

$\therefore \mathrm{\hspace{0.17em}}{\mathrm{X}}_{\mathrm{A}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{solvent}}}$

${\mathrm{P}}_{\mathrm{A}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}{\mathrm{K}}_{\mathrm{H}}\mathrm{\hspace{0.17em}}\times \hspace{0.17em}\hspace{0.17em}\frac{{\mathrm{\eta }}_{\mathrm{A}}}{{\mathrm{\eta }}_{\mathrm{solvent}}}$
$7.2 \mathrm{bar}=40\times {10}^3\times \frac{{\mathrm{n}}_{\mathrm{Argon}}}{{\displaystyle \frac{1000}{18}}}$
${\mathrm{n}}_{\mathrm{Argon}}=\frac{7.2}{720}=0.01$
$\frac{\mathrm{w}}{40}=0.01$
Weight of Argon gas dissolved in 1 kg of water = 0.4g