Calculate the pressure of gas if the solubility of gas in water at $25^{\circ} \mathrm{C}$ is $6.85 \times…
Calculate the pressure of gas if the solubility of gas in water at $25^{\circ} \mathrm{C}$ is $6.85 \times 10^4 \mathrm{~mol} \mathrm{dm}^{-3}$ (Henry's law constant is $6.85 \times 10^4 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{bar}^{-1}$ )
0.853 bar
$1.5 \mathrm{bar}$
0.5 bar
1 bar
Solution
We know
$\mathrm{S}=\mathrm{K}_{\mathrm{H}} \mathrm{P}$
In $1 \mathrm{dm}^3$ or 1 lit moles of gas dissolved is $6.85 \times 10^4$ $\therefore \mathrm{p}=6.85 \times 10^4 / 6.85 \times 10^4=1.0$ bar.