According to Henry's law, the solubility of a gas in a liquid under constant temperature is directly proportional to the partial pressure of the gas above the liquid but is inversely proportional to the Henry's law constant $$ (\mathrm{K}_\mathrm{H}) $$ for the gas. Henry's law can be expressed as:

$$\mathrm{c} = \frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}$$

where $ \mathrm{c} $ is the concentration (or solubility) of the gas in the liquid, $ \mathrm{P} $ is the partial pressure of the gas, and $$ \mathrm{K}_{\mathrm{H}} $$ is Henry's law constant.

From the equation, it is clear that the solubility of the gas is inversely related to $$ \mathrm{K}_{\mathrm{H}} $$; if $$ \mathrm{K}_{\mathrm{H}} $$ increases, the solubility decreases, and vice versa.

In the given problem, you have the $$ \mathrm{K}_{\mathrm{H}} $$ values of the gases A, B, and C as follows:

• A: 145 kbar
• B: $$ 2 \times 10^{-5} \mathrm{~kbar} $$
• C: 35 kbar

Comparing the $$ \mathrm{K}_{\mathrm{H}} $$ values:

• Gas B has the lowest $$ \mathrm{K}_{\mathrm{H}} $$ and hence the highest solubility.
• Gas A, with the highest $$ \mathrm{K}_{\mathrm{H}} $$ among the three, will have the lowest solubility.
• Gas C has a $$ \mathrm{K}_{\mathrm{H}} $$ value less than A but greater than B, so its solubility will be lower than B but higher than A.

Therefore, the order of solubility of the gases in water from the highest to the lowest is:

B > C > A

Thus, the correct option is:

Option B: B > C > A