Use Henry’s law

For gases dissolved in water, $p = K_H x$ where $p$ is the partial pressure of the gas, $K_H$ is Henry’s constant, and $x$ is mole fraction of dissolved gas.

Hence, at a given pressure, $x = \frac{p}{K_H}$ So, smaller $K_H$ means higher solubility.

Given values

$K_H(\alpha)=50\ \text{kbar}$ $K_H(\beta)=2\ \text{kbar}$ $K_H(\gamma)=2\times 10^{-5}\ \text{kbar}$ $K_H(\delta)=0.5\ \text{kbar}$

Check option C

Since solubility is inversely proportional to $K_H$, the gas with the smallest Henry’s constant has the highest solubility.

Among the given values, $2\times 10^{-5} < 0.5 < 2 < 50$ so $\gamma$ has the highest solubility, not $\alpha$.

Therefore, Option C is false.

Check option A

Dissolution of gases in water generally decreases with increase in temperature, so Henry’s constant increases with temperature and solubility decreases.

Therefore, for $\gamma$, solubility at $308\,\text{K}$ is lower than at $298\,\text{K}$.

So, Option A is true.

Check options B and D using 55.5 molal solution

A $55.5$ molal solution means:

• $55.5$ mol gas per $1\,\text{kg}$ water.

Number of moles of water in $1\,\text{kg}$: $n_{\text{water}}=\frac{1000}{18}=55.5\overline{5}\approx 55.5$

Thus, moles of solute gas $\approx 55.5$ and moles of water $\approx 55.5$.

So mole fraction of gas is approximately $x = \frac{55.5}{55.5+55.5}=\frac12$

More accurately, with $55.5$ mol water approximation used in such problems, $x \approx 0.5$

Option B: gas $\delta$

$p = K_H x = (0.5\,\text{kbar})(0.5)=0.25\,\text{kbar}$

Since $1\,\text{kbar}=1000\,\text{bar}$ we get $0.25\,\text{kbar}=250\,\text{bar}$

Therefore, Option B is true.

Option D: gas $\gamma$

$p = K_H x = (2\times 10^{-5}\,\text{kbar})(0.5)=1\times 10^{-5}\,\text{kbar}$

Convert to bar: $1\times 10^{-5}\,\text{kbar}=10^{-5}\times 1000\,\text{bar}=10^{-2}\,\text{bar}=0.01\,\text{bar}$

This is not $1\,\text{bar}$.

Therefore, Option D is false.

Final result

Correct options are: $\boxed{A,\ B}$

Comparison with stored answer

Stored correct answer: $A, B$

This matches exactly.