Correct Option: (B) greater for gases with lower solubility
Step 1 (Formula / Governing Law): [0.5 Mark]
According to Henry's Law, the partial pressure of the gas in vapour phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution. The mathematical expression is:
$$p = K_H \cdot x$$Where $K_H$ is the Henry's law constant.
Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
Rearranging the formula to express the mole fraction ($x$), which represents the solubility of the gas:
$$x = \frac{p}{K_H}$$From this relationship, it is evident that for a given partial pressure ($p$), the solubility ($x$) is inversely proportional to the Henry's constant ($K_H$):
$$x \propto \frac{1}{K_H}$$Step 3 (Final Concluded Answer): [0.25 Mark]
Since $x$ and $K_H$ are inversely proportional, a higher value of $K_H$ results in a lower value of $x$ (solubility). Therefore, gases with lower solubility possess a higher value of $K_H$ at a given temperature.
💡 EL10 Concept Insight:
Think of $K_H$ as a "resistance" factor; the higher the $K_H$, the more the gas "resists" dissolving into the liquid, which is why gases that don't like to dissolve (low solubility) have a high $K_H$ value.