The values of Henry’s law constant for the given gases in water, at 293 K, are
I. He - 144.97 kbar II. H2 - 69.16 kbar
III. N2 - 76.48 kbar IV. O2 - 34.86 kbar
Which gas has the lowest solubility in water?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → He
By Henry’s law, p = KHx, so at a given partial pressure the mole fraction of gas dissolved is x = p/KH. The higher the value of KH, the lower the solubility of the gas.
Comparing the given values (kbar, 293 K):
He (144.97) > \( \ce{N2} \) (76.48) > \( \ce{H2} \) (69.16) > \( \ce{O2} \) (34.86)
Helium has the largest KH, so it has the lowest solubility in water; \( \ce{O2} \), with the smallest KH, is the most soluble of the four.
Shortcut: Lowest solubility means largest KH; just pick the biggest number, 144.97 for He.
NCERT (2026-27 edition):
Solubility of a Gas in a Liquid – …
