The \(K_H\) values of \(HCHO\), \(CH_4\), \(Ar\) and \(CO_2\) at 298 K are \(1.83 × 10^{-5}\) kbar, \(0.413\) kbar, \(40\) kbar and \(1.67\) kbar respectively. Arrange them in the order of their decreasing solubility at constant pressure:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(HCHO > CH_4 > CO_2 > Ar\)
By Henry’s law, p = KH x, so the mole fraction of a gas dissolved at a given pressure is x = p/KH. The higher the value of KH, the lower is the solubility of the gas. Decreasing solubility therefore means increasing KH:
HCHO (1.83 × 10−5 kbar) < \( \ce{CH4} \) (0.413 kbar) < \( \ce{CO2} \) (1.67 kbar) < Ar (40 kbar)
NCERT (2026-27 edition):
Solubility of a Gas in a Liquid – …
