NEET 2024 Chemistry Question 77 Solution
Question
The Henry's law constant (K_H) values of three gases (A, B, C) in water are 145, 2×10⁻⁵ and 35 kbar, respectively. The solubility of these gases in water follow the order:
Dynamic 180-question full-length simulator mixed dynamically from all 6 years for 2027 & 2026 aspirants.
The Henry's law constant (K_H) values of three gases (A, B, C) in water are 145, 2×10⁻⁵ and 35 kbar, respectively. The solubility of these gases in water follow the order:
Solubility x ∝ (1 / K_H)! Lowest K_H means HIGHEST solubility! K_H order: B (2e-5) < C (35) < A (145) -> Solubility: B > C > A!
Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution:
From the rearranged equation, at a given constant pressure p, the mole fraction x (which measures gas solubility) is inversely proportional to the Henry’s law constant KH: Higher KH -> Lower solubility Lower KH -> Higher solubility
Listing the given KH values in increasing numerical order: - Gas B: KH = 2 × 10⁻⁵ kbar (exceptionally small KH -> maximum solubility) - Gas C: KH = 35 kbar (intermediate KH -> moderate solubility) - Gas A: KH = 145 kbar (largest KH -> minimum solubility)
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Assumes direct proportionality between solubility and KH (order of KH values, not solubility).
Completely incorrect ordering of KH values.
Places A above C in solubility even though KH(A) > KH(C).
Exact inverse ordering of Henry’s law constants: B (lowest KH) > C > A (highest KH).
Students frequently treat KH like an equilibrium solubility constant (Ksp) where bigger means more soluble. For Henry’s constant, BIGGER KH MEANS LESS SOLUBLE!
"Solubility of a gas is inversely proportional to its Henry’s law constant KH at constant pressure: x = p / KH."
Test your speed and accuracy in the authentic 200-minute timed NTA simulator.
Spotted a mistake in this solution? Report an error ([email protected])
Cover all 1,140 official NEET questions in 285 days at your own pace. Progress is saved on your device. No login needed.