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 follows the order:

Chemistry · Solutions and Colligative Properties

NEET (UG) 2024 — Question 65

The Henry's law constant $K_H$ values of three gases $(A, B, C)$ in water are $145$ , $2 \times 10^{-5}$ and $35\, \text{kbar}$ , respectively. The solubility of these gases in water follows the order:

Option A:
B $>$ A $>$ C
Option B:
B $>$ C $>$ A
Correct
Option C:
A $>$ C $>$ B
Option D:
A $>$ B $>$ C

Answer: B

Step-by-step solution

Henry's law states that at a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The Henry's law constant $K_H$ is inversely proportional to solubility: higher $K_H$ means lower solubility. Given $K_H$ values: $A = 145\,\text{kbar}$ , $B = 2\times10^{-5}\,\text{kbar}$ , $C = 35\,\text{kbar}$ . Order of $K_H$ : $A > C > B$ . Therefore, order of solubility (inverse of $K_H$ ): $B > C > A$ . Thus, the correct order is $B > C > A$ , which corresponds to option B.

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Exam: NEET (UG) 2024
Paper: 2024 paper
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Gas in Liquid Solutions (Henry's Law)

The plot of osmotic pressure ( Pi ) vs concentration ( mol L^-1 ) for a solution gives a straight line with slope 25.73 L bar mol^-1 . The…

Answer: B

Step-by-step solution

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