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Question

The Henry's law constant KHK_H values of three gases (A, B, C) in water are 145, 2×1052 \times 10^{-5}, and 35 kbar, respectively. Determine the order of solubility of these gases in water from highest to lowest:

A

B > C > A

B

A > C > B

C

A > B > C

D

B > A > C

Step-by-Step Solution

According to Henry's law, the partial pressure of a gas in the vapour phase (pp) is directly proportional to its mole fraction (xx) in the solution, expressed as p=KHxp = K_H x. This implies that at a constant pressure, the higher the value of Henry's law constant (KHK_H), the lower the solubility of the gas in the liquid .

Given the KHK_H values: KH(A)=145 kbarK_H(\text{A}) = 145 \text{ kbar} KH(B)=2×105 kbarK_H(\text{B}) = 2 \times 10^{-5} \text{ kbar} KH(C)=35 kbarK_H(\text{C}) = 35 \text{ kbar}

The decreasing order of KHK_H values is A > C > B. Since solubility is inversely proportional to KHK_H, the decreasing order of solubility will be exactly the reverse: B > C > A.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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