The Henry’s law constant (KH) 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 :
- B > A > C
- B > C > A
- A > C > B
- A > B > C
The Correct Option is B
Solution and Explanation
Step 1: Recall Henry’s Law
Henry’s law states that:
$$ C = \frac{P}{K_H} $$
\( C \) = Concentration (solubility) of the gas
\( P \) = Partial pressure of the gas
\( K_H \) = Henry’s law constant
Step 2: Analyze the Relationship Between \( K_H \) and Solubility
A lower \( K_H \) value corresponds to higher solubility, and a higher \( K_H \) value corresponds to lower solubility.
Step 3: Compare the \( K_H \) Values
For Gas A: \( K_H = 145 \) kbar (highest \( K_H \), least soluble).
For Gas B: \( K_H = 2 \times 10^{-5} \) kbar (lowest \( K_H \), most soluble).
For Gas C: \( K_H = 35 \) kbar (moderate \( K_H \)).
Step 4: Arrange in Decreasing Solubility
Since solubility is inversely proportional to \( K_H \), the order of solubility is:
$$ B > C > A $$
Conclusion
The correct order of solubility is: B > C > A.
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