Question:
Calculate the pressure at which solubility of a certain gas in water is 0.165 mol L\({}^{-1}\) at \(25\ ^{\circ }\)C. (Henry's law constant for this gas in water is 0.034 mol L\({}^{-1}\)bar\({}^{-1}\) at \(25\ ^{\circ }\)C.)
Calculate the pressure at which solubility of a certain gas in water is 0.165 mol L\({}^{-1}\) at \(25\ ^{\circ }\)C. (Henry's law constant for this gas in water is 0.034 mol L\({}^{-1}\)bar\({}^{-1}\) at \(25\ ^{\circ }\)C.)
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Henry's law says solubility is proportional to pressure, so pressure equals solubility divided by the constant.
Updated On: Nov 12, 2026
- \(2.85\) bar
- \(4.85\) bar
- \(6.85\) bar
- \(1.85\) bar
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Henry's law in solubility form: \(S=K_H\,P\).
Step 2: Rearrange
\(P=\dfrac{S}{K_H}\).
Step 3: Calculation
\(P=\dfrac{0.165}{0.034}=4.85\text{ bar}\).
Final Answer:
The required pressure is about 4.85 bar. \[ \boxed{\text{(B) }4.85\text{ bar}} \]
Henry's law in solubility form: \(S=K_H\,P\).
Step 2: Rearrange
\(P=\dfrac{S}{K_H}\).
Step 3: Calculation
\(P=\dfrac{0.165}{0.034}=4.85\text{ bar}\).
Final Answer:
The required pressure is about 4.85 bar. \[ \boxed{\text{(B) }4.85\text{ bar}} \]
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