Question

Observe the following data given in the table. (\(K_H\) = Henry's law constant) 

Gas	CO₂	Ar	HCHO	CH₄
\(K_H\) (k bar at 298 K)	1.67	40.3	\(1.83 \times 10^{-5}\)	0.413

The correct order of their solubility in water is

• A. CO₂>CH₄>HCHO>Ar
• B. Ar>HCHO>CH₄>CO₂
• C. HCHO>CH₄>CO₂>Ar
• D. CO₂>HCHO>CH₄>Ar

Hint:

Remember the inverse relationship: \textbf{High \(K_H\)} = \textbf{Low Solubility}. A simple way to think about it is that a high \(K_H\) means a high pressure is needed to dissolve even a small amount of gas. Formaldehyde (HCHO) is extremely soluble in water (forming formalin), so it must have a very low \(K_H\), which is consistent with the data.

Answer 1

The Correct Option is C

Step 1: Understanding Henry's Law: Henry's Law states that the partial pressure of a gas in the vapour phase (\( p \)) is proportional to the mole fraction of the gas (\( \chi \)) in the solution. \[ p = K_H \cdot \chi \] Here, \( K_H \) is the Henry's law constant.
Step 2: Relationship between \( K_H \) and Solubility: From the formula, mole fraction (solubility) \( \chi = \frac{p}{K_H} \). For a given pressure, the solubility (\( \chi \)) is inversely proportional to \( K_H \). \[ \text{Solubility} \propto \frac{1}{K_H} \] Therefore, a lower value of \( K_H \) indicates higher solubility.
Step 3: Comparing \( K_H \) values: Let's list the \( K_H \) values (in k bar): - HCHO: \( 1.83 \times 10^{-5} \) (Lowest) - \( \text{CH}_4 \): 0.413 - \( \text{CO}_2 \): 1.67 - Ar: 40.3 (Highest)
Step 4: Arranging in Order of Solubility: Order of \( K_H \): HCHO<\( \text{CH}_4 \)<\( \text{CO}_2 \)<Ar Order of Solubility (Reverse of \( K_H \) order): HCHO>\( \text{CH}_4 \)>\( \text{CO}_2 \)>Ar