If the solubility of the given gases in water decreases in the order of formaldehyde, methane, vinyl chloride, argon.
Match List-I with List-II.
|
List-I Gas |
List-II $K_H/k$ bar |
|
(A) Formaldehyde |
(I) 40.3 |
|
(B) Vinyl Chloride |
(II) 0.413 |
|
(C) Argon |
(III) $1.83 × 10^{-5}$ |
|
(D) Methane |
(IV) 0.611 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A)-(III), (B)-(IV), (C)-(I), (D)-(II)
|
List-I Gas |
List-II $K_H/k$ bar |
|
(A) Formaldehyde |
(III) $1.83 × 10^{-5}$ |
|
(B) Vinyl Chloride |
(IV) 0.611 |
|
(C) Argon |
(I) 40.3 |
|
(D) Methane |
(II) 0.413 |
Lower Solubility Higher $K_{H}$ (Henry's Law Constant)
The problem states: Solubility decreases in the order: Formaldehyde > Methane > Vinyl chloride > Argon.
Therefore, the $K_{H}$ values should increase in the same order.
Order of increasing $K_{H}$ values: Formaldehyde < Methane < Vinyl chloride < Argon
Given $K_{H}$ values (in increasing order):
1. $1.83\times10^{-5}$ (Smallest $K_{H}$, so Highest Solubility)
2. 0.413
3. 0.611
4. 40.3 (Largest $K_H$, so Lowest Solubility)