Question

The K_H value (K bar) of Argon (I), Carbondioxide (II) formuldehyde (III) and methane (IV) are respectively 40.3, 167, 1.83×10^-5 and 0.413 at 298 K. The increasing order of solubility of gas in liquid is

• A. I < II < IV < III
• B. III < IV < II < I
• C. I < III < II < IV
• D. I < IV < II < III

Answer 1

The Correct Option is A

Given the Henry's law constant (\( K_H \)) values (in bar) for the gases:
Argon (I): 40.3
Carbon dioxide (II): 1.67
Formaldehyde (III): \( 1.83 \times 10^{-5} \)
Methane (IV): 0.413

According to Henry's law, the solubility of a gas in a liquid is inversely proportional to its \( K_H \) value. Hence, the solubility increases as \( K_H \) decreases.

Argon (I) has the highest \( K_H \) value, so it has the lowest solubility.
Carbon dioxide (II) has a moderate \( K_H \) value, so it has a moderate solubility.
Formaldehyde (III) has the lowest \( K_H \) value, so it has the highest solubility.
Methane (IV) has a \( K_H \) value higher than formaldehyde but lower than argon, so its solubility lies between carbon dioxide and argon.

Therefore, the increasing order of solubility is:
\( \text{I} < \text{IV} < \text{II} < \text{III} \)

The correct answer is (A) : I < II < IV < III.