Question

The value of Henry's law constant for argon $(Ar)$, carbon dioxide $\left( CO _{2}\right)$, methane $\left( CH _{4}\right)$ and formaldehyde $( HCHO )$ are respectively $40.3\, K$ bar, $1.67\, K$ bar, $0.413\, K$ bar and $1.83 \times 10^{-5} K$ bar at $298\, K$. The correct order of their solubility is

• A. $Ar < CH _{4}< CO _{2}< HCHO$
• B. $HCHO < CH _{4}< CO _{2}< Ar$
• C. $Ar < CO _{2}< CH _{4}< HCHO$
• D. $Ar < CO _{2}< HCHO < CH _{4}$

Answer 1

Answer: (C)

Higher the value of Henry's law constant, lower is the solubility of gas in water (or in liquid solvent).
The order of $K_{ H }$ value is:
$\underset{40.3}{Ar}\,\, \underset{> 1.67}{CO _{2}}\,\, \underset{> 0.413}{CH _{4}}\,\, \underset{> 1.83 \times 10^{-5}}{HCHO}$
Thus, the order of solubility is:
$Ar < CO _{2}< CH _{4}< HCHO$