The Henry’s law constant (KH) values of three gases (A, B, C) in water are 145, 2 × 10^–5 and 35 kbar, respectively. The solubility of these gases in water follow the order:

The correct option is B > C > A.

Understanding Henry's Law:
Henry's Law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it is expressed as:
$p=K_{\text{H}}·x$
where:
• $p$ is the partial pressure of the gas,
• $K_{\text{H}}$ is the Henry's law constant, and
• $x$ is the mole fraction of the gas in the solution (which is a measure of its solubility).

Relationship between solubility ($x$) and Henry's law constant ($K_{\text{H}}$):
From the equation, at a given partial pressure ($p$), the relationship between solubility and Henry's law constant can be rearranged as:
$x=\frac{p}{K_{\text{H}}}$
This shows that solubility ($x$) is inversely proportional to the Henry's law constant ($K_{\text{H}}$) for a given pressure. That is, a higher $K_{\text{H}}$ value means lower solubility of the gas in the solvent, and a lower $K_{\text{H}}$ value means higher solubility.

Analyzing the given data:
We are given the Henry's law constant ($K_{\text{H}}$) values for three gases (A, B, and C) in water:
• For Gas A: $K_{\text{H}}=145\text{kbar}$
• For Gas B: $K_{\text{H}}=2\times {10}^{-5}\text{kbar}$
• For Gas C: $K_{\text{H}}=35\text{kbar}$

Comparing the $K_{\text{H}}$ values:
Let's arrange the gases in increasing order of their $K_{\text{H}}$ values:
$2\times {10}^{-5}\text{kbar}<35\text{kbar}<145\text{kbar}$
Therefore:
$K_{\text{H}}\text{(B)}<K_{\text{H}}\text{(C)}<K_{\text{H}}\text{(A)}$

Determining the solubility order:
Since solubility is inversely proportional to $K_{\text{H}}$, the order of solubility will be the exact reverse of the order of $K_{\text{H}}$ values:
Solubility of B > Solubility of C > Solubility of A
Or simply:
B > C > A