Question

The variation of solubility of four different gases $(G_1,G_2,G_3,G_4)$ in a given solvent with pressure at a constant temperature is given below. The gas having lowest value of Henry's law constant is:

• A. $G_1$
• B. $G_2$
• C. $G_4$
• D. $G_3$

Answer 1

The correct option is C $G_4$

The variation of solubility of four different gases $(G_1,G_2,G_3,G_4)$ in a given solvent with pressure at constant temperature is shown in the plot.
According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution
In mathematical term,
$p=K_h\chi$
Where, $p=\text{Partial pressure of the gas}$
$\chi =\text{Mole fraction of solute}$
Again more is the mole fraction more is the solubility.
$\therefore \chi =\frac{1}{K_H}p$
Which is analogous to $y=mx+C$
So, slope will be $\frac{1}{K_H}$
Since $G_4$ has highest solubility at a particular pressure, hence, $K_H$ will be lowest for $G_4$.