Question

The value of Henry's law constant for some gases at 293 K is given below. Arrange the gases in the increasing order of their solubility.

He: 144.97 kbar, $H_{2}$ : 69.16 kbar,

$N_{2}$ : 76.48 kbar, $O_{2}$ : 34.86 kbar

• A. $He < N_{2} < H_{2} < O_{2}$
• B. $O_{2} < H_{2} < N_{2} < He$
• C. $H_{2} < N_{2} < O_{2} < He$
• D. $He < O_{2} < N_{2} < H_{2}$

Answer 1

Answer: (A)

Higher the value of $K_{H}$ , lower is the solubility of gas in the liquid.

The correct order of solubility of these gases is as follows

$\underset{144.97}{H e} < \underset{76.48}{N_{2}} < \underset{69.16}{H_{2}} < \underset{34.86}{O_{2}}$

$K_{H}=144.97,76.48,69.16,34.86 \, kbar$