Question

The Henry's law constant for the solubility of oxygen in water is $3.3\times {10}^{-4}$ M/atm at 12${}^o$C. Air is 21 mol% oxygen. How many grams of oxygen can be dissolved in one litre of a trout stream at 12${}^o$C at an air pressure of 1.00 atm?

• A. $2.22$ mg
• B. $0.11222$ mg
• C. $0.333$ g
• D. $0.422$ g

Answer 1

The correct option is A $2.22$ mg

The total pressure is $1$ atm. The mole fraction of oxygen is $0.21$.

Hence, the partial pressure of oxygen is $0.21\times 1=0.21atm$

According to Henry's law, $S=K_{H}P$

$S$ is the solubility in moles per litre

$K_H$ is Henry's law constant

$P$ is the pressure in atm.

Substitute values in the above expression:

$S=3.3\times {10}^{-4}Lat{m}^{-1}\times 0.21atm=6.93\times {10}^{-5}M$

Thus $6.93\times {10}^{-5}$ moles of oxygen are dissolved in $1$ L of water.

This corresponds to $6.93\times {10}^{-5}\times 32=2.22mg$ of oxygen.

Hence, the correct answer is option $\text{A}$.