Question

The Henry’s law constant for the solubility of oxygen in water is $3.3 \times {10^{ - 4}}M\,at{m^{ - 1}}$ at ${12^0 }C$. Air is $21\;mol\%$ oxygen. How many grams of oxygen can be dissolved in one litre of a trout stream at ${12^0}C$ at an air pressure of $1.00atm$?
(A) $0.222g$
(B) $0.11222g$
(C) $0.333g$
(D) $0.422g$

Answer 1

As we know that Henry's law states that the solubility of the gas given as $S$ is directly proportional to the mole fraction of the gas given as $\chi$. We also know that the amount of any substance can be obtained by the product of its moles and the molecular mass.

Complete step-by-step solution: As we know that Henry’s law describes the solubility of the gas and it states that partial pressure of the gas in vapour phase $({p^0})$ is directly proportional to the mole fraction of that gas and is expressed as show below:
${p^0 } = {K_H}\chi$, where ${K_H}$ is Henry's law constant.
Now we are given that the total pressure is $1.00atm$ and mole fraction of oxygen is given as $21\;mol\%$ or $0.21$. So we can identify the partial pressure from these given values and we will get:
${p^0 } = 0.21 \times 1 = 0.21\;atm$
Now, we are given with the Henry’s law constant of oxygen as $3.3 \times {10^{ - 4}}M\,at{m^{ - 1}}$ and we calculated the partial pressure of oxygen as $0.21\;atm$. So, we can calculate the solubility of oxygen by substituting these values in the formula and we will get:
$S = 3.3 \times {10^4} \times 0.21$
$S = 6.93 \times {10^{ - 3}}M$
We also know that the solubility of any substance is basically its maximum amount that can be dissolved in a specified amount of solvent. So we can say that $6.93 \times {10^{ - 3}}$ moles of oxygen are dissolved in $1L$ of water.
Now we have moles and we know the molecular mass of oxygen is $32g$, so we can easily calculate the amount of mass of oxygen that can be dissolved in one litre of trout stream. Hence, we will get:
$moles = \dfrac{{mass}}{{molecular\;mass}}$
$\Rightarrow mass = moles \times molecular\;mass$
$\Rightarrow mass = 6.93 \times {10^{ - 3}} \times 32$
$\therefore mass = 0.222g$
Thus, $0.222g$ of oxygen can be dissolved in one litre.

Therefore, the correct answer is (A).

Note: Always remember that the solubility depends upon the nature of solute and solvent as well as temperature and pressure. As the temperature increases the solubility decreases as the Henry’s law constant value increases with increase in temperature and solubility increases with increase in pressure.