# Question 635d7

Dec 16, 2015

${\text{0.030 g O}}_{2}$

#### Explanation:

Your strategy here will be to use Henry's Law to determine the molarity of the solution with regard to the dissolved oxygen, then use the definition of molarity to find how many moles of oxygen would be present in this solution.

As you know, the Henry's Law constant, ${k}_{H}$, is defined as

color(blue)(k_H = c_"aq"/P)" ", where

${c}_{\text{aq}}$ - the concentration of the gas dissolved in solution
$P$ - the partial pressure of the gas above solution

Now, the Henry's law constant for oxygen gas has a value of $1.3 \cdot {10}^{- 3} {\text{mol L"^(-1)"atm}}^{- 1}$ at ${25}^{\circ} \text{C}$. This means that you can use the partial pressure of the oxygen gas at equilibrium to determine the concentration of the dissolved oxygen.

Do not forget to convert the pressure from mmHg to atm, since that is the unit used for the value of ${k}_{H}$!

${k}_{H} = {c}_{\text{aq"/P implies c_"aq}} = {k}_{H} \times P$

${c}_{{O}_{2}} = 1.3 \cdot {10}^{- 3} {\text{mol L"^(-1) color(red)(cancel(color(black)("atm"^(-1)))) * 920/760color(red)(cancel(color(black)("atm"))) = "0.001574 mol L}}^{- 1}$

As you know, molarity is defined as moles of solute, which in your case is oxygen gas, divided by liters of solution. This means that the number of moles of oxygen gas dissolved in solution will be equal to

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{{O}_{2}} = {\text{0.001574 mol" color(red)(cancel(color(black)("L"^(-1)))) * 600.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0009444 moles O}}_{2}$

Finally, use oxygen's molar mass to determine how many grams of oxygen would contain that many moles

0.0009444 color(red)(cancel(color(black)("moles"))) * "31.9988 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.030 g O"_2)#

The answer is rounded to two sig figs.