Question

Calculate the mass of oxygen gas (O2) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.210 and the Henry's law constant for air in water at this temperature to be 1.30 × 10-3 M/atm. Report your answer in mg.

Answer 1

Answer:

The mass of oxygen gas dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air is 0.04936 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{gas}=K_H\times p_{gas}$

where,

$K_H$ = Henry's constant =

$p_{O_2}$ = partial pressure of oxygen

We have :

Pressure of the air = P

Mole fraction of oxygen in air = $\chi_{O_2}=0.210$

$p_{O_2}=P\times \chi_{O_2}$

$=0.210\times 1.13 atm= 0.2373 atm$

$K_H$ = Henry's constant = $1.30\times 10^{-3}M/atm$

Putting values in above equation, we get:

$C_{O_2}=1.30\times 10^{-3}M/atm\times 0.2373 atm\\\\C_{O_2}=0.003085 M$

Moles of oxygen gas = n

Volume of water = V = 5 L

$Molarity = \frac{Moles}{Volume(L)}$

$0.003085 M=\frac{n}{5 L}$

$n = 0.003085 M\times 5 L=0.001542 mol$

Mass of 0.001542 moles of oxygen gas:

0.001542 mol × 32 g/mol = 0.04936 g

The mass of oxygen gas dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air is 0.04936 grams.