Question

The Henry’s law constant for O₂ is 1.3 × 10^−3 M/atm at 25 °C. Assuming ideal solution behavior, what mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O₂ is 0.21 atm?

a) 0.54 g
b) 0.37 g
c) 0.28 g
d) 0.45 g

Answer 1

b) 0.37 g

The mass of oxygen that would be dissolved in a 40-L aquarium at 25 °C, with an atmospheric pressure of 1.00 atm and a partial pressure of O₂ of 0.21 atm, is approximately 0.35 g, the closest option to this is 0.37 g.

Step-by-step explanation:

To calculate the mass of oxygen dissolved in a 40-L aquarium at 25 °C with an atmospheric pressure of 1.00 atm and a partial pressure of O₂ of 0.21 atm using Henry's law constant for O₂, we can follow these steps:

• Use Henry's law to find the solubility of O₂ in moles per liter (M), which is the product of the Henry's law constant (k) and the partial pressure of O₂ (P₂): Solubility (M) = k × P₂.
• Then, convert the solubility to moles of O₂ by multiplying by the volume of the solution in liters.
• Finally, convert the moles of O₂ to grams using the molar mass of O₂, 32.00 g/mol.

Let's apply these steps:

Solubility of O₂ = Henry's law constant × partial pressure of O₂ = (1.3 × 10⁻³ M/atm) × (0.21 atm) = 2.73 × 10⁻⁴ M.Total moles of O₂ = Solubility × Volume = (2.73 × 10⁻⁴ M) × (40 L) = 0.01092 mol.Mass of O₂ = moles × molar mass = (0.01092 mol) × (32.00 g/mol) = 0.34944 g.

The mass of oxygen that would be dissolved in the aquarium is 0.34944 g, which we can round to 0.35 g, selecting the closest option which is b) 0.37 g