Question

Calculate the mass of oxygen (in mg) 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 0.21 and the Henry's law constant for oxygen in water at this temperature to be 1.3 x 10-3 M/atm. Group of answer choices

Answer 1

Answer:

50 mg

Explanation:

First, we have to calculate the partial pressure of O₂ (pO₂) using the following expression.

pO₂ = P × X(O₂) = 1.13 atm × 0.21 = 0.24 atm

where,

P: total pressure

X(O₂): mole fraction of oxygen

Then, we can calculate the concentration of O₂ in water (C) using Henry's law.

C = k × pO₂ = (1.3 × 10⁻³ M/atm) × 0.24 atm = 3.1 × 10⁻⁴ M

where,

k: Henry's constant for O₂

The mass of oxygen in a 5.00 L bucket with a concentration of 3.1 × 10⁻⁴ M is: (MM 32.0 g/mol)

$5.00L.\frac{3.1 \times 10^{-4}mol}{L} .\frac{32.0 \times 10^{3}mg}{mol} =50mg$