Question

Calculate the mass of oxygen (in mg) dissolved in a 4.97 L bucket of water exposed to a pressure of 1.08 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 × 10-3 M/atm O2. (Enter your value using three significant figures.)

Answer 1

46.9mg of oxygen

Step-by-step explanation:

From Henry's law,

Concentration of oxygen (C) = Henry's constant (K) × partial pressure of oxygen in air (p)

K = 1.3×10^-3M/atm O2, p = mole fraction of oxygen in air × pressure of air = 0.21×1.08atm = 0.2268atm

C = K×p = 1.3×10^-3 × 0.2268 = 0.00029484M of O2

Concentration (C) = number of moles of oxygen (n)/volume of water (V)

Volume of water (V) = 4.97L

n = CV = 0.00029484 × 4.97 = 0.001465mole

number of moles (n) = mass of O2/MW of O2

mass of O2 = number of moles of O2 × MW of O2 = 0.001465mole × 32g/mole = 0.0469g = 0.0469×1000mg = 46.9mg (to three significant figures)