If water contains about 42 mg of oxygen per liter at 20°C and 1.0 atm, what would be the value of Henry's law constant for oxygen dissolving in water? The mole fraction of oxygen in air is 0.21. A) 0.25 mol/(L atm) B) 1.3 *10^-³ mol/(L atm) C) 1.9 *10^-² mol/(L atm) D) 6.3 * 10^-³ mol/(L atm) E) 0.010 mol/(L atm)

Question

Quizplus · Accepted Answer

Henry's law states that the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, it can be expressed as:

C = kH * P

where C is the concentration of the gas in the liquid, PH is the partial pressure of the gas above the liquid, and kH is Henry's law constant.

In this case, we know that at 20°C and 1.0 atm, water contains about 42 mg of oxygen per liter. We also know that the mole fraction of oxygen in air is 0.21.

First, we need to convert the concentration of oxygen in water from milligrams per liter to moles per liter. The molar mass of oxygen is 32 g/mol, so 42 mg/L is equivalent to 1.31 x 10^-3 mol/L.

Next, we can use the mole fraction of oxygen in air to calculate the partial pressure of oxygen:

PO2 = 0.21 * 1.0 atm = 0.21 atm

Finally, we can use Henry's law to solve for kH:

kH = C/PO2 = (1.31 x 10^-3 mol/L)/(0.21 atm) = 6.3 x 10^-3 mol/(L atm)

Therefore, the answer is D: 6.3 x 10^-3 mol/(L atm).

If Water Contains About 42 Mg of Oxygen Per Liter