The solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at 4.0 atm and 25 °C is ________ g/L.
A)0.041
B)0.014
C)0.31
D)0.16
E)4.0

Question

Quizplus · Accepted Answer

To solve this problem, we can use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically, this can be expressed as:

S = kP

where S is the solubility, P is the partial pressure, and k is the proportionality constant (Henry's law constant).

We can use this equation to calculate the solubility of oxygen at 4.0 atm:

S2 = kP2

We can rearrange the equation to solve for k:

k = S1/P1

where S1 and P1 are the solubility and partial pressure at the given conditions (25°C and 1.0 atm).

Then, we can substitute in the values for k and P2 to find S2:

S2 = (S1/P1) x P2 = (0.041 g/L / 1.0 atm) x 4.0 atm = 0.164 g/L

Therefore, the solubility of oxygen in water at 4.0 atm and 25°C is 0.164 g/L, which is closest to choice D (0.16).

The Solubility of Oxygen Gas in Water at 25 °C