In 1.00 atm of pure oxygen,the solubility of O₂(g)in water is 1.26 × 10−³ M at 25.0°C.The mole fraction of oxygen in air is 0.210.If the atmospheric pressure is 0.979 atm,what is the solubility of oxygen in air at 25.0°C?

Question

Quizplus · Accepted Answer

The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. This is described by Henry's law. Given the solubility of oxygen in water under a specific pressure, we can calculate its solubility under a different pressure using the formula: $S_2 = S_1 \times \frac{P_2}{P_1}$, where $S_1$ is the initial solubility, $P_1$ is the initial pressure, $S_2$ is the solubility under the new conditions, and $P_2$ is the new pressure. The partial pressure of oxygen in air is $0.210 \times 0.979 \, \text{atm}$, and the initial solubility is given as $1.26 \times 10^{-3} \, M$ at 1.00 atm. Therefore, the new solubility is calculated based on the partial pressure of oxygen in the new conditions, leading to the correct answer.

In 100 Atm of Pure Oxygen,the Solubility of O2(g)in Water Is

In 100 Atm of Pure Oxygen,the Solubility of O₂(g)in Water Is