Oxygen gas makes up 21 % of the atmosphere by volume. What is the solubility of O₂(g) in water at 25^$\circ$C if the atmospheric pressure is 741 mmHg The Henry's law constant for oxygen gas at 25^$\circ$C is 1.3 * 10^-3 mol/L·atm. A) 2.7 * 10^-4 M B) 1.3 * 10^-3 M C) 6.2 * 10^-3 M D) 9.6 * 10^-3 M E) 0.96 M

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Quizplus · Accepted Answer

The solubility of oxygen in water can be calculated using Henry's law, which states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid. The formula for Henry's law is $S = kH \cdot P$, where $S$ is the solubility of the gas (in mol/L), $kH$ is the Henry's law constant (in mol/L·atm for oxygen at 25°C, given as $1.3 \times 10^{-3}$ mol/L·atm), and $P$ is the partial pressure of the gas (in atm). The partial pressure of oxygen can be calculated from the total atmospheric pressure (741 mmHg) and the percentage of oxygen in the atmosphere (21%). First, convert the atmospheric pressure to atm: $741 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.975 \, \text{atm}$. Then, calculate the partial pressure of oxygen: $0.975 \, \text{atm} \times 0.21 = 0.20475 \, \text{atm}$. Finally, apply Henry's law: $S = (1.3 \times 10^{-3} \, \text{mol/L·atm}) \times 0.20475 \, \text{atm} = 2.66175 \times 10^{-4} \, \text{mol/L}$, which rounds to $2.7 \times 10^{-4} \, \text{M}$, making option A correct.

Oxygen Gas Makes Up 21 % of the Atmosphere by Volume