The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air is 2.67 x 10–4 M at sea level and 25°C, what is the solubility of O2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm? Assume the temperature is 25°C, and that the mole fraction of O2 in air is 0.209 at both 12,000 ft and at sea level.Group of answer choices4.06 x 10–4 MNone of the above.1.75 x 10–4 M 3.66 x 10–5 M2.67 x 10–4 M
Question
The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air is 2.67 x 10–4 M at sea level and 25°C, what is the solubility of O2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm? Assume the temperature is 25°C, and that the mole fraction of O2 in air is 0.209 at both 12,000 ft and at sea level.Group of answer choices4.06 x 10–4 MNone of the above.1.75 x 10–4 M 3.66 x 10–5 M2.67 x 10–4 M
Solution
To solve this problem, we need to use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The formula for Henry's Law is:
S1/P1 = S2/P2
where: S1 is the initial solubility (2.67 x 10–4 M at sea level) P1 is the initial pressure (1 atm at sea level) S2 is the final solubility (what we're trying to find) P2 is the final pressure (0.657 atm at 12,000 ft)
We can rearrange the formula to solve for S2:
S2 = S1 * (P2/P1)
Substituting the given values:
S2 = 2.67 x 10–4 M * (0.657 atm / 1 atm)
S2 = 1.75 x 10–4 M
So, the solubility of O2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm is 1.75 x 10–4 M.
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