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Calculate the mass of oxygen dissolved in 120.0 L

of water at 25 ∘C
at a total pressure of 1 atm
and a mole fraction for oxygen of 0.210. The Henry's law constant for oxygen at 25 ∘C
is 1.3×10−3M/atm
.

1 Answer

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Answer: Henry's law states that the solubility of a gas in a liquid is proportional to the partial pressure of the gas over the solution. The relationship can be represented by the equation:

C = kP

where C is the concentration of the gas (in mol/L), k is the Henry's law constant, and P is the partial pressure of the gas.

Given that the mole fraction of oxygen in water is 0.210, the partial pressure of oxygen in the solution is:

P = 0.210 * 1 atm = 0.210 atm

Since the Henry's law constant for oxygen at 25 ∘C is 1.3 × 10^(-3) M/atm, the concentration of oxygen in water can be calculated as:

C = 1.3 × 10^(-3) M/atm * 0.210 atm = 0.000276 M

Finally, the mass of oxygen dissolved in 120.0 L of water can be calculated as:

mass = C * volume * molecular weight of oxygen

= 0.000276 M * 120.0 L * 32.0 g/mol

= 10.57 g

Step-by-step explanation:

User Paul Podgorsek
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