What is the solubility of oxygen (in units of grams per liter) in water at 25 °C, when the O2 has a partial pressure of 0.200 atm (typical of oxygen in air under normal conditions)?

Question

What is the solubility of oxygen (in units of grams per liter) in water at 25 °C, when the O[latex]_{2}[/latex] has a partial pressure of 0.200 atm (typical of oxygen in air under normal conditions)?

Accepted Answer

You are asked to calculate the solubility of a gas in water at a given temperature.
You are given the identity of the gas and the partial pressure of the gas at a given temperature.
The solubility of O[latex]_{2}[/latex] is given by Henry’s law

S(O[latex]_{2}[/latex]) = k[latex]_{H}[/latex](O[latex]_{2}[/latex]) × P(O[latex]_{2}[/latex])

S(O[latex]_{2}[/latex]) = (1.28 × [latex]10^{-3}[/latex] mol/L · atm)(0.200 atm) = 2.56 × [latex]10^{-4}[/latex] mol/L

2.56 × [latex]10^{-4}[/latex] mol/L [latex]\frac{32.00 ext{ g}}{1 ext{ mol O}_{2}}=8.19 imes 10^{-3}[/latex]g/L

Chapter 13

Q. 13.3.1

Q. 13.3.1

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