Question

Calculate the mass of oxygen (in mg) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.21 given that kH for O2 is 1.3 × 10-3 M/ atm at this temperature.

Answer 1

Explanation:

It is known that relation between partial pressure, mole fraction and pressure is as follows.

      Partial pressure of gas = mole fraction of gas × Pressure of gas

Therefore, putting the given values into the above formula as follows.

   Partial pressure of gas = mole fraction of gas × Pressure of gas

                                          = $0.21 \times 1.13 atm$

                                           = 0.237 atm

According to Henry's law,

       Concentration of oxygen = Henry's law constant × partial pressure of oxygen

             = $1.3 \times 10^{-3} M/atm \times 0.2373 atm$

             = $3.08 \times 10^{-4}$ M

Therefore, calculate moles of oxygen in 5.00 L present as follows.

   Moles of oxygen in 5.00 L = volume × concentration

                                                 = $5.00 \times 3.0849 \times 10^{-4}$

                                                 = $1.542 \times 10^{-3}$ mol

Now, we will calculate the mass of oxygen as follows.

        Mass of oxygen = moles × molar mass of oxygen

                                    = $1.542 \times 10^{-3} mol \times 32 g/mol$ mol    

                                    = 0.0494 g

or,                                 = 49.4 mg           (As 1 g = 1000 mg)

thus, we can conclude that the mass of given oxygen (in mg) is 49.4 mg.