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.

Respuesta :

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

                                          = [tex]0.21 \times 1.13 atm[/tex]

                                           = 0.237 atm

According to Henry's law,

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

             = [tex]1.3 \times 10^{-3} M/atm \times 0.2373 atm[/tex]

             = [tex]3.08 \times 10^{-4}[/tex] M

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

   Moles of oxygen in 5.00 L = volume × concentration

                                                 = [tex]5.00 \times 3.0849 \times 10^{-4}[/tex]

                                                 = [tex]1.542 \times 10^{-3}[/tex] mol

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

        Mass of oxygen = moles × molar mass of oxygen

                                    = [tex]1.542 \times 10^{-3} mol \times 32 g/mol[/tex] 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.

The mass of oxygen dissolved in water has been 49.4 mg.

The partial pressure of oxygen in the air:

Partial pressure = Mole fraction [tex]\times[/tex] Pressure of gas

Partial pressure = 0.21 [tex]\times[/tex] 1.13 atm

Partial pressure of oxygen = 0.237 atm.

The concentration of oxygen can be given by Henry's law.

Concentration of Oxygen = Henry's constant ([tex]\rm k_H[/tex]) [tex]\times[/tex] Partial pressure

Concentration = 1.3 [tex]\rm \times\;10^-^3[/tex] M/atm [tex]\times[/tex] 0.237 atm

Concentration of oxygen = 3.08 [tex]\rm \times\;10^-^4[/tex] M

Moles can be given by:

Moles = Molarity [tex]\times[/tex] Volume

Moles of oxygen =  3.08 [tex]\rm \times\;10^-^4[/tex] M [tex]\times[/tex] 5

Moles of oxygen = 1.542  [tex]\rm \times\;10^-^3[/tex] mol

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Weight of oxygen = Moles [tex]\times[/tex] Molecular weight

Weight of oxygen =  1.542  [tex]\rm \times\;10^-^3[/tex] mol [tex]\times[/tex] 32 g/mol

Weight of oxygen = 0.0494 grams

Weight of oxygen = 49.4 mg.

The mass of oxygen dissolved in water has been 49.4 mg.

For more information about the mass of gas, refer to the link:

https://brainly.com/question/2216031