Calculate the solubility of oxygen in water at 25c when the total external pressure is 1 at and the mole fractionof oxygen in the air is 0.2

The given question is incomplete. The complete question is as follows.

The value of Henry's law constant for oxygen in water at is M/torr.

Calculate the solubility of oxygen in water at when the total external pressure is 1 atm and the mole fraction of oxygen in the air is 0.20 atm.

Explanation:

Formula to calculate partial pressure of a gas is as follows.

   Partial pressure of oxygen = mole fraction of oxygen x total pressure

Putting the given values into the above equation as follows.

       = = 152 torr

Therefore, solubilty (concentration) of oxygen in water  will be calculated as follows.

        Solubility = Henry's law constant x partial pressure of oxygen

                       =

                       = M

Thus, we can conclude that solubility of given oxygen is M.

the answer is the first option

explanation:

just googled it m8. not that hard

lol

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%