What is the concentration of O2 in a freshwater stream in eq

What is the concentration of O₂ in a freshwater stream in equilibrium with air at 25^oC and 1.0 bar?
Express the answer in grams of O₂ per kg of solvent. (Henry’s law constant for O₂ is 1.3x10^-3 mol/kgxbar)

Solution

To answer the question, we know that the concentration of O₂ in atmosphere is 21%.

So, the mole fraction of O₂ in air becomes 21/100 = 0.21

therefore the partial pressure(P_gas) of O₂ becomes 0.21bar in air (because at equilibrium with air it is given in question that pressure is 1.0 bar)

Now, Using Henry\'s law i.e.C=kP_gas

where

CC = solubility of a gas

P_gas = the partial pressure of the gas

k = Henry\'s law constant

Now, placing values inC=kP_gas  ,we get C=1.3x10^-3 x 0.21=2.7 x 10^-4 mol/l

Now, converting this value of C in grams of O₂ per kg of solvent, we know that 1 mole of O₂ equals 32gram

therefore, concentration of O₂=2.7 x 10^-4  x 32 = 0.0087 g/Kg

Answer = Concentration of O₂ in a freshwater stream in equilibrium with air at 25^oC and 1.0 bar is 0.0087 g/Kg