What would be the saturation concentration (mole/L) of oxygen (O2) in a river in winter when the air temperature is 0°C if the Henry's law constant at this temperature is 2.28 × 10-3 mole/L-atm? What would the answer be in units of mg/L?

20.21*10^-4 mg/L is the concentration.

Explanation:

Given:  

Pressure=1 atm.

Temperature=  0 degrees

Volume percent of air is 21%

Henry's Law constant K = 2.28 x 10^-3 mole/L-atm

The partial pressure of oxygen is 0.21 atm.

By Henry's law:

Concentration= K X Partial pressure

                        =  2.28*10^-3 *0.21

                          =  4.79*10^-4 moles/litre

Since, at STP 1 mole of oxygen occupies volume of 22.4L

concentration =mass/volume

     mass                  =  4.79*10^-4-4*22.4

                                 = 20.21*10^-4 mg

    20.21*10^-4 mg/L is the concentration.