Question

In: Chemistry

Nitrogen gas has a Henry's law constant of 8.42×10-7 M/mmHg at 40.0 °C when dissolving in...

Nitrogen gas has a Henry's law constant of 8.42×10-7 M/mmHg at 40.0 °C when dissolving in water. If the total pressure of gas (N2 gas plus water vapor) over water is 1.00 atm, what is the concentration of N2 in the water in grams per milliliter?

Pressure of the water vapor at 40.0 °C = 55.3 mmHg.

____g/mL?

Solutions

Expert Solution

Total pressure = 1atm = 760mmHg

pressure of N2 = total pressure - water vapor pressure

                         = 760-55.3   = 704.7mmHg

C    = KH * P

       = 8.42*10^-7 * 704.7

       = 0.000594M = 0.000594mole/L

     = 0.000594mole/L * 28g/mole

    = 0.016632g/L   = 0.016632g/1000ml   = 1.66*10^-5 g/ml

the concentration of N2 in the water in grams per milliliter = 1.66*10^-5g/ml >>>>answer

  


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