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

queen_honey_blossom answered 1 year ago

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