In: Chemistry
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?
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