Oxygen gas has a Henry's law constant of 1.66*10^-6 M/mm Hg at 40 degree C when dissolving in water. If the total pressure of gass (O2 gas plus water vapor) over water is 1.0 atm, what is the concentration of O2 in the water in grams per milliliter? Pressure of the water vapor at 40 degree C = 55.3 torrs.
C = k*p
C = 1.66E-6M/mm x mm
(760 mm - 55.3 = 704.7)
C = 1.66E-6 M/mm*704.7 mm = ? M
Convert that to grams/L and to g/mL.
thank you!!
To find the concentration of O2 in the water in grams per milliliter, we need to use Henry's Law.
Henry's Law relates the concentration of a gas in a liquid to the partial pressure of the gas above the liquid. The formula for Henry's Law is:
C = k * P
Where:
C is the concentration of the gas in the liquid (in this case, O2 concentration in grams per milliliter)
k is the Henry's Law constant (given as 1.66*10^-6 M/mm Hg)
P is the partial pressure of the gas (in this case, the partial pressure of O2 gas)
First, we need to determine the partial pressure of O2 gas. To do this, subtract the water vapor pressure (55.3 torrs) from the total pressure of the gases over water (1.0 atm):
Partial pressure of O2 gas = Total pressure - Water vapor pressure
Partial pressure of O2 gas = (1.0 atm - 55.3 torrs) * (1 atm / 760 torrs)
Partial pressure of O2 gas = (1.0 atm - 0.0725 atm) = 0.9275 atm
Now, we can substitute the values into the Henry's Law equation:
C = k * P
C = (1.66*10^-6 M/mm Hg) * (0.9275 atm)
To convert from moles per milliliter (M) to grams per milliliter (g/mL), we need to multiply by the molar mass of O2 (32 g/mol):
C (in grams per milliliter) = (1.66*10^-6 M/mm Hg) * (0.9275 atm) * (32 g/mol)
Calculating this expression will give you the concentration of O2 in grams per milliliter.