Oxygen has a Henry’s law constant of 1.3*10^-3 mol/kg*bar at 25 °C when dissolving in water. If the total pressure of gas ( gas plus water vapor) over water is 1.00 bar, what is the concentration of O2 in the water in grams per milliliter? (The vapor pressure of water at 25 °C is 23.8 torr.)
To find the concentration of oxygen in the water, we need to use Henry's law and the given information.
Henry's law states that the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid, at a constant temperature.
The formula for Henry's law is:
C = k * P
Where:
C is the concentration of the gas in the liquid (in mol/kg)
k is Henry's law constant (in mol/kg*bar)
P is the partial pressure of the gas above the liquid (in bar)
Given:
Henry's law constant (k) = 1.3 * 10^-3 mol/kg*bar
Total pressure (P_total) = 1.00 bar
Vapor pressure of water (P_vapor) = 23.8 torr = 23.8/760 bar
To calculate the partial pressure of oxygen (P_O2), we subtract the vapor pressure of water from the total pressure:
P_O2 = P_total - P_vapor
P_O2 = 1.00 bar - (23.8/760) bar
Now we can use Henry's law to find the concentration of oxygen in the water:
C = k * P_O2
Convert the concentration from mol/kg to g/mL:
Convert the molar mass of oxygen (O2) to grams: 2 * 16.00 g/mol = 32.00 g/mol
To convert from mol/kg to g/mL, we need to know the density of water at 25 °C. At 25 °C, the density of water is approximately 0.997 g/mL.
To find the concentration in g/mL, we use the following formula:
Concentration (g/mL) = (C * molar mass of oxygen) / density of water
Substitute the values into the formula and calculate the concentration of oxygen in g/mL.