A mixture of 0.100 mol of N2 and 0.200 mol of O2 is collected over H2O at an atmospheric pressure of 750. mm Hg and a temperature of 22 ºC. What is the partial pressure (in mmHg) of O2 in this mixture? (Vapor pressure of water @ 22 ºC is 22 mm Hg)
485
I would do this.
Ptotal = PH2O + PN2 + PO2
750 mm = 22 + pN2 + PO2.
PN2 + PO2 = 728 mm
Then you have 1/3 of 728 for N2 and 2/3 of 728 for O2.
To find the partial pressure of O2 in the mixture, we need to consider the Dalton's law of partial pressures.
According to Dalton's law, the total pressure of a gas mixture is equal to the sum of the partial pressures of each gas in the mixture.
So, to find the partial pressure of O2, we first need to find the total pressure of the mixture.
Given:
Moles of N2 = 0.100 mol
Moles of O2 = 0.200 mol
Atmospheric pressure = 750 mmHg
Vapor pressure of water @ 22 ºC = 22 mmHg
To find the total pressure of the mixture, we add the atmospheric pressure and the vapor pressure of water:
Total pressure = Atmospheric pressure + Vapor pressure of water
Total pressure = 750 mmHg + 22 mmHg
Total pressure = 772 mmHg
Now we can find the partial pressure of O2 by using the mole ratio of O2 to the total moles of gases in the mixture:
Partial pressure of O2 = (Moles of O2 / Total moles of gases) * Total pressure
Total moles of gases = Moles of N2 + Moles of O2
Total moles of gases = 0.100 mol + 0.200 mol
Total moles of gases = 0.300 mol
Partial pressure of O2 = (0.200 mol / 0.300 mol) * 772 mmHg
Partial pressure of O2 = 0.667 * 772 mmHg
Partial pressure of O2 ≈ 515.524 mmHg
Therefore, the partial pressure of O2 in this mixture is approximately 515.524 mmHg.
To determine the partial pressure of O2 in the mixture, we need to consider the total pressure and subtract the vapor pressure of water.
Step 1: Convert the given atmospheric pressure to atm.
The given atmospheric pressure is 750. mm Hg. To convert this to atm, divide by the conversion factor 760 mm Hg/atm:
750. mm Hg ÷ 760 mm Hg/atm = 0.987 atm
Step 2: Subtract the vapor pressure of water from the total pressure.
The vapor pressure of water at 22 ºC is given as 22 mm Hg. Convert this to atm by dividing by 760 mm Hg/atm:
22 mm Hg ÷ 760 mm Hg/atm = 0.0289 atm
Now, subtract the vapor pressure of water from the total pressure to get the partial pressure of the gases:
0.987 atm - 0.0289 atm = 0.9581 atm
Step 3: Convert the partial pressure back to mmHg.
To convert atm to mmHg, multiply by the conversion factor 760 mm Hg/atm:
0.9581 atm × 760 mm Hg/atm = 728.236 mm Hg
Therefore, the partial pressure of O2 in the mixture is approximately 728.236 mmHg.