Final answer:
The partial pressure of oxygen in the mixture can be calculated by multiplying the total pressure by the mole fraction of oxygen. Given the total number of moles and the mole fraction of oxygen, the partial pressure of oxygen is approximately 0.3552 atm. When collected over water, the water's vapor pressure must be subtracted from the atmospheric pressure to find the oxygen's partial pressure.
Step-by-step explanation:
The student is asking how to calculate the partial pressure of oxygen in a gaseous mixture. Given that the mixture contains 64 g of oxygen and 70 g of nitrogen with a total pressure of 0.80 atm, the student seeks to determine the partial pressure exerted by oxygen.
First, we can use the ideal gas law to find the moles of each gas.
The molar masses of oxygen (O2) and nitrogen (N2) are 32 g/mol and 28 g/mol, respectively.
Therefore, we have 64 g / 32 g/mol = 2 moles of oxygen, and 70 g / 28 g/mol = 2.5 moles of nitrogen.
The total number of moles is 2 + 2.5 = 4.5 moles.
Now, the mole fraction of oxygen is 2 moles of O2/4.5 moles total = 0.444.
The partial pressure of oxygen can be determined by multiplying the total pressure by the mole fraction of oxygen (PO2 = total pressure × mole fraction of oxygen).
Therefore, PO2 will be 0.80 atm × 0.444, which equals approximately 0.3552 atm.
If we are considering the pressure of oxygen collected over water, we need to take into account the vapor pressure of water, which will subtract from the total pressure to give the partial pressure of oxygen.
So, for instance, if the atmospheric pressure is 760 mmHg at 15°C, and the vapor pressure of water is 13 mmHg, then the partial pressure of oxygen would be the total pressure minus the vapor pressure of water.
Hence, PO2 = 760 mmHg - 13 mmHg
= 747 mmHg.