A 70.8mL sample of "wet" O2(g) is collected over water at 29.9 degrees Celsius at a barometric pressure of 743 mmHg. The vapour pressure of water at 29.9 degrees Celsius is 31 mmHg. What is the mole fraction of O2 in the gas collected?
So i found the partial pressure of O2 to be 712 mmHg, and then I found the number of moles of O2 to be 0.00267 mol. Butttt I'm not really sure how to establish the mole fraction? Help!
mole fraction O2 = (moles O2/total moles)
where total moles = moles O2 + moles H2O
tgt
To find the mole fraction of O2 in the gas collected, you need to know the total pressure of the gas mixture.
The total pressure is the sum of the partial pressures of the individual gases present. In this case, you have the partial pressure of O2 as 712 mmHg and the vapor pressure of water as 31 mmHg. So, the total pressure is:
Total Pressure = Partial Pressure of O2 + Vapor Pressure of Water
Total Pressure = 712 mmHg + 31 mmHg
Total Pressure = 743 mmHg
Now that you know the total pressure, you can calculate the mole fraction of O2.
Mole fraction is defined as the ratio of the moles of a specific component to the total moles of all components in the mixture.
Mole fraction of O2 = (moles of O2) / (total moles in the mixture)
Since you have already calculated the moles of O2 as 0.00267 mol, you need to calculate the moles of water vapor.
To calculate the moles of water vapor, you can use the ideal gas law:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 29.9 + 273.15
T(K) = 303.05 K
Now, we can use the ideal gas law to calculate the moles of water vapor:
(31 mmHg) / (743 mmHg) = n(water) / (0.0821 L·atm/mol·K * 303.05 K)
Solving for n(water):
n(water) = (31 mmHg) * (0.0821 L·atm/mol·K * 303.05 K) / (743 mmHg)
n(water) = 0.001282 mol
Now that you have the moles of O2 and water vapor, you can calculate the mole fraction of O2:
Mole fraction (O2) = (0.00267 mol) / (0.00267 mol + 0.001282 mol)
Mole fraction (O2) = 0.675
Therefore, the mole fraction of O2 in the gas collected is approximately 0.675.