Final answer:
The student's question on gas collection over water involves calculating moles of oxygen using the ideal gas law, determining the water vapor's partial pressure, correcting oxygen volume to STP, and finding the mass of oxygen.
Step-by-step explanation:
To answer the student's question, we first need to use the ideal gas law to calculate moles of oxygen. The ideal gas law is given as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Part a: Before we can use the ideal gas law, we need to convert Celsius to Kelvin, which is done by adding 273.15 to the Celsius temperature. Also, we must calculate the partial pressure of oxygen by subtracting the water vapor pressure from the total pressure. The vapor pressure of water at 23 °C is approximately 21.1 torr or 0.0278 atm. Therefore, the partial pressure of oxygen is 1.5 atm - 0.0278 atm = 1.4722 atm.
Moles of oxygen (n) can now be calculated:
n = PV / RT = (1.4722 atm)(0.193 L) / (0.0821 L·atm/mol·K)(296.15 K) ≈ 0.01231 mol
Part b: The partial pressure of water vapor at 23.0 °C is 0.0278 atm.
Part c: To calculate the volume of oxygen corrected to STP (standard temperature and pressure which is 0 degree Celsius or 273.15 K and 1 atm), we can use the ideal gas law again:
VSTP = nRT / PSTP = (0.01231 mol)(0.0821 L·atm/mol·K)(273.15 K) / 1 atm ≈ 0.276 L
Part d: To find the mass of oxygen collected, we use the fact that one mole of oxygen (O2) weighs approximately 32.00 g/mol.
Mass = n x molar mass = 0.01231 mol x 32.00 g/mol ≈ 0.394 g