Final Answer:
The volume of oxygen collected is 0.154 liters.
Step-by-step explanation:
The electrolysis of water is a redox reaction that produces oxygen and hydrogen gas according to the following equation:
2H2O(l) → 2H2(g) + O2(g)
At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 liters. Therefore, to find the volume of oxygen collected, we need to determine the number of moles of oxygen produced.
The number of moles of oxygen produced can be calculated using the following formula:
n = V / Vm
where:
n is the number of moles of gas
V is the volume of gas (in liters)
Vm is the molar volume of gas (in liters/mol)
At STP, Vm = 22.4 L/mol. We are given that the volume of water used is 129 ml, which is equivalent to 0.129 liters. Substituting these values into the formula, we get:
n = 0.129 L / 22.4 L/mol = 0.00576 moles of oxygen
We are also given that the temperature of the oxygen gas is 28°C. To convert this to Kelvin, we add 273.15 to the Celsius temperature:
T = 28°C + 273.15 = 301.15 K
We are also given that the pressure of the oxygen gas is 98 kPa. We need to convert this to atmospheres (atm) because the molar volume of a gas is defined at STP, which is 1 atm and 273.15 K.
1 atm = 101.325 kPa
Therefore, 98 kPa is equivalent to:
98 kPa / 101.325 kPa/atm = 0.969 atm
Now we can calculate the molar volume of oxygen gas at the given temperature and pressure using the ideal gas law:
PV = nRT
where:
P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles of gas
R is the ideal gas constant (0.08206 L·atm/mol·K)
T is the temperature (in Kelvin)
Solving for V, we get:
V = nRT / P = (0.00576 moles) × (0.08206 L·atm/mol·K) × (301.15 K) / (0.969 atm) = 0.154 L
Therefore, the volume of oxygen collected is 0.154 liters.