Final answer:
By using the ideal gas law and the given half-reaction, it was calculated that approximately 0.00308 moles of O2 were produced, which corresponds to about 0.01232 moles of electrons. The closest answer choice to the calculated value is 0.0030 mol if the moles of O2 were rounded incorrectly to two significant figures.
Step-by-step explanation:
To determine how many moles of electrons passed through the solution during the electrolysis of water, we first need to use the ideal gas law to find out how many moles of oxygen gas were produced. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature.
The student collected 0.076 L of O2 at 755 mmHg, which is equivalent to 0.993 atm (since 1 atm = 760 mmHg). The temperature is 25°C, which is 298 K (since absolute temperature in Kelvin is Celsius temperature + 273.15).
We use the ideal gas constant R = 0.0821 L·atm/(mol·K). Plugging the values into the ideal gas law equation:
PV = nRT → n = PV / RT
n = (0.993 atm * 0.076 L) / (0.0821 L·atm/(mol·K) * 298 K) → n ≈ 0.0031 moles of O2
From the given half-reaction, we know that 1 mole of O2 gas produces 4 moles of electrons. Therefore, the number of moles of electrons is:
0.0031 moles of O2 * 4 moles e−/mole of O2 = 0.0124 moles of e−
However, since this number (0.0124) is not one of the options provided in the question, it's possible that there might have been a misunderstanding or typo in the question or answer choices. Let's go through the calculation one more time to confirm:
0.993 atm * 0.076 L = n * 0.0821 L·atm/(mol·K) * 298 K
n = (0.993 * 0.076) / (0.0821 * 298) → n = 0.075348 / 24.4858 → n ≈ 0.00308 moles of O2
Thus, 0.00308 moles of O2 * 4 moles e−/mole of O2 = 0.01232 moles of e−, which, when rounded to two significant figures, gives us 0.012 moles of e−, suggesting that the closest answer provided is (d) 0.0030 mol if the moles of O2 were incorrectly rounded to two significant figures.