a) A dilute aqueous solution of K2SO4 is electrolyzed between platipum electrodes for 8.00 bours with a current of 2.00 A. UShat volume of gas, saturated with water vapor at 25^circ C and at a total pressure of 745 ropplig, would be collected at the anode? (Vapor pressure of water at 25^circ C is 23.8mmHg) (5 marks)

To solve this problem, we need to use Faraday's law of electrolysis and the ideal gas law.First, let's calculate the total charge (Q) passed through the solution during electrolysis using the formula:Q = I * twhere I is the current (2.00 A) and t is the time (8.00 hours converted to seconds).Q = 2.00 A * (8.00 hours * 3600 s/hour) = 57600 CNext, we need to determine the number of moles of electrons (n) using the formula:n = Q / (F * z)where F is the Faraday constant (96485 C/mol) and z is the number of electrons transferred per molecule of (2 electrons).n = 57600 C / (96485 C/mol * 2) = 0.298 molNow, we can use the ideal gas law to calculate the volume of gas collected at the anode. The total pressure is given as 745 mmHg, and the vapor pressure of water at is 23.8 mmHg. Therefore, the partial pressure of the gas is:Partial pressure of gas = Total pressure - Vapor pressure of water= 745 mmHg - 23.8 mmHg= 721.2 mmHgWe need to convert this pressure to atm:Partial pressure of gas in atm = 721.2 mmHg / 760 mmHg/atm = 0.946 atmNow, we can use the ideal gas law to find the volume of gas:PV = nRTwhere P is the partial pressure of the gas (0.946 atm), V is the volume of gas, n is the number of moles of gas (0.298 mol), R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).V = nRT / P= (0.298 mol * 0.0821 L·atm/mol·K * 298.15 K) / 0.946 atm= 7.34 LTherefore, the volume of gas collected at the anode is 7.34 liters.