Ethanol, CH.0, is common beverage alcohol. At its boiling point of 78.5 °C, the enthalpy of vaporization of ethanol is 38.6 kJ/mol. How much heat is required to vaporize 250 g of ethanol at 78.5 °C? Ans = 209.8 KJ

To find out how much heat is required to vaporize 250 g of ethanol, we can use the given enthalpy of vaporization and the molar mass of ethanol.

Step 1: Calculate the number of moles of ethanol. The molar mass of ethanol, C2H5OH, is given by the sum of the atomic masses of carbon, hydrogen, and oxygen: Molar mass of C2H5OH = (2 * 12.01 g/mol) + (6 * 1.008 g/mol) + (16.00 g/mol) = 46.07 g/mol.

Step 2: Divide the mass of ethanol by its molar mass to find the number of moles: Number of moles = mass / molar mass = 250 g / 46.07 g/mol ≈ 5.43 mol.

Step 3: Use the enthalpy of vaporization to determine the total heat required: Heat required (q) = number of moles * enthalpy of vaporization q = 5.43 mol * 38.6 kJ/mol ≈ 209.8 kJ.

Thus, approximately 209.8 kJ of heat is required to vaporize 250 g of ethanol at 78.5 °C.

Extra: The enthalpy of vaporization is a thermodynamic quantity that represents the amount of energy needed to change one mole of a substance from the liquid phase to the vapor phase at constant temperature and pressure. It's also a measure of the strength of the intermolecular forces in the liquid, because these forces must be overcome for the molecules to escape into the vapor phase.