To calculate the standard enthalpy change for the vaporization of 25.0 g of water, we first need to determine the number of moles of water present. We can do this using the molar mass of water:moles of water = mass of water / molar mass of water moles of water = 25.0 g / 18.015 g/mol moles of water 1.387 molesNow that we have the number of moles of water, we can use the molar enthalpy of vaporization to calculate the standard enthalpy change:H = moles of water molar enthalpy of vaporization H = 1.387 moles 40.7 kJ/mol H 56.4 kJTherefore, the standard enthalpy change for the vaporization of 25.0 g of water at 100C is approximately 56.4 kJ.