To calculate the standard enthalpy change for the vaporization of 25.0 grams of water, we first need to determine the number of moles of water present. The molar mass of water H2O is 18.015 g/mol.Number of moles of water = mass of water / molar mass of water Number of moles of water = 25.0 g / 18.015 g/mol = 1.387 molesNow, we can use the molar enthalpy of vaporization to calculate the standard enthalpy change for the vaporization of 25.0 grams of water.Standard enthalpy change = number of moles of water molar enthalpy of vaporization Standard enthalpy change = 1.387 moles 40.7 kJ/mol = 56.5 kJThe standard enthalpy change for the vaporization of 25.0 grams of water at 100C is 56.5 kJ.