The standard enthalpy change for the vaporization of water at its boiling point 100C and standard pressure 1 atm is 40.7 kJ/mol. To calculate the enthalpy change for 50 grams of water, we need to convert grams to moles and then multiply by the enthalpy change per mole.First, we need to find the number of moles of water:50 grams / 18.015 g/mol = 2.775 moles of waterNow, we can calculate the enthalpy change:2.775 moles * 40.7 kJ/mol = 112.94 kJThe standard enthalpy change for the vaporization of 50 grams of water at its boiling point 100C and standard pressure 1 atm is approximately 112.94 kJ.