To calculate the standard enthalpy change for the vaporization of 50.0 g of water, we first need to determine the number of moles of water present. The molar mass of water H2O is approximately 18.015 g/mol. Number of moles = mass / molar massNumber of moles = 50.0 g / 18.015 g/mol = 2.776 molesNow, we can use the molar enthalpy of vaporization to find the total enthalpy change.Standard enthalpy change = number of moles molar enthalpy of vaporizationStandard enthalpy change = 2.776 moles 40.7 kJ/mol = 112.98 kJThe standard enthalpy change for the vaporization of 50.0 g of water at a temperature of 298.15 K and a pressure of 1 atm is approximately 112.98 kJ.