First, we need to convert the mass of water 100 g into moles. The molar mass of water HO is 18.015 g/mol.moles of water = mass / molar massmoles of water = 100 g / 18.015 g/molmoles of water 5.55 molNow, we can use the enthalpy of vaporization Hvap to calculate the standard enthalpy change for the vaporization of 100 g of water.H = moles HvapH = 5.55 mol 40.7 kJ/molH 225.89 kJThe standard enthalpy change for the vaporization of 100 g of water at 100C is approximately 225.89 kJ.