To calculate the standard enthalpy change for the vaporization of 10 grams of water, we first need to determine the number of moles of water present. The molar mass of water HO is approximately 18.015 g/mol 1.008 g/mol for hydrogen and 15.999 g/mol for oxygen .Number of moles n = mass m / molar mass M n = 10 g / 18.015 g/moln 0.555 moles of waterNow, we can calculate the standard enthalpy change H using the enthalpy of vaporization H_vap and the number of moles n .H = n * H_vapH = 0.555 moles * 40.7 kJ/molH 22.59 kJThe standard enthalpy change for the vaporization of 10 grams of water at 100C is approximately 22.59 kJ.