To calculate the standard enthalpy change for the vaporization of 15 grams of water, we first need to determine the number of moles of water present. The molar mass of water is 18.015 g/mol.Number of moles = mass of water / molar mass of water Number of moles = 15 g / 18.015 g/mol = 0.8326 molNow, we can use the molar enthalpy of vaporization to find the standard enthalpy change for the vaporization of 15 grams of water.Standard enthalpy change = number of moles molar enthalpy of vaporization Standard enthalpy change = 0.8326 mol 40.7 kJ/mol = 33.89 kJThe standard enthalpy change for the vaporization of 15 grams of water at its boiling point is approximately 33.89 kJ.