Question

What is the standard enthalpy change for the vaporization of 50 grams of water at 100°C and 1 atm pressure? Given the molar enthalpy of vaporization of water is 40.7 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization of 50 grams of water, we first need to determine the number of moles of water present. The molar mass of water  H2O  is 18.015 g/mol.Number of moles = mass / molar massNumber of moles = 50 g / 18.015 g/mol = 2.775 molesNow, we can use the molar enthalpy of vaporization to calculate the standard enthalpy change.Standard enthalpy change = number of moles  molar enthalpy of vaporizationStandard enthalpy change = 2.775 moles  40.7 kJ/mol = 112.9415 kJThe standard enthalpy change for the vaporization of 50 grams of water at 100C and 1 atm pressure is approximately 112.94 kJ.