Question

What is the standard enthalpy change for the vaporization of 50 grams of liquid water at 100°C and 1 atm pressure? Given that the molar enthalpy of vaporization of water is 40.6 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the vaporization of 50 grams of liquid water, we need to first determine the number of moles of water and then use the molar enthalpy of vaporization.1. Calculate the number of moles of water: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 = 2.776 moles2. Calculate the standard enthalpy change:Standard enthalpy change = number of moles  molar enthalpy of vaporizationStandard enthalpy change = 2.776 moles  40.6 kJ/mol = 112.71 kJThe standard enthalpy change for the vaporization of 50 grams of liquid water at 100C and 1 atm pressure is 112.71 kJ.