To calculate the enthalpy of vaporization for water, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization Hvap to the boiling point Tb and the change in vapor pressure with temperature:Hvap = -R * Tb^2 * d ln P /dT However, we don't have enough information to use this equation directly. Instead, we can use the specific heat capacity and heat of fusion to estimate the enthalpy of vaporization.First, let's calculate the energy required to heat 1 mole of water from 0C to 100C:q = mcTwhere q is the heat energy, m is the mass of water, c is the specific heat capacity, and T is the change in temperature.Since we are dealing with 1 mole of water, the mass of water m is equal to its molar mass:m = 18.015 g/molThe change in temperature T is 100C - 0C = 100C.Now, we can calculate the heat energy q :q = 18.015 g/mol * 4.18 J/gC * 100C = 7533.57 J/molNext, we need to account for the heat of fusion Hfus to convert water from liquid to vapor. The heat of fusion for water is given as 40.7 kJ/mol, which is equal to 40700 J/mol.Now, we can estimate the enthalpy of vaporization Hvap by adding the heat energy required to heat the water to its boiling point and the heat of fusion:Hvap = q + Hfus = 7533.57 J/mol + 40700 J/mol = 48233.57 J/molTherefore, the estimated enthalpy of vaporization for water is approximately 48.2 kJ/mol.