Final answer:
To estimate the enthalpy of vaporization of water at 100°C for 685 grams, first calculate the moles of water and then multiply by the molar heat of vaporization. Approximately 1548.442 kJ of heat is required to vaporize 685 g of H₂O at 100°C.
Step-by-step explanation:
The enthalpy of vaporization of water at 100°C can be estimated using the given molar heat of vaporization and the mass of water to be vaporized. Since 1 mol of water has a mass of approximately 18 grams, you can first determine the number of moles in 685 grams of water by dividing 685 g by 18 g/mol. Then, the total heat absorbed, or the heat of vaporization, can be calculated by multiplying the number of moles by the molar heat of vaporization (40.7 kJ/mol or 540 cal/g).
To calculate the heat required to vaporize 685 g of water at 100°C:
- Calculate the moles of water: 685 g ÷ 18 g/mol ≈ 38.06 mol.
- Calculate the total heat absorbed by multiplying the moles of water (38.06 mol) by the molar heat of vaporization (40.7 kJ/mol): 38.06 mol × 40.7 kJ/mol ≈ 1548.442 kJ.
Therefore, approximately 1548.442 kJ of heat is necessary to vaporize 685 g of H₂O at 100°C.