Final answer:
The energy required to vaporize 25.0 g of water at 100°C is calculated as 56.44 kJ when using the heat of vaporization (∆Hvap) of water, which is 40.7 kJ/mol.
Step-by-step explanation:
To calculate the energy required to vaporize 25.0 g of H2O at 100°C, given that the ∆Hvap of water is 40.7 kJ/mol, we must first determine the number of moles of water in 25.0 g. The molar mass of water (H2O) is approximately 18.015 g/mol. Using the formula for the number of moles (n = mass / molar mass), we find that n = 25.0 g / 18.015 g/mol = 1.387 mol.
Next, we use the given heat of vaporization (∆Hvap) to calculate the total energy required. The energy (Q) required for vaporizing water is calculated as Q = n × ∆Hvap. Therefore, Q = 1.387 mol × 40.7 kJ/mol = 56.44 kJ.
Thus, the energy needed to vaporize 25.0 g of water at 100°C is 56.44 kJ.