Answer:
56.4 kJ
Step-by-step explanation:
First, let's convert the mass of water from grams to moles. We can do this by dividing the mass by the molar mass of water, which is approximately 18 g/mol.
25 g ÷ 18 g/mol ≈ 1.39 mol
So we have 1.39 moles of water that we want to vaporize.
Next, we need to use the molar heat of vaporization to calculate how much energy is required to vaporize one mole of water. The molar heat of vaporization tells us how much energy is needed to vaporize one mole of a substance at a constant temperature and pressure. In this case, the molar heat of vaporization for water is 40.6 kJ/mol.
So, to vaporize 1 mole of water, we need 40.6 kJ of energy.
Finally, we can use this information to calculate how much energy is required to vaporize 1.39 moles of water. We can multiply the energy required to vaporize one mole of water by the number of moles we have:
40.6 kJ/mol × 1.39 mol ≈ 56.4 kJ
Therefore, it would take approximately 56.4 kJ of energy to vaporize 25 g of liquid water at 100°C.
I hope this explanation helps!