How much energy is required to convert 50.0 g of water at 100°C to 50.0 g of steam at 100°C? (Given: The enthalpy of vaporization of water at 100°C is 40.7 kJ/mol.)

• A. 22.6 kJ
• B. 40.7 kJ
• C. 113 kJ
• D. 203.5 kJ

Answer: (C)

Verified step by step guidance

1. Identify the process: converting liquid water at 100°C to steam at 100°C involves a phase change (vaporization) without a temperature change.
2. Recall that the energy required for a phase change is given by the formula: \(Q = n \times \Delta H_{vap}\), where \(Q\) is the heat energy, \(n\) is the number of moles, and \(\Delta H_{vap}\) is the enthalpy of vaporization.
3. Calculate the number of moles of water using the molar mass of water (approximately 18.0 g/mol): \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0\,\text{g}}{18.0\,\text{g/mol}}\).
4. Multiply the number of moles by the enthalpy of vaporization to find the total energy required: \(Q = n \times 40.7\,\text{kJ/mol}\).
5. Express the final answer in kilojoules, which represents the energy needed to convert 50.0 g of water at 100°C to steam at 100°C.