How much energy is required to vaporize 10.00 grams of water at its boiling point (100 °C)? (The enthalpy of vaporization of water at 100 °C is 40.7 kJ/mol.)

• A. 4.07 kJ
• B. 1.80 kJ
• C. 407 kJ
• D. 22.6 kJ

Answer: (D)

Verified step by step guidance

1. Identify the given data: mass of water = 10.00 g, enthalpy of vaporization (\( \Delta H_{vap} \)) = 40.7 kJ/mol, and boiling point = 100 °C.
2. Calculate the number of moles of water using the molar mass of water (\( H_2O \)), which is approximately 18.02 g/mol, using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{10.00\, \text{g}}{18.02\, \text{g/mol}} \]
3. Use the enthalpy of vaporization to find the energy required to vaporize the given amount of water by multiplying the moles by \( \Delta H_{vap} \): \[ q = \text{moles} \times \Delta H_{vap} \]
4. Make sure the units are consistent and the final energy will be in kJ since \( \Delta H_{vap} \) is given in kJ/mol.
5. Interpret the result as the amount of energy required to vaporize 10.00 grams of water at 100 °C.