15. Alice made a large pitcher of lemonade and placed several ice cubes in to cool the lem However, 100 grams of water vapour from the air condensed onto the outside of the p much energy did the condensing water vapour add to the pitcher of lemonade? (Heat for watcr is 40.66kJ/mol. A. 226 kJ B. 7.32 kJ C. 1.37 kJ D. 5.55 kJ

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15. Alice made a large pitcher of lemonade and placed several ice cubes in to cool the lem However, 100 grams of water vapour from the air condensed onto the outside of the p much energy did the condensing water vapour add to the pitcher of lemonade? (Heat for watcr is 40.66kJ/mol. A. 226 kJ B. 7.32 kJ C. 1.37 kJ D. 5.55 kJ

Answer 1

Here's how to solve this problem:1. **Calculate the moles of water:** The molar mass of water (H₂O) is approximately 18 g/mol. So, 100 g of water is equal to 100 g / 18 g/mol = 5.56 moles.2. **Calculate the energy released:** The heat of condensation is equal to the heat of vaporization but with the opposite sign. Since the problem gives the heat of vaporization as 40.66 kJ/mol, the heat of condensation is -40.66 kJ/mol. Therefore, the energy released by the condensation of 5.56 moles of water is 5.56 mol * -40.66 kJ/mol = -226 kJ.3. **Consider the energy added to the lemonade:** The question asks how much energy the condensing water *added* to the lemonade. Since the condensation *releases* 226 kJ of energy, this is the amount of energy *added* to the lemonade.Therefore, the answer is

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