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Question
Chemistry
Posted 6 months ago
The heat of vaporization of water at 100C100^{\circ} \mathrm{C} is 40.66 kJ/mol40.66 \mathrm{~kJ} / \mathrm{mol}. Calculate the quantity of heat that is absorbed/released when 9.00 g9.00 \mathrm{~g} of steam condenses to liquid water at 100C100^{\circ} \mathrm{C}.
20.3 kJ20.3 \mathrm{~kJ} of heat are released.
81.3 kJ81.3 \mathrm{~kJ} of heat are absorbed.
81.3k81.3 \mathrm{k} of heat are released.
20.3 kJ20.3 \mathrm{~kJ} of heat are absorbed.
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Answer from Sia
Posted 6 months ago
B
Key Concept
Heat of vaporization and mass-to-mole conversion
Explanation
To find the heat absorbed or released during the condensation of steam, you need to convert the mass of water to moles using its molar mass (approximately 18.02mathrm g/mol18.02 \\mathrm{~g/mol} for water), and then multiply by the heat of vaporization. The calculation is as follows: frac9.00mathrm g18.02mathrm g/moltimes40.66mathrm kJ/mol=20.3mathrm kJ\\frac{9.00 \\mathrm{~g}}{18.02 \\mathrm{~g/mol}} \\times 40.66 \\mathrm{~kJ/mol} = 20.3 \\mathrm{~kJ}. Since condensation is the opposite of vaporization, the heat is released, not absorbed. However, the question asks for the heat absorbed, which is the same magnitude but opposite in sign, hence 20.3mathrm kJ20.3 \\mathrm{~kJ} of heat are absorbed.

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