How much heat is required to vaporize 10.0 g of water at 100 °C? ΔHvap = 40.7 kJ/mol; M(H2O) = 18.015 g/mol

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Multiple Choice

How much heat is required to vaporize 10.0 g of water at 100 °C? ΔHvap = 40.7 kJ/mol; M(H2O) = 18.015 g/mol

Explanation:
Vaporizing a liquid at its boiling point requires energy equal to the number of moles you’re converting multiplied by the molar enthalpy of vaporization. Here, the mass is 10.0 g and the molar mass of water is 18.015 g/mol, so moles = 10.0 / 18.015 ≈ 0.555 mol. Multiply by the molar enthalpy of vaporization: 0.555 mol × 40.7 kJ/mol ≈ 22.6 kJ. This is the amount of heat needed to turn 0.555 moles of liquid water at 100 °C into steam at the same temperature. The result is 22.6 kJ (three significant figures). The reason this is the correct value is that the heat for a phase change scales with the amount of substance, not just the per-mole energy.

Vaporizing a liquid at its boiling point requires energy equal to the number of moles you’re converting multiplied by the molar enthalpy of vaporization. Here, the mass is 10.0 g and the molar mass of water is 18.015 g/mol, so moles = 10.0 / 18.015 ≈ 0.555 mol. Multiply by the molar enthalpy of vaporization: 0.555 mol × 40.7 kJ/mol ≈ 22.6 kJ. This is the amount of heat needed to turn 0.555 moles of liquid water at 100 °C into steam at the same temperature. The result is 22.6 kJ (three significant figures). The reason this is the correct value is that the heat for a phase change scales with the amount of substance, not just the per-mole energy.

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