This question tests the calculation of energy released during freezing (liquid to solid transition). When benzene freezes, it releases energy equal to its enthalpy of fusion multiplied by the number of moles. The energy released = ΔHfus × moles = 9.95 kJ/mol × 0.80 mol = 7.96 kJ. A common error is to use the enthalpy value directly without accounting for the partial mole (choice C: 9.95 kJ), treating the molar enthalpy as if it were the total energy for any amount. For phase transitions, always multiply the per-mole enthalpy value by the actual number of moles present.

This question tests the ability to calculate the energy absorbed during melting using the enthalpy of fusion and the mass of the substance. Convert 27.0 g of ice to moles with 18.0 g/mol, yielding 1.50 moles. Multiply by 6.01 kJ/mol to get 9.02 kJ absorbed, reflecting the endothermic nature of breaking hydrogen bonds in ice. This matches choice B, the energy for complete melting at 0°C. Choice C, 6.01 kJ, tempts those who omit the mole conversion, confusing the per-mole value with the total energy. A transferable strategy is to identify whether the phase change is endothermic or exothermic by considering if it's increasing or decreasing molecular disorder.

This question tests the ability to calculate the energy absorbed during sublimation using the enthalpy of sublimation and the mass of the substance. Convert 10.0 g of iodine to moles using 254 g/mol, resulting in about 0.0394 moles. Multiply by 62.4 kJ/mol to find approximately 2.46 kJ absorbed, as sublimation is endothermic, directly transitioning solid to gas and requiring energy to break bonds. This corresponds to choice B, the energy input for the phase change. Choice D, 62.4 kJ, attracts those who skip the mole calculation and use the enthalpy value alone, mistakenly treating it as per gram instead of per mole. A transferable strategy is to remember that enthalpies of phase changes are molar values, so always scale by the number of moles involved.

This question tests the ability to calculate the energy absorbed during melting using the enthalpy of fusion and the mass of the substance. Start by converting the 36.0 g of water to moles using its molar mass of 18.0 g/mol, resulting in exactly 2.00 moles. Multiply this by the enthalpy of fusion, 6.01 kJ/mol, to find 12.0 kJ absorbed, as melting is endothermic and energy is needed to overcome lattice forces in the solid. This corresponds to choice A, indicating the total energy for the phase transition from solid to liquid at constant temperature. Choice E, 24.0 kJ, is a common distractor from doubling the correct value, perhaps from mistakenly using twice the moles or confusing fusion with vaporization enthalpies. A transferable strategy is to ensure the sign of energy reflects whether the process is endothermic (absorbed) or exothermic (released) based on the phase change direction.

This question tests the ability to calculate the energy absorbed during vaporization using the enthalpy of vaporization and the mass of the substance. Divide 40.0 g of methanol by 32.0 g/mol to obtain 1.25 moles. Multiply by 35.3 kJ/mol, resulting in 44.1 kJ absorbed, as vaporization demands energy to separate liquid molecules into gas. This is choice A, quantifying the endothermic process at the boiling point. Choice D, 35.3 kJ, is a distractor from forgetting to multiply by moles and using the raw enthalpy, underestimating the energy for the given mass. A transferable strategy is to use dimensional analysis to confirm that units cancel correctly to kJ.

This question tests the ability to calculate the energy released during freezing using the enthalpy of fusion and the mass of the substance. Divide the 15.0 g of benzene by its molar mass of 78.0 g/mol to get approximately 0.192 moles. Multiply by the enthalpy of fusion, 9.95 kJ/mol, yielding about 1.91 kJ released, since freezing is exothermic as molecules form a more ordered solid structure. This matches choice B, the energy liberated in the liquid-to-solid phase change. Choice C, 9.95 kJ, is a distractor for those who neglect the mole conversion and apply the enthalpy to the mass directly, confusing molar quantities with mass-based ones. A transferable strategy is to double-check unit consistency, ensuring mass is converted to moles when using molar enthalpies.

This question tests the ability to calculate the energy released during condensation using the enthalpy of vaporization and the mass of the substance. Convert the 22.4 g to moles by dividing by 56.0 g/mol, giving 0.400 moles. Since condensation is the reverse of vaporization, the energy released is the same magnitude, so multiply 0.400 moles by 28.0 kJ/mol to get 11.2 kJ released, as it's exothermic and strengthens intermolecular forces. This aligns with choice A, representing the heat given off during the gas-to-liquid transition. Choice B, 28.0 kJ, tempts those who forget to convert to moles and use the enthalpy directly, misunderstanding that enthalpy is per mole, not per gram. A transferable strategy is to recognize that for reverse processes like condensation or freezing, the energy magnitude is identical to the forward process but opposite in sign.

This question tests the skill of calculating the energy released during freezing using the enthalpy of fusion. The energy released by the 12.0 g liquid sample is 12.0 g × 400 J/g = $4.80×10^3$ J, as freezing is exothermic. This represents the heat given off as the substance solidifies and forms a more ordered structure. Choice A is the correct answer. Choice B, $1.20×10^3$ J, could tempt due to the misconception of using one-tenth of the mass or dividing instead of multiplying, yielding a smaller value. A transferable strategy for phase change energy problems is to identify the process as endothermic or exothermic and use the absolute value of ΔH for magnitude calculations.

This question tests the skill of calculating energy absorbed during ice melting using enthalpy of fusion. Energy absorbed is calculated as mass times enthalpy of fusion: Energy = 24 g × 334 J/g = 8,016 J. When ice melts, energy is absorbed to break the hydrogen bonds holding water molecules in the rigid ice structure, allowing them to move more freely as liquid. The temperature remains at 0°C throughout melting as all energy goes into the phase change. Answer D (334 J) shows the common mistake of using only the enthalpy value without considering the 24 g mass, forgetting that ΔHfus represents energy per gram not total energy. To solve phase change problems systematically, always identify the phase change direction, determine if energy is absorbed or released, then multiply the given mass by the appropriate enthalpy value.

This question tests the skill of calculating energy changes during phase transitions using enthalpy of fusion. To find the energy absorbed during melting, we multiply the mass of the sample by the enthalpy of fusion: Energy = mass × ΔHfus = 25 g × 200 J/g = 5,000 J. During melting, energy is absorbed to break intermolecular forces while the temperature remains constant at the melting point. A common misconception leading to answer B (200 J) is using only the enthalpy value without multiplying by mass, forgetting that enthalpy of fusion is given per gram. When solving phase change problems, always multiply the given enthalpy per gram by the total mass of the sample to find the total energy change.