Final answer:
To find the heat required to vaporize 2.58 kg of water, convert the mass to grams and calculate the number of moles. Multiply the number of moles by the heat of vaporization to find the total heat required. The correct answer, given the data, is not listed among the options provided.
Step-by-step explanation:
The question asks to calculate the amount of heat required to vaporize 2.58 kg of water at its boiling point, given the heat of vaporization of water (∆Hvap) at the boiling point is 40.7 kJ/mol. To find the heat Q required for vaporization, we can use the formula Qv = m * Lv, where m is the mass of the water and Lv is the latent heat of vaporization.
First, we need to convert the mass of water from kg to grams because the heat of vaporization provided is per mole and the molar mass of water is expressed in grams. Thus, 2.58 kg of water is equivalent to 2580 g. The molar mass of water (H2O) is approximately 18.015 g/mol. To find the number of moles of water, we divide the mass in grams by the molar mass:
n = 2580 g / 18.015 g/mol = 143.2 mol
Now, using the heat of vaporization (40.7 kJ/mol), we can calculate the heat required to vaporize the water:
Q = n * ∆Hvap = 143.2 mol * 40.7 kJ/mol = 5825.44 kJ
None of the provided answers (A) 58.7 kJ, (B) 104.66 kJ, (C) 141.53 kJ, or (D) 165.2 kJ match the correct calculation, which is 5825.44 kJ.