Question

How many kilojoules of heat are absorbed when 0.38 g of chloroethane (C₂H₅Cl, bp 12.3°C) vaporizes at its normal boiling point? (The molar heat of vaporization of chloroethane is 26.4 kJ/mol.)

Options:
A) 0.62 kJ
B) 1.20 kJ
C) 2.08 kJ
D) 4.48 kJ

Answer 1

When 0.38 g of chloroethane vaporizes, approximately 0.1556 kJ of heat is absorbed, given the substance's molar heat of vaporization of 26.4 kJ/mol. This value doesn't match any of the provided options perfectly, but the closest is option A (0.62 kJ).

Step-by-step explanation:

The question asks how many kilojoules of heat are absorbed when 0.38 g of chloroethane vaporizes at its normal boiling point, given that the molar heat of vaporization of chloroethane is 26.4 kJ/mol.

To find the answer, we first need to calculate the number of moles of chloroethane in 0.38 g. The molar mass of chloroethane (C2H5Cl) is approximately 64.5 g/mol (12*2 + 1*5 + 35.5). Moles of chloroethane in 0.38 g = 0.38 g / 64.5 g/mol ≈ 0.00589 mol. Now, using the molar heat of vaporization, we multiply the number of moles by the heat per mole to find the total heat absorbed:

Heat absorbed = 0.00589 mol * 26.4 kJ/mol ≈ 0.1556 kJ.

Therefore, option A (0.62 kJ) is too high; option B (1.20 kJ) is also too high; option C (2.08 kJ) is even higher and not correct; and option D (4.48 kJ) is far from correct. The closest approximation to the calculated value of 0.1556 kJ would be option A (0.62 kJ) if we had to choose from the given options, keeping in mind that none of them are precise.