Answer:
Option 1 is correct.
ΔH for the condensation of 1 kg of Acetone is - 501 KJ.
Explanation:
Molar mass of acetone = 58.08 g/mol
Number of moles of acetone that condenses = mass/molar mass = 1000/58.08 = 17.22 moles
1 mole of Acetone vaporizer with an enthalpy change of 29.1 KJ.
1 mole of Acetone will condense with an enthalpy change of - 29.1 KJ (since condensation is the exact reverse of vaporization)
That is, vaporization is given by,
Acetone(l) ----> Acetone(g) ΔH = 29.1 KJ
Then condensation will be given by,
Acetone(g) ----> Acetone(l) ΔH = -29.1 KJ
So,
1 mole of Acetone will condense with an enthalpy change of - 29.1 KJ
17.22 moles of Acetone will condense with an enthalpy change of 17.22 × - 29.1 KJ = - 501 KJ.
The first option is correct.
