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Answer :
Option (a), The heat required to vaporize 1.00 mol of benzene at its boiling point is 34.1 kJ/mol.
The heat of vaporization is the amount of heat required to transform a substance from the liquid state to the gas state.
The formula for the heat required to vaporize the benzene can be given as:
Q = n*ΔHvap
Where,
Q = heat required to vaporize the benzene
ΔHvap = heat of vaporization = 34.1 kJ/mol
n = number of moles = 1.00 mol
Now, substitute the values in the above equation:
Q = 1.00 mol x 34.1 kJ/mol
Q = 34.1 k
Option A is the correct answer.
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