In steps. First convert 702 g C6H6 to mols. mols = grams/molar mass.
q1 to melt benzene @ 5.5 C = mols C6H6 x molar heat fusion
q2 to raise C6H6 liquid to it's boiling point = mols C6H6 x heat capacity of liquid x (Tfinal-Tinitial) = ? Where Tfinal is boiling point Temperature and Tinitial is melting point temperature.
q3 to vaporize the C6H6 at the boiling point = mols C6H6 x heat vaporization = ?
q4 to raise T to 100 C = mols C6H6 x specific heat vapor x (Tfinal - Tinitial) = ? where Tfinal is 100 C and Tinitial is boiling point temperature.
qtotal = q1 + q2 + q3 + q4.
Note that some of the units are in J/mol and others in kJ/mol. Keep the units straight and remember that when adding, all of the units must be the same.
Post your work if you get stuck.
The molar heat capacity of C6H6(ℓ) is 136J/mol · ◦C and of C6H6(g) is 81.6 J/mol · ◦C. The molar heat of fusion for benzene is 9.92 kJ/mol and its molar heat of vaporization is 30.8 kJ/mol. The melting point of benzene is 5.5◦C, its boiling point is 80.1◦C, and its molecular weight 78.0 g/mol. How much heat would be required to convert 702 g of solid benzene (C6H6(s)) at 5.5◦C into benzene vapor (C6H6(g)) at 100.0◦C?
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