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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  1. 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.

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