The molar heat capacity of C6H6(ℓ) is 136 J/mol ·◦C and of C6H6(g) is 81.6 J/mol · ◦C.

Question

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 624 g of solid
benzene (C6H6(s)) at 5.5 ◦C into benzene vapor (C6H6(g)) at 100.0 ◦C?

Answer 1

You do all of the problems the same way.
First, since the heat capacity is listed in J/mol or kJ/mol, convert 624 g benzene to moles. mol = grams/molar mass.Then
within a phase, q is
mass x specific heat in that phase x (Tfinal-Tinitial).

Converting a phase (solid to liquid or reverse, liquid to vapor or reverse), q is
mass x heat fusion at melting point or
mass x heat vaporization at boiling point.
Then add all of the qs together.

Answer 2

what is the answer

Answer 3

bruh just do the math for the ut quest