chem help please?

Question

The molar heat of fusion of benzene is 9.92 kJ/mol. Its molar heat of vaporization is 30.7 kJ/mol. Calculate the heat required to melt 8.01 g of benzene at its normal melting point.
kJ
Calculate the heat required to vaporize 8.01 g of benzene at its normal boiling point
kJ
Why is the heat of vaporization more than three times the heat of fusion?

Answer 1

Use H = mL and H = ms del T