# Problem: The molar heat of fusion of benzene (C 6H6) is 9.92 kJ/mol. Its molar heat of vaporization is 30.7 kJ/mol. Calculate the heat required to melt 8.25 g benzene at its normal melting point. Calculate the heat required to vaporize 8.25 g benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

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###### FREE Expert Solution

The heat needed to melt:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{n}}{\mathbf{∆}}{{\mathbf{H}}}_{{\mathbf{fusion}}}}$

###### Problem Details

The molar heat of fusion of benzene (C 6H6) is 9.92 kJ/mol. Its molar heat of vaporization is 30.7 kJ/mol. Calculate the heat required to melt 8.25 g benzene at its normal melting point. Calculate the heat required to vaporize 8.25 g benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?