# 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?

###### FREE Expert Solution
89% (352 ratings)
###### FREE Expert Solution

The heat needed to melt:

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

89% (352 ratings) ###### 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?

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Based on our data, we think this problem is relevant for Professor Collins' class at OSU.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.