# 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?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Heating and Cooling Curves concept. You can view video lessons to learn Heating and Cooling Curves. Or if you need more Heating and Cooling Curves practice, you can also practice Heating and Cooling Curves practice problems.

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Our tutors rated the difficulty ofThe molar heat of fusion of benzene (C 6H6) is 9.92 kJ/mol. ...as medium difficulty.

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What professor is this problem relevant for?

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

What textbook is this problem found in?

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