   Chapter 10, Problem 100E

Chapter
Section
Textbook Problem

The molar heat of fusion of benzene (C6H6) 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?

Interpretation Introduction

Interpretation: The heat of vaporization and heat required to melt 8.25 g benzene should be calculated.

Concept Introduction:

Heat of vaporization:

The energy required for a quantity of liquid substance converted in to into a gases substance is called heat of vaporization.

Explanation

Explanation

Record the given data:

Molar heat of fusion of benzene = 9.92 kJ/mol.

Molar heat of vaporization = 30.7 kJ/mol.

The given molar heat of fusion and molar heat of vaporization of benzene are recorded as shown above.

To calculate the heat required to melt 8.25 g benzene at its normal melting point.

Molecular weight of benzene is 78.11g .

=8.25gbenzene×1molbenzene78.11g×9.92kJ1molebenzene=1.05kJ

• The given heat of fusion of benzene and molecular weight of benzene are plugging in the equation to give heat required to melt 8.25 g benzene.
• The required heat for melt 8.25 g benzene is 1.05kJ

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