Chapter 15, Problem 64E

Interpretation Introduction

Interpretation:

The amount of kilojoules of energy required to vaporize a $\text{28}\text{.6-g}$ sample of liquid benzene at its normal boiling point of $\text{80}°\text{C}$ is to be stated.

Concept Introduction:

Heat of vaporization is defined as the amount of heat required to vaporize $\text{1 g}$ of substance. If “q” is the amount of heat used to vaporize and “m” is the gram of the substance, then the heat of vaporization will be equal to the $\frac{\text{q}}{\text{m}}$.