Chapter 17, Problem 80A

Interpretation Introduction

Interpretation: The heat required to vaporize 25.0 g of ethanol at its boiling point should be determined.

Concept Introduction: The energy required to change the amount of a substance from liquid to gaseous state is said to be heat of vaporization.

Answer to Problem 80A

The heat required to vaporize 25.0 g of ethanol at its boiling point is $\text{28}\text{.9 kJ}$ .

Explanation of Solution

Given Information:

The molar heat of vaporization of ethanol is $38.6\text{ kJ/mol}$ .

The energy required to vaporize one mole of a liquid at its boiling point is known as the molar heat of vaporization, ${\text{ΔH}}_{\text{vaporization}}$ .

First, convert the mass of ethanol to mol, as:

  $\begin{array}{l}\text{mol of ethanol = }\frac{\text{mass of ethanol}}{\text{Molar mass of ethanol}}\\ \text{mol of ethanol = }\frac{\text{25}\text{.0 g}}{\text{46}\text{.07 g/mol}}\\ \text{mol of ethanol = 0}\text{.542 mol}\end{array}$

The given molar heat of vaporization of ethanol is $38.6\text{ kJ/mol}$ which means to vaporize one mol of ethanol $38.6\text{ kJ}$ of heat is required. So, to vaporize 0.542 mol of ethanol the required heat is:

  $\begin{array}{l}{\text{ΔH = ΔH}}_{\text{vaporization}}\times \text{mol of ethanol}\\ \text{ΔH = 38}\text{.6 kJ/mol}\times \text{0}\text{.542 mol}\\ \text{ΔH = 28}\text{.9 kJ}\end{array}$

Hence, the heat required to vaporize 25.0 g of ethanol at its boiling point is $\text{28}\text{.9 kJ}$ .