Chapter 17.3, Problem 25SP

Interpretation Introduction

Interpretation: The amount of heat absorbed when a certain amount of chloroethane vaporizes at its normal boiling point.

Concept Introduction: The molar enthalpy of vaporization is the amount of energy that must be added to 1 mole of liquid substance to transform it into vapor at its normal boiling point.

Answer to Problem 25SP

The amount of heat absorbed when $\text{0}\text{.46 g}$ of chloroethane vaporizes at its normal boiling point is $\text{0}\text{.175 kJ}$

Explanation of Solution

The molar heat of vaporization of chloroethane at $12.3\text{°C}$ is $24.7\text{ kJ/mol}$ .

This means to convert $1\text{ mole}$ of chloroethane at $12.3\text{°C}$ to vaporize, $24.7\text{ kJ}$ of heat must be absorbed.

Amount of chloroethane ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl)}$ given = $\text{0}\text{.46 g}$

Moles of substance is calculated by dividing the amount of substance by the molar mass of the substance.

Moles of chloroethane = $\frac{\text{0}\text{.46 g }}{\text{64}\text{.51 g/mol}}=0.0071\text{ mol }$

  $1\text{ mole}$ of chloroethane at $12.3\text{°C}$ requires = $\text{24}\text{.7 kJ}$ of heat

Thus, $0.0071\text{ mol}$ of chloroethane at $12.3\text{°C}$ requires = $\frac{\text{24}\text{.7 kJ}}{\text{1 mol}}\times 0.0071\text{ mol= 0}\text{.175 kJ}$