Chapter 11, Problem 11.29CP

The following phase diagram shows part of the liquid/vapor phase-transition boundaries for pure ether and a solution of a nonvolatile solute in ether.

1. (a) What is the approximate normal boiling point of pure ether?
2. (b) What is the approximate molal concentration of the solute? [K_b for ether is 2.02 (°C ·kg)/mol.]

Interpretation Introduction

Interpretation:

The approximate normal boiling point of pure ether has to be determined

Answer to Problem 11.29CP

The approximate normal boiling point of pure ether is $\sim {\text{37}}^{\text{o}}\text{C}$

Explanation of Solution

Let us consider the given phase diagrams

The given phase diagram shows part of the liquid and vapour phase-transition boundaries for pure ether and a solution of a non-volatile solute in ether.

Figure 1

The normal boiling point for ether is the temperature where the upper curve intersects so the approximate normal boiling point of pure ether is $\sim {\text{37}}^{\text{o}}\text{C}$

Interpretation Introduction

Interpretation:

The approximate molal concentration of the solute has to be calculated.

Concept introduction:

• Elevation of boiling point:

  The boiling point of the solution is increases when the solute is dissolved in the solvent is called Elevation of boiling point.  It is one of the colligative Properties thus,

$\begin{array}{l}\text{ΔT}\text{=}{\text{iK}}_{\text{b}}{\text{m}}_{\text{solute}}\mathrm{......}(1)\\ \text{ΔT is boiling-point elevation}\\ {\text{K}}_{\text{b}}\text{is}\text{molal boiling-point elevation constant}\\ \text{m}\text{is }\text{molality of the solute}\\ \text{i}\text{is}\text{the}\text{van't Hoff factor}\end{array}$

Answer to Problem 11.29CP

The approximate molal concentration of the solute is $\text{1}\text{.5}\text{m}$

Explanation of Solution

Given data,

${\text{K}}_{\text{b}}\text{for}\text{ether}\text{is}\text{2}\text{.02}{\text{(}}^{\text{o}}\text{C}\text{.kg)/mol}$

The approximate molal concentration of the solute can be calculated as

$\begin{array}{l}{\text{ΔT}}_{\text{b}}\text{=}{\text{K}}_{\text{b}}\text{.m}\\ \\ \text{m}\text{=}\frac{{\text{ΔT}}_{\text{b}}}{{\text{K}}_{\text{b}}}\\ \text{boiling-point elevation}{\text{ΔT}}_{\text{b}}\text{is}{\text{3}}^{\text{o}}\text{C}\\ \text{m}\text{=}\frac{{\text{3}}^{\text{o}}\text{C}}{\text{2}\text{.02}\frac{{}^{\text{o}}\text{C}\text{•}\text{kg}}{\text{mol}}}\text{1}\text{.5}\text{mol/kg}\\ \text{m}\text{=}\text{1}\text{.5}\text{m}\end{array}$

The approximate molal concentration of the solute is $\text{1}\text{.5}\text{m}$