Chapter 14, Problem 104IAE

Interpretation Introduction

(a)

Interpretation:

Equivalence of the magnitude of molality and the molarity in a dilute aqueous solution should be demonstrated.

Concept introduction:

The molarity, molality and density of a solution can be calculated as follows:

$\begin{array}{l}\text{Molarity }\left(\text{M}\right)\text{ = }\frac{\text{amount of solute (moles)}}{\text{volume of solution (liters)}}\\ \text{Molality (m) = }\frac{\text{amount of solute (moles)}}{\text{mass of solvent (kilograms)}}\\ \text{Density = }\frac{\text{Mass}}{\text{Volume}}\end{array}$

Interpretation Introduction

(b)

Interpretation:

In a dilute solution, proportionality of the solute mole fraction to the molality should be demonstrated.

Concept introduction:

In a dilute solution, solvent concentration is larger than the solute concentration.

$\begin{array}{l}\text{Molality (m) = }\frac{\text{amount of solute (moles)}}{\text{mass of solvent (kilograms)}}\\ {\text{Solute mole fraction = X}}_B=\frac{{\text{n}}_B}{{\text{n}}_A+{\text{n}}_B}\\ {\text{Solvent mole fraction = X}}_A=\frac{{\text{n}}_A}{{\text{n}}_A+{\text{n}}_B}\\ \text{Number of moles = }\frac{\text{Mass (W)}}{\text{Molecular mass (M)}}\end{array}$

Interpretation Introduction

(c)

Interpretation:

In a dilute aqueous solution, proportionality of the solute mole fraction to the molarity should be demonstrated.

Concept introduction:

In an aqueous solution, solvent is water. A dilute aqueous solution, water concentration is larger than the solute concentration.

$\text{Molarity }\left(\text{M}\right)\text{ = }\frac{\text{amount of solute (moles)}}{\text{volume of solution (liters)}}$

$\begin{array}{l}{\text{Solute mole fraction = X}}_B=\frac{{\text{n}}_B}{{\text{n}}_A+{\text{n}}_B}\\ {\text{Solvent mole fraction = X}}_A=\frac{{\text{n}}_A}{{\text{n}}_A+{\text{n}}_B}\\ \text{Number of moles = }\frac{\text{Mass (W)}}{\text{Molecular mass (M)}}\end{array}$