Chapter 11, Problem 9ALQ

Interpretation Introduction

Interpretation: The molality or molarity dependent on temperature is should be explained and reason for the usage of molality in freezing-point depression and boiling point elevation calculations are explained.

Concept introduction:

• Elevation of boiling point:

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

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

• Depression in freezing point:

  The freezing point the solution is decreases when the solute is dissolved in the solvent is called Elevation of boiling point. it is one of the colligative Properties thus

                                 $\begin{array}{l}\text{ΔT}\text{=}{\text{iK}}_{\text{f}}{\text{m}}_{\text{solute}}\mathrm{......}(2)\\ \text{ΔT is boiling-point elevation}\\ {\text{K}}_{\text{f}}\text{is}\text{molal freezing-point depression constant}\\ \text{m}\text{is }\text{molality of the solute}\\ \text{i}\text{is}\text{the}\text{van't Hoff factor}\end{array}$

• Molarity:

                  The gram moles of solute in liter of solvent is called molarity and in is the term of  concentration.

                            $\text{Molarity}\text{=}\frac{\text{solute}\text{mass}\text{g}}{\text{solvent}\text{volume}\text{L}}$

• Molality:

                  The gram moles of solute in kilogram of solvent is called molality and in is the term of  concentration.

                            $\text{Molality}\text{=}\frac{\text{solute}\text{mass}\text{g}}{\text{solvent}\text{mass}\text{kg}}$