Chapter 13, Problem 66QRT

Interpretation Introduction

Interpretation:

The boiling point and the freezing point of the solution have to be calculated.

Concept introduction

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$. That is water changes from liquid phase to gas phase at temperature ${\text{100}}^{\text{o}}\text{C}$.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}$.

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point.

  ${\text{T}}_{\text{b}}-$ Boiling point of the solution.

  ${\text{T}}_{\text{b}}^{\text{°}}-$ Boiling point of pure solvent.

TheBoiling point elevation${\text{(ΔT}}_{\text{b}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point.

  ${\text{K}}_{\text{b}}-$ Molal boiling point constant.

  $m-$ Molality of the solution.

Freezing point is the temperature at which liquid turns into solid.

TheFreezing point depression${\text{(ΔT}}_{\text{f}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

${\text{ΔT}}_{\text{f}}-$ Change in freezing point.

${\text{K}}_{\text{f}}-$ Molal freezing point constant.

$m-$ Molality of the solution.

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.