Chapter 12, Problem 12.69QP

(a)

Interpretation Introduction

Interpretation:

Which of the given aqueous solution has higher boiling point, higher freezing point and the lower vapor pressure have to be explained.

Concept introduction

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

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

Where,

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

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

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

Boiling point elevation$(\Delta T_b)$is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

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.

Freezing point depression${\text{(ΔT}}_{\text{f}}\text{)}$ is distinction between freezing point of the pure solvent ${\text{(T}}_{\text{f}}^{\text{°}}\text{)}$ and the freezing point of the solution ${\text{(T}}_{\text{f}}\text{)}$.

${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

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

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

m- Molality of the solution

Vapor pressure is the pressure excreted by the molecules at the surface of the liquid in the closed container.

Answer to Problem 12.69QP

${\text{CaCl}}_{\text{2}}$Solution has higher boiling point

Explanation of Solution

Explanation of${\text{CaCl}}_{\text{2}}$solution has higher boiling point than urea

${\text{CaCl}}_{\text{2}}$is an ionic compound and thus strong electrolyte in water. We know that strong electrolytes completely dissociates in water (not form ion pair, van’t Hoff factor i=3). Hence the total concentration should be $\text{1}\text{.05m}$ ($\text{3× 0}\text{.35m}$) which larger value compared to concentration of Urea ( $\text{0}\text{.90m}$ ). More concentration shows that more particles of the solute in the solution. Therefore ${\text{CaCl}}_{\text{2}}$ shows higher boiling point.

(b)

Interpretation Introduction

Interpretation:

Which of the given aqueous solution has higher boiling point, higher freezing point and the lower vapor pressure have to be explained.

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.

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

Where,

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

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

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

Boiling point elevation$(\Delta T_b)$is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

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.

Freezing point depression${\text{(ΔT}}_{\text{f}}\text{)}$ is distinction between freezing point of the pure solvent ${\text{(T}}_{\text{f}}^{\text{°}}\text{)}$ and the freezing point of the solution ${\text{(T}}_{\text{f}}\text{)}$.

${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

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

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

m- Molality of the solution

Vapor pressure is the pressure excreted by the molecules at the surface of the liquid in the closed container.

Answer to Problem 12.69QP

${\text{CaCl}}_{\text{2}}$has higher freezing point

Explanation of Solution

Explanation of Urea has higher freezing point depression than ${\text{CaCl}}_{\text{2}}$

Freezing point depends on the molality and van’t Hoff factor of the solute. In this case,

${\text{CaCl}}_{\text{2}}$has more concentration( i = 3, $\text{3× 0}\text{.35m = 1}\text{.05m}$) compare to Urea. So ${\text{CaCl}}_{\text{2}}$has highest freezing point depression.

(c)

Interpretation Introduction

Interpretation:

Which of the given aqueous solution has higher boiling point, higher freezing point and the lower vapor pressure have to be explained.

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.

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

Where,

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

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

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

Boiling point elevation$(\Delta T_b)$is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

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.

Freezing point depression${\text{(ΔT}}_{\text{f}}\text{)}$ is distinction between freezing point of the pure solvent ${\text{(T}}_{\text{f}}^{\text{°}}\text{)}$ and the freezing point of the solution ${\text{(T}}_{\text{f}}\text{)}$.

${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

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

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

m- Molality of the solution

Vapor pressure is the pressure excreted by the molecules at the surface of the liquid in the closed container.

Answer to Problem 12.69QP

${\text{CaCl}}_{\text{2}}$solution has lower vapor pressure

Explanation of Solution

Explanation of${\text{CaCl}}_{\text{2}}$ has lower vapour pressure than urea

${\text{CaCl}}_{\text{2}}$has lower vapour pressure because ${\text{CaCl}}_{\text{2}}$ is an ionic compound and thus strong electrolyte in water. We know that strong electrolytes completely dissociates in water (not form ion pair, van’t Hoff factor i=3). Hence the total concentration should be $\text{1}\text{.05m}$ ($\text{3× 0}\text{.35m}$) which larger value compared to concentration of Urea ( $\text{0}\text{.90m}$ ) which is non-electrolyte.