Chapter 13, Problem 77QRT

(a)

Interpretation Introduction

Interpretation:

The mass of the glycol that have been added has to be determined.

Concept introduction

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

The Freezing point depression ${\text{(ΔT}}_{\text{f}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}{\text{m}}_{solute}{\text{i}}_{solute}$.

Where,

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

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

$m-$ Molality of the solution.

${\text{i}}_{solute}$- van’t Hoff factor.

Answer to Problem 77QRT

The mass of the glycol that have been added is $2500g{C}_2{H}_6{O}_2.$

Explanation of Solution

Using Freezing point of the solution:

    $\begin{array}{c}{\text{ΔT}}_{\text{f }}{\text{=T}}_{\text{f }}^{\text{°}}\left(\text{solution}\right){\text{-T}}_{\text{f }}\left(\text{solvent}\right)\\ \\ \text{= -15}{\text{.0}}^{\circ }\text{C}\text{+}\left(\text{-0}{\text{.0}}^{\circ }\text{C}\right)\\ \\ \text{= -15}{\text{.0}}^{\circ }\text{C}\text{.}\end{array}$

Given data is as follows:

Using the formula as shown below,

  $\begin{array}{l}{\text{ΔT}}_{\text{f }}{\text{= K}}_{\text{f}}{\text{m}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{i}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}\\ \\ {\text{m}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}\text{=}\frac{{\text{ΔT}}_{\text{f }}}{{\text{ K}}_{\text{f}}}\text{=}\frac{\text{-15}{\text{.0}}^{\text{o}}\text{C}}{\left(\text{-1}{\text{.86}}^{\text{o}}\text{C}\text{.}\text{kg}{\text{.mol}}^{\text{-1}}\right)}\text{=}\text{8}\text{.06}\text{mol/kg}\text{.}\end{array}$

Mass of the glycol:

  $\text{5}\text{.0}\text{kg}\text{water}\text{×}\frac{\text{8}\text{.06}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{1}\text{kg}\text{water}}\text{×}\frac{\text{62}\text{.07}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}\text{=}2500g{C}_2{H}_6{O}_2.$

Therefore, the mass of the glycol that have been added is $2500g{C}_2{H}_6{O}_2.$

(b)

Interpretation Introduction

Interpretation:

The boiling point of the coolant mixture has 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.

The Boiling 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.

Answer to Problem 77QRT

Boiling point of the solution is $104.12^{o}C.$

Explanation of Solution

Given data as follows: Mass of ethylene glycol from part (a) $\text{=2500 g}$.

Mass of water is $\text{100 g}$.

Value of ${\text{K}}_{\text{b}}\left(water\right)\text{=}\text{0}{\text{.512}}^{\text{o}}\text{C kg/mol}$

Calculate mole of solute (urea):

  $\begin{array}{c}\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}\\ \\ \text{=}\frac{\text{2500}\cancel{\text{g}}}{62.07\cancel{\text{g}}\text{/mol}}\text{= 40}\text{.3mol}\\ \\ \left[\because \text{ethylene glycol}\text{Molecular}\text{mass}=62.07\text{ g/mol}\right].\end{array}$

Substituting the values of ethylene glycol mass and its molecular mass into the moles–mass conversion formula, the moles of the substance was obtained.

Calculate molality of the solution:

  $\begin{array}{c}\text{molality =}\frac{\text{40}\text{.3mol}}{\text{5}\text{.0 kg water}}\\ \\ \text{=}\text{8}\text{.06}\text{m}\text{.}\end{array}$

Substituting in the value of moles of the solute and mass of solvent in kilogram; molality of the solution was obtained.

Calculate Boiling point of the solution:

Using the boiling point formula as follows,

  $\begin{array}{c}{\text{ΔT}}_{\text{b }}{\text{= K}}_{\text{b}}{\text{m}}_{solute}\text{= (0}\text{.512°C/m)}\left(\text{8}\text{.06}\text{m}\text{.}\right)\text{ = 4}{\text{.12}}^{\circ }\text{C}\\ \\ {\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}\end{array}$

  $\begin{array}{c}{\text{T}}_{\text{b}}={\text{ΔT}}_{\text{b}}+{\text{T}}_{\text{b}}^{\text{°}}\\ \\ ={100}^{\circ }\text{C}+\text{ 4}{\text{.12}}^{\circ }\text{C}\\ \\ =104.12^{o}C.\end{array}$

Substituting the values of molality, the boiling point was calculated.

Therefore, the Boiling point of the solution is $104.12^{o}C.$