Chapter 13, Problem 64QP

How many liters of the antifreeze ethylene glycol $\text{[C}{\text{H}}_{\text{2}}\text{(OH)C}{\text{H}}_{\text{2}}\text{(OH)]}$ would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is - $\text{20°C}$ ? Calculate the boiling point of this water- ethylene glycol mixture. (The density of ethylene glycol is 1.11 g/mL)

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Interpretation Introduction

Interpretation:

The volume (in liters) of the antifreeze ethylene glycol and the boiling point of the mixture are to be evaluated.

Concept introduction:

The expression for the freezing point depression is as follows:

$\begin{array}{c}\Delta T_{\text{f}}=i{K}_{\text{f}}m\\ T_{\text{f}}^{°}-T_{\text{f}}=i{K}_{\text{f}}m\end{array}$

Here, $\Delta T_{\text{f}}$ denotes the freezing point depression of solute, $i$ denotes van’tHoff factor, $K_{\text{f}}$ denotes the freezing point depression constant $\left({1.86}^{\circ }\text{C/}m\right)$ and $m$ is the molality of the solute.

The freezing point of pure solute is represented as $T_{\text{f}}^{°}$ and the freezing point of the solution is represented as $T_{\text{f}}$.

The expression for the boiling point elevation is as follows:

$T_{\text{b}}-T_{\text{b}}^{\circ }=i{K}_{\text{b}}m$

Here, the boiling point of pure solute is represented as $T_{\text{b}}^{°}$, the boiling point of solution is representedas $T_{\text{b}}$, $i$ denotes van’t Hoff factor, $K_{\text{b}}$ denotes the boiling point elevation constant $\left(0.52°\text{C/}m\right)$, and $m$ is the molality of the solute.

Answer to Problem 64QP

Solution: $3.93\text{ L}$ and ${105.6}^{\circ }\text{C}$.

Explanation of Solution

Given information:

$\begin{array}{l}{V}_{water}=6.50\text{ L}\\ T_{\text{f}}^{°}=0°\text{C}\\ T_{\text{f}}=-20°\text{C}\\ \text{Density of ethylene glycol =1}\text{.11 g/mL}\end{array}$

Ethylene glycol is non-ionizing, so van’t Hoff factor, $i=1$.

Evaluate molality of ethylene glycol by using the equation given below:

$T_{\text{f}}^{°}-T_{\text{f}}=i{K}_{\text{f}}m$

Substitute $-{20}^{\circ }\text{C}$ for $T_f$, ${1.86}^{\circ }\text{C/}m$ for $K_f$, and $1$ for $i$,

$\begin{array}{c}{0}^{\circ }\text{C}-\left(-{20}^{\circ }\text{C}\right)=1\times \left({1.86}^{\circ }\text{C/}m\right)\times m\\ {20}^{\circ }\text{C}=\left({1.86}^{\circ }\text{C/}m\right)\times m\\ m=10.8m\end{array}$

Therefore, the molality of ethylene glycolis $10.8m$.

Density of water is $1\text{ kg}/\text{L, so, 6}\text{.50 L }\times \text{ 1 kg/1L=}\text{6}\text{.5 kg}$

Hence,

$\begin{array}{c}10.75\text{mol/ kg = moles of solute}/6.5\text{kg}\\ \text{mole of solute =10}\text{.75 mol/kg }\times 6.5\text{ kg}\\ \text{=69}\text{.9 mol}\end{array}$

The molar mass of ethylene glycol is $62.07g/mol$

The mass of ethylene glycol required to add in $6.5\text{ kg}$ of water can be evaluated as follows:

$\begin{array}{c}\text{mass of ethylene glycol}=69.9\cancel{\text{mol}}\left(\text{62}\text{.07 }\frac{\text{g}}{\cancel{\text{mol}}}\right)\\ =4.36\times {10}^3\text{ g}\end{array}$

The density of ethylene glycol is $\text{1}\text{.11 g/mL}$.

The volume of ethylene glycol can be evaluated as follows:

$d=\frac{m}{V}$

Rearrange the equation and substitute $4.36\times {10}^3\text{ g}$ for $m$ and $\text{1}\text{.11 g/mL}$ for $d$.

$\begin{array}{c}V=\frac{m}{d}\\ =\frac{\text{4}\text{.36}\times {\text{10}}^3\text{ g}}{\text{1}\text{.11 g/mL}}\\ =\left(\text{4}\text{.36}\times {\text{10}}^3\cancel{\text{g}}\right)\left(\frac{1\cancel{\text{mL}}}{1.11\cancel{\text{g}}}\right)\left(\frac{1\text{ L}}{1000\cancel{\text{mL}}}\right)\\ =3.93\text{ L}\end{array}$

So, the volume of ethylene glycol is $3.93\text{ L}$.

Calculate the boiling point of the solution by using the equation given below:

$T_{\text{b}}-T_{\text{b}}^{\circ }=i{K}_{\text{b}}m$

Substitute ${100}^{\circ }\text{C}$ for $T_{\text{b}}^{\circ }$, ${0.52}^{\circ }\text{C/}m$ for $K_{\text{b}}$, $10.8m$ for $m$, and $1$ for $i$,

$\begin{array}{c}{T}_{\text{b}}-{100}^{\circ }\text{C}=\left(1\right)\left({0.52}^{\circ }\text{C/}m\right)\left(10.8m\right)\\ T_{\text{b}}={105.6}^{\circ }\text{C}\end{array}$

Therefore, the boiling point of the solution is ${105.6}^{\circ }\text{C}$.

Conclusion

The volume of ethylene glycol is $3.93\text{ L}$ and the boiling point of the solution is ${105.6}^{\circ }\text{C}$.