Chapter 13, Problem 13.77P

(a)

Interpretation Introduction

Interpretation:

The volume percent of ethylene glycol is to be calculated.

Concept introduction:

The volume percent is defined as the volume of the component divided by the volume of solution, multiplied by 100.

The formula to calculate the volume percent of the solution is as follows:

  $\text{Volume percent}=\left(\frac{\text{Volume of component}}{\text{Volume of solution}}\right)\left(100\text{ \%}\right)$        (1)

The formula to calculate the density is as follows:

  $\text{Density}\left(\rho \right)=\frac{\text{Mass }\left(M\right)}{\text{Volume }\left(V\right)}$        (2)

Answer to Problem 13.77P

The volume percent of ethylene glycol is $50.61\%$.

Explanation of Solution

Rearrange equation (2) to calculate the mass as follows:

  $\text{Mass}=\left(\text{Density}\right)\left(\text{Volume}\right)$        (3)

Consider the volumes of ethylene glycol and water to be $1\text{L}$.

Substitute $1.114\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of ethylene glycol.

  $\begin{array}{c}\text{Mass of ethylene glycol}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1.114\text{g}}{1\text{mL}}\right)\\ =1114\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\\ =1\times {10}^3\text{g}\end{array}$

The formula to calculate the moles of the compound is as follows:

  $\text{Moles of compound}=\frac{\text{Given mass}}{\text{Molar mass}}$        (4)

Substitute $1114\text{g}$ for the given mass and $62.07\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  $\begin{array}{c}\text{Moles of ethylene glycol}=\left(1114\text{g}\right)\left(\frac{1\text{mol}}{62.07\text{g}}\right)\\ =17.94747865\text{mol}\end{array}$

Substitute $1\times {10}^3\text{g}$ for the given mass and $18.02\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of water.

  $\begin{array}{c}\text{Moles of water}=\left(1\times {10}^3\text{g}\right)\left(\frac{1\text{mol}}{18.02\text{g}}\right)\\ =55.49389567\text{mol}\end{array}$

Rearrange equation (2) to calculate the volume as follows:

  $\text{Volume}=\frac{\text{Mass}}{\text{Density}}$        (5)

The formula to calculate the mass of the solution is as follows:

  $\text{Mass of solution}=\left(\text{Mass of solute}+\text{Mass of solvent}\right)$        (6)

Substitute $1114\text{g}$ for the mass of solute and $1\times {10}^3\text{g}$ for the mass of solvent in equation (6).

  $\begin{array}{c}\text{Mass of solution}=\left(1114\text{g}\right)+\left(1\times {10}^3\text{g}\right)\\ =\text{2114}\text{g}\end{array}$

Substitute $\text{2114}\text{g}$ for the mass and $1.070\text{g/mL}$ for the density in equation (5) to calculate the volume of the solution.

  $\begin{array}{c}\text{Volume of solution}=\left(2114\text{g}\right)\left(\frac{1\text{mL}}{1.070\text{g}}\right)\left(\frac{1\text{L}}{{10}^3\text{mL}}\right)\\ =1.97570\text{L}\end{array}$

Substitute $1\text{L}$ for the volume of component and $1.97570\text{L}$ for the volume of solution in equation (1) to calculate the volume percent of ethylene glycol.

  $\begin{array}{c}\text{Volume \% of ethylene glycol}=\left(\frac{1\text{L}}{1.97570\text{L}}\right)\left(100\text{ \%}\right)\\ =50.61497\text{\%}\\ \approx 50.61\%.\end{array}$

Conclusion

The volume percent is a concentration term that changes with the change in temperature as it includes the volumes of component and volume of solution.

(b)

Interpretation Introduction

Interpretation:

The mass percent of ethylene glycol is to be calculated.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture, multiplied by 100.

The formula to calculate the mass percent is as follows:

  $\text{Mass}\text{percent}=\left(\frac{\text{mass}\text{of}\text{solute}}{\text{mass}\text{of}\text{solution}}\right)\left(100\text{ \%}\right)$        (7)

The formula to calculate the density is as follows:

  $\text{Density}\left(\rho \right)=\frac{\text{Mass }\left(M\right)}{\text{Volume }\left(V\right)}$        (2)

Answer to Problem 13.77P

$52.7\%$ is the volume percent of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be $1\text{L}$.

Substitute $1.114\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of ethylene glycol.

  $\begin{array}{c}\text{Mass of ethylene glycol}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1.114\text{g}}{1\text{mL}}\right)\\ =1114\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\\ =1\times {10}^3\text{g}\end{array}$

Substitute $1114\text{g}$ for the mass of solute and $1\times {10}^3\text{g}$ for the mass of solvent in equation (6).

  $\begin{array}{c}\text{Mass of solution}=1114\text{g}+1\times {10}^3\text{g}\\ =\text{2114}\text{g}\end{array}$

Substitute $1114\text{g}$ for the mass of solute and $2114\text{g}$ for the mass of solution in equation (7).

  $\begin{array}{c}\text{Mass \% of ethylene glycol}=\left(\frac{1114\text{g}}{2114\text{g}}\right)\left(100\text{ \%}\right)\\ =52.6963\%\\ \approx 52.7\%.\end{array}$

Conclusion

The mass percent is a concentration term that does not depend on the temperature as it depends only on the masses of solute and solution.

(c)

Interpretation Introduction

Interpretation:

The molarity of the ethylene glycol is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

  $\text{Molarity}\left(M\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$        (8)

The formula to calculate the density is as follows:

  $\text{Density}\left(\rho \right)=\frac{\text{Mass }\left(M\right)}{\text{Volume }\left(V\right)}$        (2)

Answer to Problem 13.77P

$9.08M$ is the molarity of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be $1\text{L}$.

Substitute $1.114\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of ethylene glycol.

  $\begin{array}{c}\text{Mass of ethylene glycol}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1.114\text{g}}{1\text{mL}}\right)\\ =1114\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\\ =1\times {10}^3\text{g}\end{array}$

Substitute $1114\text{g}$ for the given mass and $62.07\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  $\begin{array}{c}\text{Moles of ethylene glycol}=\left(1114\text{g}\right)\left(\frac{1\text{mol}}{62.07\text{g}}\right)\\ =17.94747865\text{mol}\end{array}$

Substitute $1114\text{g}$ for the mass of solute and $1\times {10}^3\text{g}$ for the mass of solvent in equation (6).

  $\begin{array}{c}\text{Mass of solution}=\left(1114\text{g}\right)+\left(1\times {10}^3\text{g}\right)\\ =\text{2114}\text{g}\end{array}$

Substitute $\text{2114}\text{g}$ for the mass and $1.070\text{g/mL}$ for the density in equation (5) to calculate the volume of the solution.

  $\begin{array}{c}\text{Volume of solution}=\left(2114\text{g}\right)\left(\frac{1\text{mL}}{1.070\text{g}}\right)\left(\frac{1\text{L}}{{10}^3\text{mL}}\right)\\ =1.97570\text{L}\end{array}$

Substitute $17.94747865\text{mol}$ for the amount of solute and $1.97570\text{L}$ for the volume of solution in equation (8) to calculate the molarity of ethylene glycol.

  $\begin{array}{c}\text{Molarity of ethylene glycol}=\frac{17.94747865\text{mol}}{1.97570\text{L}}\\ =9.08411M\\ \approx 9.08M.\end{array}$

Conclusion

Molarity is a concentration term that changes with the change in temperature as it includes a volume term which is temperature dependent.

(d)

Interpretation Introduction

Interpretation:

The molality of ethylene glycol is to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $\text{m}$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the molality of the solution is as follows:

  $\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$        (9)

The formula to calculate the density is as follows:

  $\text{Density}\left(\rho \right)=\frac{\text{Mass }\left(M\right)}{\text{Volume }\left(V\right)}$        (2)

Answer to Problem 13.77P

$17.9m$ is the molality of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be $1\text{L}$.

Substitute $1.114\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of ethylene glycol.

  $\begin{array}{c}\text{Mass of ethylene glycol}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1.114\text{g}}{1\text{mL}}\right)\\ =1114\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\\ =1\times {10}^3\text{g}\end{array}$

Substitute $1114\text{g}$ for the given mass and $62.07\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  $\begin{array}{c}\text{Moles of ethylene glycol}=\left(1114\text{g}\right)\left(\frac{1\text{mol}}{62.07\text{g}}\right)\\ =17.94747865\text{mol}\end{array}$

Substitute $17.94747865\text{mol}$ for the amount of solute and $1\times {10}^3\text{g}$ for the mass of solvent in equation (9) to calculate the molality of ethylene glycol.

  $\begin{array}{c}\text{Molarity of ethylene glycol}=\left(\frac{17.94747865\text{mol}}{1\times {10}^3\text{g}}\right)\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =17.94747865\text{}m\\ \approx 17.9m.\end{array}$

Conclusion

Molality is a concentration term that does not change with the change in temperature as it depends on the amount of solute and mass of the solvent.

(e)

Interpretation Introduction

Interpretation:

The mole fraction of ethylene glycol is to be determined.

Concept introduction:

The mole fraction is defined as the ratio of the number of moles of solute to the total number of moles in the mixture. It is represented by $X$.

The formula to calculate the mole fraction is as follows:

  $\text{Mole}\text{fraction}\left(X\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\left(\begin{array}{l}\text{amount}\left(\text{mol}\right)\text{of}\text{solute}+\\ \text{amount}\left(\text{mol}\right)\text{of}\text{solvent}\end{array}\right)}$        (10)

The formula to calculate the density is as follows:

  $\text{Density}\left(\rho \right)=\frac{\text{Mass }\left(M\right)}{\text{Volume }\left(V\right)}$        (2)

Answer to Problem 13.77P

$0.244$ is the mole fraction of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be $1\text{L}$.

Substitute $1.114\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of ethylene glycol.

  $\begin{array}{c}\text{Mass of ethylene glycol}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1.114\text{g}}{1\text{mL}}\right)\\ =1114\text{g}\end{array}$

Substitute $1\text{g/mL}$ for the density and $1\text{L}$ for the volume in equation (3) to calculate the mass of water.

  $\begin{array}{c}\text{Mass of water}=\left(1\text{L}\right)\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\\ =1\times {10}^3\text{g}\end{array}$

Substitute $1114\text{g}$ for the given mass and $62.07\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  $\begin{array}{c}\text{Moles of ethylene glycol}=\left(1114\text{g}\right)\left(\frac{1\text{mol}}{62.07\text{g}}\right)\\ =17.94747865\text{mol}\end{array}$

Substitute $1\times {10}^3\text{g}$ for the given mass and $18.02\text{g/mol}$ for the molar mass in equation (4) to calculate the moles of water.

  $\begin{array}{c}\text{Moles of water}=\left(1\times {10}^3\text{g}\right)\left(\frac{1\text{mol}}{18.02\text{g}}\right)\\ =55.49389567\text{mol}\end{array}$

Substitute $17.94747865\text{mol}$ for the amount of solute and $55.49389567\text{mol}$ for the amount of solvent in equation (10) to calculate the mole fraction of ethylene glycol.

  $\begin{array}{c}\text{Mole fraction of ethylene glycol}=\frac{17.94747865\text{mol}}{\left(17.94747865\text{mol}+55.49389567\text{mol}\right)}\\ =0.244378\\ \approx \mathrm{0.244.}\end{array}$

Conclusion

Mole fraction is a concentration term that does not depend on the temperature as it depends only on the amounts of solute and solvent.