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)

(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)

(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)

(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)

(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)