Chapter 13, Problem 13.70P

(a)

Interpretation Introduction

Interpretation:

The mole fraction of isopropanol 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 $\text{X}$.

The formula to calculate the mole fraction is as follows:

$\text{Mole}\text{fraction}\left(\text{X}\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}+\text{amount}\left(\text{mol}\right)\text{of}\text{solvent}}$ (1)

(b)

Interpretation Introduction

Interpretation:

The mass percent of isopropanol is to be determined.

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:

$\text{Mass}\text{percent}=\frac{\text{mass}\text{of}\text{solute}}{\text{mass}\text{of}\text{solution}}\times 100$ (3)

(c)

Interpretation Introduction

Interpretation:

The molality of isopropanol is to be determined.

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:

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$ (6)