Chapter 13, Problem 72GQ

Interpretation Introduction

Interpretation:

Acetic acid’s mole fraction, molality, concentration in parts per million has to be determined and the reason for not able to calculate the molarity of this solution from the information provided also has to be given.

Concept introduction:

• Mass percentage is one way to represent the concentration of an element in a compound or a component in a mixture. Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  $\begin{array}{l}\\ {\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}\end{array}$

• $\text{Mass}\text{of}\text{the}\text{substance}=\text{Number}\text{of}\text{moles}\text{×}\text{Molar}\text{mass}\text{of}\text{the}\text{substance}$
• $\text{Number}\text{of}\text{mole}=\frac{\text{}\text{Mass}\text{of}\text{the}\text{substance}\text{in}\text{gram}}{\text{Molar}\text{mass}}$
• Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

• Molarity (M): Molarity is number of moles of the solute present in the one litter of the solution.

  $\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

• Parts per million (ppm) concentrations: It is the ratio of the number of grams of solute for every one million grams of solution.

    $Partspermillion(ppm)=\frac{Massofsolute}{Massofsolution}\times {10}^6$