Chapter 15, Problem 25A

Interpretation Introduction

Interpretation:

The molarity of each ion present in FeCl₃ solution has to be calculated.

Concept Introduction:

The composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methodsare usedto describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one liter of the solution.

In expression,

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

Answer to Problem 25A

Molarity of $\text{F}{\text{e}}^{\text{3}+}$ ion is 0.62 M and molarity of $\text{C}{\text{l}}^{-}$ ion is 1.86 M.

Explanation of Solution

  

Data given: Mass of $\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}$ = 1.01 gm

Volume of $\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}$ solution = 10.0 mL = 0.01 L

  $\begin{array}{l}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}}\end{array}$

   $\begin{array}{c}=\frac{\text{1.01}\text{g}\text{m}}{\text{1}\text{62.2}\text{g}\text{m}/\text{m}\text{o}\text{l}}=\text{0}.\text{0}\text{0}\text{6}\text{2}\text{m}\text{o}\text{l}\end{array}$

   

    $\begin{array}{c}=\frac{\text{0}.\text{0}\text{0}\text{6}\text{2}\text{m}\text{o}\text{l}\text{e}}{ \text{0}.\text{0}\text{1}\text{l}\text{i}\text{t}\text{r}\text{e}}=\text{0}.\text{6}\text{2}\text{M}\end{array}$

  $\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}\to \text{F}{\text{e}}^{\text{3}+}+\text{3}\text{C}{\text{l}}^{-}$

$\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}$ gives 1 $\text{F}{\text{e}}^{\text{3}+}$ ion and 3 $\text{C}{\text{l}}^{-}$ ions in the solution. 1 mole of $\text{F}\text{e}\text{C}{\text{l}}_{\text{3}}$ gives 1 mole of $\text{F}{\text{e}}^{\text{3}+}$ ion and 3 mol of $\text{C}{\text{l}}^{-}$ ions. Therefore, the molarity of $\text{F}{\text{e}}^{\text{3}+}$ ion is 0.62 M and the molarity of $\text{C}{\text{l}}^{-}$ ion is 1.86 M.