Chapter 20, Problem 1ASA

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:

$5\mathrm{mL}4.0\mathrm{M\; acetone} +10\mathrm{mL}1.0\mathrm{M\; HCl}\mathrm{ +}10\mathrm{mL}0.0050\mathrm{M}{\mathrm{ I}}_2 +25\mathrm{mL}{\mathrm{H}}_2\mathrm{O}$

a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles $A=M_A\times V$, where $M_A$ is the molarity of A and V is the volume in liters of the solution of A that was used.

__________ moles acetone

b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again,

$M_A=\frac{\mathrm{moles\; of }A}{V\mathrm{of\; soln.\; in\; liters}}$

__________ M acetone

c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of ${\mathrm{H}}^{+}$ ion and ${\mathrm{I}}_2$ as in the original mixture?

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

  $\text{5mL 4}\text{.0 M acetone + 10 mL 1}\text{.0 M HCl + 10 mL 0}{\text{.0050 M I}}_{\text{2}}{\text{+ 25 mL H}}_{\text{2}}\text{O}$

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

  $\begin{array}{l}\text{Total volume = 5 + 10 + 10 + 25}\hfill \\ \text{ = 50 mL}\hfill \\ \text{ = 0}\text{.05 L}\hfill \end{array}$

The calculation of moles of acetone is shown below:

  $\begin{array}{c}\text{moles}\text{=}\text{molarity}\text{×}\text{volume}\\ \text{=}\text{4}{\text{molL}}^{\text{-1}}\text{×}\text{0}\text{.005}\text{L}\\ \text{=}\text{0}\text{.02}\text{mol}\end{array}$

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

  $\text{5mL 4}\text{.0 M acetone + 10 mL 1}\text{.0 M HCl + 10 mL 0}{\text{.0050 M I}}_{\text{2}}{\text{+ 25 mL H}}_{\text{2}}\text{O}$

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

  $\begin{array}{l}\text{Total volume = 5 + 10 + 10 + 25}\hfill \\ \text{ = 50 mL}\hfill \\ \text{ = 0}\text{.05 L}\hfill \end{array}$

The calculation of molarity of acetone is shown below:

  $\begin{array}{c}\text{Molarity}\text{=}\frac{\text{moles}}{\text{volume}\left(\text{L}\right)}\\ \text{=}\frac{\text{0}\text{.02}\text{mol}}{\text{0}\text{.05}\text{L}}\\ \text{=}\text{0}\text{.4}\text{M}\end{array}$

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

  $\text{5mL 4}\text{.0 M acetone + 10 mL 1}\text{.0 M HCl + 10 mL 0}{\text{.0050 M I}}_{\text{2}}{\text{+ 25 mL H}}_{\text{2}}\text{O}$

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

  $\text{20}\text{mL}\text{of}\text{4M}\text{acetone}\text{+}\text{10}\text{mL}\text{of}\text{1M}\text{HCl}\text{+}\text{10}\text{mL}\text{of}\text{0}\text{.005}\text{M}{\text{I}}_{\text{2}}\text{+}\text{10}\text{mL}{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

  $\text{20}\text{mL}\text{of}\text{4M}\text{acetone}\text{+}\text{10}\text{mL}\text{of}\text{1M}\text{HCl}\text{+}\text{10}\text{mL}\text{of}\text{0}\text{.005}\text{M}{\text{I}}_{\text{2}}\text{+}\text{10}\text{mL}{\text{H}}_{\text{2}}\text{O}$