Chapter 16, Problem 89A

(a)

Interpretation Introduction

Interpretation: The amount of hydrogen peroxide, ${\text{H}}_2{\text{O}}_2$ present in a 3% $250\text{mL}$ solution of hydrogen peroxide is to be calculated.

Concept Introduction: The concentration of solute in an aqueous solution is expressed by using the term volume percentage. Volume percentage is the ratio of the volume of a solute to the total volume of the solution and then multiplied by a hundred to make the value in percentage.

The formula used for the determination of concentration or volume percentage is,

  $\text{Percentage volume}=\frac{\text{volume of solute}}{\text{total volume of solution}}$

Answer to Problem 89A

The amount of hydrogen peroxide present in a $250\text{mL}$ 3% solution of hydrogen peroxide is $7.5\text{g}$ .

Explanation of Solution

The volume percentage of a solution details the ratio of the volume of solute and the total volume of solution.

So, a 3% ${\text{H}}_2{\text{O}}_2$ solution contains $3\text{g}$ of the solute, ${\text{H}}_2{\text{O}}_2$ in $100\text{mL}$ of the solution.

For a $250\text{mL}$ , the amount of ${\text{H}}_2{\text{O}}_2$ is,

  $3\times \frac{250}{100}=7.5\text{g}$

That is, $7.5\text{g}$ of ${\text{H}}_2{\text{O}}_2$ is present in $250\text{mL}$ of the solution.

(b)

Interpretation Introduction

Interpretation: The molarity of a $250\text{mL}$ bottle of 3% ${\text{H}}_2{\text{O}}_2$ solution is to be calculated.

Concept Introduction: Molarity is the ratio of the number of moles of solute to the volume of solution in liters. It is denoted by $M$ .

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solvent}}$   ...... (1)

The expression for molarity is as follows:

  $\text{M}=\frac{\text{moles of solute}}{\text{Volume of solution(L)}}$   ...... (2)

Answer to Problem 89A

The molarity of the $250\text{mL}$ bottle of 3% ${\text{H}}_2{\text{O}}_2$ solution is $0.88\text{M}$ .

Explanation of Solution

Molar mass of ${\text{H}}_2{\text{O}}_2$ is $\text{34}\text{.0147 g/mol}$ .

Substitute, $7.5\text{g}$ for the mass of ${\text{H}}_2{\text{O}}_2$ and $\text{34}\text{.0147 g/mol}$ for molar mass of ${\text{H}}_2{\text{O}}_2$ in equation (1).

  $\begin{array}{c}\text{moles}=\frac{7.5\text{g}}{\text{34}\text{.0147 g/mol g/mol}}\\ =0.22\text{ mol}\end{array}$

Convert $\text{mL}$ to $\text{L}$ .

  $\begin{array}{c}250\text{mL}=250\times \frac{1\text{L}}{1000\text{mL}}\\ =0.25\text{L}\end{array}$

Substitute, $0.22moles$ for ${\text{H}}_2{\text{O}}_2$ and $\text{0}\text{.25}\text{L}$ for the volume of solution in equation (2).

  $\begin{array}{c}\text{M}=\frac{0.22\text{ moles }}{\text{0}\text{.25L}}\\ =0.88\text{ M}\end{array}$

So, the molarity of the $250\text{mL}$ bottle of 3% ${\text{H}}_2{\text{O}}_2$ solution is $0.88\text{M}$ .