Chapter 7, Problem 7.86AP

(a)

Interpretation Introduction

Interpretation:

The volume (in milliliters) of sucrose that is required to prepare $\text{45mL}$ of $\text{4}\text{.0M}$ solution has to be calculated.

Concept Introduction:

Molarity:  Molarity is defined as the mass of solute in one liter of solution.  Molarity is the preferred concentration unit for stoichiometry calculations.  The formula is,

  $\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

  ${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}$

${\text{M}}_{\text{1}}\text{=}$ Initial molarity of the solution

${\text{V}}_{\text{1}}\text{=}$ Initial volume of the solution

${\text{M}}_{\text{2}}\text{=}$ Final molarity of the solution

${\text{V}}_{\text{2}}\text{=}$ Final volume of the solution

Answer to Problem 7.86AP

The volume (in milliliters) of sucrose that is required to prepare $\text{45mL}$ of $\text{4}\text{.0M}$ solution is $\text{36mL}\text{.}$

Explanation of Solution

Given,

  ${\text{M}}_{\text{1}}\text{=5}\text{.0M}$

  ${\text{M}}_{\text{2}}\text{=4}\text{.0M}$

  ${\text{V}}_{\text{2}}\text{=45mL}$

The initial volume of the solution is calculated as,

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{1}}\text{=}\frac{{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{M}}_{\text{1}}}\\ {\text{V}}_{\text{1}}\text{=}\frac{\left(\text{4M}\right)\left(\text{45mL}\right)}{\text{5M}}\\ {\text{V}}_{\text{1}}\text{=36mL}\end{array}$

The volume (in milliliters) of sucrose that is required to prepare $\text{45mL}$ of $\text{4}\text{.0M}$ solution is $\text{36mL}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The volume (in milliliters) of sucrose that is required to prepare $\text{150mL}$ of $\text{0}\text{.5M}$ solution has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.86AP

The volume (in milliliters) of sucrose that is required to prepare $\text{150mL}$ of $\text{0}\text{.5M}$ solution is $\text{15mL}\text{.}$

Explanation of Solution

Given,

  ${\text{M}}_{\text{1}}\text{=5}\text{.0M}$

  ${\text{M}}_{\text{2}}\text{=0}\text{.5M}$

  ${\text{V}}_{\text{2}}\text{=150mL}$

The initial volume of the solution is calculated as,

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{1}}\text{=}\frac{{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{M}}_{\text{1}}}\\ {\text{V}}_{\text{1}}\text{=}\frac{\left(\text{0}\text{.5M}\right)\left(\text{150mL}\right)}{\text{5M}}\\ {\text{V}}_{\text{1}}\text{=15mL}\end{array}$

The volume (in milliliters) that is required to prepare $\text{150mL}$ of $\text{0}\text{.5M}$ solution is $\text{15mL}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The volume (in milliliters) of sucrose that is required to prepare $\text{1}\text{.2L}$ of $\text{0}\text{.025M}$ solution has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.86AP

The volume (in milliliters) that is required to prepare $\text{1}\text{.2L}$ of $\text{0}\text{.025M}$ solution is $\text{6mL}\text{.}$

Explanation of Solution

Given,

  ${\text{M}}_{\text{1}}\text{=5}\text{.0M}$

  ${\text{M}}_{\text{2}}\text{=0}\text{.025M}$

  ${\text{V}}_{\text{2}}\text{=1}\text{.2L}$

Liters is converted to milliliters as,

  $\begin{array}{l}\text{mL=1}\text{.2L×}\frac{\text{1000mL}}{\text{1L}}\\ \text{mL=1200mL}\end{array}$

The initial volume of the solution is calculated as,

  $\begin{array}{l}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{1}}\text{=}\frac{{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{M}}_{\text{1}}}\\ {\text{V}}_{\text{1}}\text{=}\frac{\left(\text{0}\text{.025M}\right)\left(\text{1200mL}\right)}{\text{5M}}\\ {\text{V}}_{\text{1}}\text{=6mL}\end{array}$

The volume (in milliliters) that is required to prepare $\text{1}\text{.2L}$ of $\text{0}\text{.025M}$ solution is $\text{6mL}\text{.}$