Chapter 7, Problem 7.69AP

(a)

Interpretation Introduction

Interpretation:

The procedure that is required to prepare $\text{4}\text{.8\%}\left(\text{w/v}\right)$ acetic acid in water has to be calculated.

Concept Introduction:

Weight/volume percent:  Weight/volume percent is defined as the concentration expressed in terms of grams of solute dissolved in volume of solution.  Weight/volume percent is given as,

  $\left(\text{w/v}\right)\text{\%=}\frac{\text{Mass}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{mL}\right)}\text{×100\%}$

Explanation of Solution

Given,

Weight/volume percent of the solution=$\text{4}\text{.8\%}\left(\text{w/v}\right)$

Volume of the solution=$\text{250mL}$

The mass of the solute is calculated as,

  $\begin{array}{l}\left(\text{w/v}\right)\text{\%=}\frac{\text{Mass}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{mL}\right)}\text{×100\%}\\ \text{4}\text{.8\%=}\frac{\text{x}}{\text{250mL}}\text{×100\%}\\ \text{x=12g}\end{array}$

The amount of solute that is required to prepare $\text{4}\text{.8\%}\left(\text{w/v}\right)$ acetic acid in water is $\text{12}\text{g}\text{.}$

Twelve gram of acetic acid is added to flask and water is added to bring volume to $\text{250mL}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The procedure that is required to prepare $\text{22\%}\left(\text{v/v}\right)$ ethyl acetate in water has to be calculated.

Concept Introduction:

Volume/volume percent:  Volume/volume percent is defined as the concentration expressed in terms of volume of solute dissolved in volume of solution.  Volume/volume percent is given as,

  $\left(\text{v/v}\right)\text{\%=}\frac{\text{Volume}\text{of}\text{solute(in}\text{mL)}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{mL}\right)}\text{×100\%}$

Explanation of Solution

Given,

Volume/volume percent of the solution=$\text{22\%}$

Volume of the solution=$\text{250mL}$

The amount of solute is calculated as,

  $\begin{array}{l}\left(\text{v/v}\right)\text{\%=}\frac{\text{Volume}\text{of}\text{solute(in}\text{mL)}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{mL}\right)}\text{×100\%}\\ \text{22\%=}\frac{\text{x}}{\text{250mL}}\text{×100\%}\\ \text{x=55mL}\end{array}$

The amount of solute that is required to prepare $\text{22\%}\left(\text{v/v}\right)$ ethyl acetate in water is $\text{55mL}\text{.}$

Fifty-five milliliter of ethyl acetate is added to the flask and water is added to bring the volume to $\text{250mL}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The procedure that is required to prepare $\text{2}\text{.5M}$ sodium chloride 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,

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

Explanation of Solution

Given,

Molarity of the solution=$\text{2}\text{.5M}$

Volume of the solution=$\text{250mL}$

Milliliters is converted to liters as,

  $\begin{array}{l}\text{L=250mL×}\frac{\text{1L}}{\text{1000mL}}\\ \text{L=0}\text{.250L}\end{array}$

The amount of the solute is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{2}\text{.5M=}\frac{\text{x}}{\text{0}\text{.250L}}\\ \text{x=0}\text{.63mol}\end{array}$

Moles of sodium chloride is converted into grams,

  $\begin{array}{l}\text{Grams=0}\text{.63Mole×}\frac{\text{58}\text{.44g}}{\text{1}\text{mole}}\\ \text{Grams=36}\text{.817g}\end{array}$

The amount of solute that is required to prepare $\text{2}\text{.5M}$ sodium chloride solution is $\text{37g}\text{.}$

Thirty-seven grams of sodium chloride is added to the flask and water is added to bring the volume to $\text{250mL}\text{.}$