Chapter 10, Problem 32E

In lab you need to prepare at least 100 mL of each of the following solutions. Explain how you would proceed using the given information.

a. 2.0 m KCl in water (density of H₂O = 1.00 g/cm³)

b. 15% NaOH by mass in water (d = 1.00 g/cm³)

c. 25% NaOH by mass in CH₃OH (d = 0.79 g/cm³)

d. 0.10 mole fraction of C₆H₁₂O₆ in water (d = 1.00 g/cm³)

Interpretation Introduction

Interpretation:

The preparation of $\text{100 mL}$ of $2.0m\text{KCl}$ in water has to be explained.

Concept introduction:

Molality:

The gram moles of solute dissolve in kilo gram of solvent is called molality of solution.

  $\text{Molality }\left(\text{g}\right)\frac{\text{moles}\text{of}\text{solute}\text{g}}{\text{kg}\text{of}\text{solvent}}(1)$

Answer to Problem 32E

$\text{15}\text{g}$ of $\text{KCl}$ dissolved in $\text{100 mL}$ of water to prepare a $\text{100 mL}$ of 2.0 $\text{m}\text{KCl}$ solution.

Explanation of Solution

To calculate the mass of $\text{KCl}$ to prepare the 100mL 2.0 $\text{m}\text{KCl}$ solution.

Given:

Molality of $\text{KCl}$ solution = $\text{2}\text{.0}\text{m}$.

  $\begin{array}{l}\text{=}\text{0}\text{.100}\text{kg}\text{water×}\frac{\text{2}\text{.0}\text{mol}\text{KCl}}{\text{kg}}\text{×}\frac{\text{74}\text{.55}\text{g}}{\text{1}\text{mol}}\\ \text{=}\text{15}\text{g}\text{KCl}\end{array}$

• The given water density $\text{1}\text{mL}\text{equal}\text{to}\text{1}\text{g}$ so the weight of water and mole of $\text{KCl}$ and molecular weight of $\text{KCl}$ are plugging in above equation to give a mass of $\text{KCl}$ to prepare a $\text{100 mL}$ 2.0 $\text{m}\text{KCl}$ solution.
• The required mass of $\text{KCl}$ to prepare a 100mL 2.0 $\text{m}\text{KCl}$ solution is $\text{15}\text{g}$.
• Take a volumetric flask and drop sum water and add $\text{15}\text{g}$ of $\text{KCl}$ and make up to $\text{100 mL}$.

Interpretation Introduction

Interpretation:

The preparation of $\text{100 mL}$ of $\text{15}\text{\%}\text{NaOH}$ by mass in water has to be explained.

Concept introduction:

Mass percent:

The relation of gram of solute to gram of solution times 100% is called mass percent.

    $\text{Mass}\text{percent}\text{=}\frac{\text{gram}\text{of}\text{solute}}{\text{gram}\text{of}\text{solution}}\text{×100}\text{(2)}$

Answer to Problem 32E

The $\text{18 g NaOH}$ dissolved in $\text{100 mL}$ of water to prepare a $\text{15\%}$ $\text{NaOH}$ solution.

Explanation of Solution

To calculate the mass of $\text{NaOH}$ to $\text{15\%}$ $\text{NaOH}$ solution.

  $\begin{array}{l}\text{given:}\\ \text{Mass}\text{\%=}\text{15}\\ \text{15}\text{\%}\text{=}\frac{\text{x}}{\text{100+x}}\\ \text{100x=}\text{15x+1500}\\ \text{=}\text{18}\text{g}\end{array}$

• The given mass percent is plugging in equation 2 and rearrange to solve to give required mass of $\text{NaOH}$ to prepared $\text{15\%}$ $\text{NaOH}$ solution.
• From the result $\text{18}\text{g}$ of $\text{NaOH}$ is dissolve in 100 ml of water to prepare a $\text{15\%}$ $\text{NaOH}$ solution.

Interpretation Introduction

Interpretation:

The preparation of $\text{100 mL}$ of $\text{25}\text{\%}\text{NaOH}$ by mass in ${\text{CH}}_{\text{3}}\text{OH}$ has to be explained.

Concept introduction:

Mass percent:

The relation of gram of solute to gram of solution times 100% is called mass percent.

    $\text{Mass}\text{percent}\text{=}\frac{\text{gram}\text{of}\text{solute}}{\text{gram}\text{of}\text{solution}}\text{×100}\text{(2)}$

Answer to Problem 32E

The $\text{26 g}$$\text{NaOH}$ dissolved in $\text{100 mL}$ of methanol to prepare a $\text{25\%}$ methanoic $\text{NaOH}$ solution.

Explanation of Solution

To calculate the mass of $\text{NaOH}$ to prepare a $\text{25\%}$ methanolic $\text{NaOH}$ solution.

Given:

  Density of methanol $\text{=}\text{0}{\text{.79 g/cm}}^{\text{3}}$

  Volume of solution $\text{= 100 mL}$

  $\begin{array}{l}\text{100}\text{mL}\text{of}\text{methanol}\text{mass,}\\ \text{100}\text{mL}{\text{CH}}_{\text{3}}\text{OH=}\text{100×}\frac{\text{0}\text{.79}\text{g}}{\text{mL}}\\ \text{=}\text{79}\text{g}\\ \text{Mass}\text{of}\text{25}\text{\%=}\frac{\text{x}}{\text{79+x}}\text{×100}\\ \text{100(x)=}\text{25x+25(79)}\\ \text{x=}\text{26}\text{g}\end{array}$

• The given density of methanol is multiplied by required volume ($\text{100 mL}$) to give mass of methanol in $\text{100 mL}$.
• The given mass percent and mass of $\text{100 mL}$ methanol are plugging in equation 2 and rearrange to solve to give  required mass of $\text{NaOH}$ to prepared $\text{25\%}$ methanolic $\text{NaOH}$ solution.
• From the result $26\text{g}$ of $\text{NaOH}$ is dissolve in 100 ml of methanol to prepare a $\text{25\%}$ $\text{NaOH}$ solution.

Interpretation Introduction

Interpretation:

The preparation of $\text{100 mL}$ of 0.10 mole fraction of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in water has to be explained.

Concept Introduction:

Mole fraction:

The ratio of given mole of solute to total mole of solvent is called mole fraction.

  $\text{Mole}\text{fraction}\text{=}\frac{\text{mole}\text{of}\text{solute (g)}}{\text{mole}\text{of}\text{solution}(g\text{)}}(3)$

Answer to Problem 32E

Dissolve $\text{110 g}$ ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in $\text{100}\text{mL}$ of ${\text{H}}_{\text{2}}\text{O}$ to prepare a 0.10 mole fraction solution of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$

Explanation of Solution

Calculate the mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ to prepare a 0.10 mole fraction solution of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.

Given:

    Mole fraction =  $\text{0}\text{.10}$

  Molecular weight of water =$\text{18}\text{.02g}$

  Molecular weight of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$$\text{=}\text{108}\text{.2g}$

  $\begin{array}{l}\text{Mole}\text{of}\text{100}\text{mL}\text{water}\text{=}\text{100×}\frac{\text{1}}{\text{18}\text{.02g}}\\ \text{=}\text{5}\text{.55}\text{mole}\\ \text{Mole}\text{fraction}\text{of}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{=}\text{0}\text{.10}\\ {\text{χ}}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\text{=}\frac{\text{x}}{\text{x+5}\text{.55}}\\ \text{x=}\text{0}\text{.10x+0}\text{.56}\\ \text{=}\text{0}\text{.62}{\text{mole}}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\\ \text{Massof}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{=}\text{0}\text{.62×}\frac{\text{180}\text{.2g}}{\text{1}\text{mol}}\\ \text{=}\text{110g}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\end{array}$

• The mole of water is calculated by dividing the volume and molecular weight of water.
• The calculated mole of water and mole fraction of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ are plugging in to equation and rearrange to give a mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.
• The mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is multiplied by molecular weight of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ to give mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ to prepare 0.10 mole fraction solution of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.
• Dissolve $\text{110 g}$ ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in $\text{100}\text{mL}$ of ${\text{H}}_{\text{2}}\text{O}$ to prepare a 0.10 mole fraction solution of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.