Chapter 13, Problem 13.24QP

(a)

Interpretation Introduction

Interpretation:

The molality and the molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution and also the volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ has to be determined.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.24QP

The molality of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is $3.83m$

Explanation of Solution

The molality of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is calculated.

Given,

$\begin{array}{c}Masspercentofethanolinaqueoussolution=0.15\%\\ \text{Density of }aqueoussolution\text{= }\text{0}\text{.984}\text{g/mL}\\ Molecularmassofethanol=46.07g\end{array}$

$\begin{array}{c}Massofethanolinthesolution=0.15\times 100=15g\\ Massofwater=100g-15g=85g=0.085kg\end{array}$

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ =\frac{15g}{46.07g}\\ =0.3256mol\end{array}$

Therefore,

The molality of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is,

$\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}\\ =\frac{0.3256mol}{0.085kg}\\ =3.83m\end{array}$

The molality of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is $3.83m$

(b)

Interpretation Introduction

Interpretation:

The molality and the molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution and also the volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ has to be determined.

Concept introduction:

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Answer to Problem 13.24QP

The molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is $3.20\text{M}$

Explanation of Solution

The molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is calculated.

Given,

$\begin{array}{c}Masspercentofethanolinaqueoussolution=0.15\%\\ \text{Density of }aqueoussolution\text{= }\text{0}\text{.984}\text{g/mL}\\ Molecularmassofethanol=46.07g\end{array}$

To convert volume from gram

$\text{100g ×}\frac{\text{1mL}}{\text{0}\text{.984}\text{g}}\text{ = 101}\text{.6}\text{mL= 0}\text{.1016}\text{L}$

$\begin{array}{c}Massofethanolinthesolution=0.15\times 100=15g\\ \text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ =\frac{15g}{46.07g}\\ =0.3256mol\end{array}$

Therefore,

The molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is,

$\begin{array}{c}\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}\\ \text{=}\frac{\text{0}\text{.3256}\text{mol}}{\text{0}\text{.1016}\text{L}}\\ =3.20\text{M}\end{array}$

The molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is $3.20\text{M}$

(c)

Interpretation Introduction

Interpretation:

The molality and the molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution and also the volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ has to be determined.

Concept introduction:

Number of moles from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Answer to Problem 13.24QP

The volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ is $78.1mL$

Explanation of Solution

The volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ is calculated.

Given,

$\begin{array}{c}Masspercentofethanolinaqueoussolution=0.15\%\\ \text{Density of }aqueoussolution\text{= }\text{0}\text{.984}\text{g/mL}\\ Molecularmassofethanol=46.07g\end{array}$

The molarity of ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ solution is $3.20\text{M}$ which means $1L$ of the solution contains $3.20\text{moles}$ of ethanol.

Therefore,

The volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ is

$\begin{array}{c}Thevolumeofsolution=0.250mol\times \frac{1L}{3.20\text{mol}} \\ =0.0781L\\ =78.1mL\end{array}$

The volume of solution containing $0.250mole$ ethanol ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH)}$ is $78.1mL$