Chapter 13, Problem 59QRT

(a)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ has to be calculated.

Concept introduction:

Mass fraction: The mass of a single solute divided by the total mass of all solutes and solvent in the solution.

  $\text{Mass fraction of A = }\frac{\text{Mass of A}}{\text{ Mass of}\text{A + Mass of}\text{B + Mass of}\text{C+ }\mathrm{...}}.$

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by $100\%$

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}\text{.}$

Answer to Problem 59QRT

The mass fraction of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $0.0586$ and weight percent of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is calculated as $5.86\%.$

Explanation of Solution

• The Mass fraction:

  $\text{Mass fraction of A = }\frac{\text{Mass of A}}{\text{ Mass of}\text{A + Mass of}\text{B + Mass of}\text{C+ }\mathrm{...}}$.

Given information as: Mass of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\text{14}\text{.0 g}$ and pure water is $\text{ 225}\text{g}\text{.}$

Mass fraction of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$$\text{= }\frac{{\text{Mass of K}}_{\text{2}}{\text{CrO}}_{\text{4}}}{\text{Mass of}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{ + Mass of}\text{water }}$

  $\text{= }\frac{\text{14}\text{.0 g}}{\left(\text{14}\text{.0g}\text{+ 225g}\right)}\text{ = }0.0586.$

Hence, the mass fraction of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $0.0586.$

• The Weight percent:

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}\text{.}$

Given information as: Mass of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\text{14}\text{.0 g}$ and pure water is $\text{ 225}\text{g}\text{.}$

  $\begin{array}{l}{\text{weight \% of K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{=}\frac{{\text{Mass of K}}_{\text{2}}{\text{CrO}}_{\text{4}}}{\text{ Mass of}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{ + Mass of}\text{water }}\text{×100\%}\\ \\ \text{=}\left(\text{0}\text{.0586}\right)\text{×100\% =}5.86\%.\end{array}$

The weight percent of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is calculated as shown above. Hence, the weight percent obtained is $5.86\%.$

(b)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of ethanol has to be calculated.

Concept introduction:

Refer to (a)

Answer to Problem 59QRT

The mass fraction of ethanol is $0.0836$ and weight percent of ethanol is calculated as $8.36\%.$

Explanation of Solution

• The Mass fraction:

  $\text{Mass fraction of A = }\frac{\text{Mass of A}}{\text{ Mass of}\text{A + Mass of}\text{B + Mass of}\text{C+ }\mathrm{...}}$.

Given information as: Mass of ethanol is $\text{4}\text{.56 g}$ ; benzene is $\text{50}\text{.0}\text{g}\text{.}$

Mass fraction of ethanol $\text{= }\frac{\text{Mass of ethanol}}{\text{Mass of}\text{ethanol + Mass of}\text{benzene }}$

  $\text{= }\frac{\text{4}\text{.56}\text{g}}{\left(\text{4}\text{.56}\text{g}\text{+ 50}\text{.0g}\right)}\text{ = 0}\text{.0836}\text{.}$

Hence, the mass fraction of $\text{ethanol}$ is $0.0836.$

• The Weight percent:

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$.

Given information as: Mass of ethanol is $\text{4}\text{.56 g}$ ; benzene is $\text{50}\text{.0}\text{g}\text{.}$

  $\begin{array}{l}\text{weight \% of ethanol}\text{=}\frac{\text{Mass of ethanol}}{\text{ Mass of}\text{ethanol}\text{+}\text{Mass of}\text{benzene }}\text{×100\%}\\ \\ \text{=}\left(\text{0}\text{.0836}\right)\text{×100\% = }8.36\%.\end{array}$

The weight percent of ethanol is calculated as shown above. Hence, the weight percent obtained is $8.36\%.$

(c)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of methanol has to be calculated.

Concept introduction:

Refer to (a)

Answer to Problem 59QRT

The mass fraction of methanol is $0.144$ and weight percent of methanol is calculated as $30.8\%.$

Explanation of Solution

• The Mass fraction:

  $\text{Mass fraction of A = }\frac{\text{Mass of A}}{\text{ Mass of}\text{A + Mass of}\text{B + Mass of}\text{C+ }\mathrm{...}}$.

Given information as: Mass of methanol is $\text{15}\text{.0 g}$ and pure ethanol $\text{89}\text{.0 }\text{g}\text{.}$

Mass fraction of methanol $\text{= }\frac{\text{Mass of methanol}}{\text{Mass of}\text{methanol+ Mass of}\text{ethanol }}$

  $\text{= }\frac{\text{15}\text{.0 g}}{\left(15.0\text{g}\text{+89}\text{.0g}\right)}\text{ = }\mathrm{0.144.}$

Hence, the mass fraction of methanol is $0.144.$

• The Weight percent:

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$.

Given information as: Mass of methanol is $\text{15}\text{.0 g}$ and pure ethanol $\text{89}\text{.0 }\text{g}\text{.}$

  $\begin{array}{l}\text{weight \% of methanol}\text{=}\frac{\text{Mass of methanol}}{\text{ Mass of}\text{methanol}+\text{Mass of}\text{ethanol }}\text{×100\%}\\ \\ \text{=}\left(0.144\right)\text{×100\% =}14.4\%.\end{array}$

The weight percent of methanol is calculated as shown above. Hence, the weight percent obtained is $14.4\%.$

(d)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of ethylene glycol has to be calculated.

Concept introduction:

Refer to (a)

Answer to Problem 59QRT

The mass fraction of ethylene glycol is $0.0744$ and weight percent of ethylene glycol is calculated as $7.4\%.$

Explanation of Solution

• The Mass fraction:

  $\text{Mass fraction of A = }\frac{\text{Mass of A}}{\text{ Mass of}\text{A + Mass of}\text{B + Mass of}\text{C+ }\mathrm{...}}$.

Given information as: Density of ethylene glycol is $\text{1}\text{.11 g/mL}$ ; volume is $\text{14}\text{.5 mL}$ and water is $\text{200}\text{g}\text{.}$

Mass of ethylene glycol: $\text{= }\text{Density}\text{×}\text{Volume}$

  $\begin{array}{l}\text{=}\left(\text{1}\text{.11}\text{g/mL}\right)\left(14.5\text{mL}\right)\\ \\ \text{=}\text{16}\text{.095}\text{g}\text{.}\end{array}$

Mass fraction of ethylene glycol $\text{= }\frac{\text{Mass of ethylene glycol}}{\text{Mass of}\text{ethylene glycol + Mass of}\text{Water }}$

  $\text{= }\frac{\text{16}\text{.095}\text{g}}{\left(\text{16}\text{.095}\text{g}\text{+ 200}\text{.0g}\right)}\text{ = }\mathrm{0.0744.}$

Hence, the mass fraction of ethylene glycol is $0.0744.$

• The Weight percent:

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$.

  $\begin{array}{l}\text{weight \% of ethylene glycol}\text{=}\frac{\text{Mass of ethylene glycol}}{\text{ Mass of}\text{ethylene glycol}\text{+}\text{Mass of}\text{water }}\text{×100\%}\\ \\ \text{=}\left(\text{0}\text{.0744}\right)\text{×100\% = }7.4\%.\end{array}$

The weight percent of ethylene glycol is calculated as shown above. Hence, the weight percent obtained is $7.4\%.$