Chapter 13, Problem 58QRT

(a)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of $\text{NaCl}$ 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 58QRT

The mass fraction of $\text{NaCl}$ is $0.106$ and weight percent of $\text{NaCl}$ is calculated as $10.6\%.$

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 is as follows: Mass of $\text{NaCl}$ is $\text{20}\text{.7 g}$ and pure water is $\text{175}\text{g}\text{.}$.

Mass fraction of $\text{NaCl}$$\text{= }\frac{\text{Mass of NaCl}}{\text{Mass of}\text{NaCl + Mass of}\text{water }}$.

$\text{= }\frac{\text{20}\text{.7 g}}{\left(\text{20}\text{.7g}\text{+ 175g}\right)}\text{ = }0.106.$

Hence, the mass fraction of $\text{NaCl}$ is $0.106.$

• 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 is as follows: Mass of $\text{NaCl}$ is $\text{20}\text{.7 g}$ ; pure water is $\text{175}\text{g}\text{.}$.

  $\begin{array}{l}\underset{\_}{\text{weight percent of NaCl:}}\\ \\ \text{weight \% of NaCl}\text{=}\frac{\text{Mass of NaCl}}{\text{ Mass of}\text{NaCl + Mass of}\text{water }}\text{×100\%}\\ \\ \text{=}\left(0.106\right)\text{×100\% =}10.6\%.\end{array}$

The weight percent of $\text{NaCl}$ is calculated as shown above. Hence, the weight percent obtained is $10.6\%.$

(b)

Interpretation Introduction

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

Concept introduction:

Refer to (a):

Answer to Problem 58QRT

The mass fraction of ethanol is $0.127$ and weight percent of ethanol is calculated as $12.7\%.$

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 is as follows: Mass of ethanol is $\text{1}\text{.45 g}$ ; pure water is $\text{10}\text{.0}\text{g}$.

Mass fraction of ethanol can be calculated as given below:

  $\begin{array}{c}\text{Mass}\text{fraction= }\frac{\text{Mass of ethanol}}{\text{Mass of}\text{ethanol + Mass of}\text{water }}\\ \\ \text{= }\frac{\text{1}\text{.45}\text{g}}{\left(\text{1}\text{.45}\text{g}\text{+ 10}\text{.0g}\right)}\text{ = }\mathrm{0.127.}\end{array}$.

Hence, the mass fraction of $\text{ethanol}$ is $0.127.$

• 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 is as follows: Mass of ethanol is $\text{1}\text{.45 g}$ ; pure water is $\text{10}\text{.0}\text{g}$.

  $\begin{array}{l}\underset{\_}{\text{weight percent of ethanol:}}\\ \\ \text{weight \% of ethanol}\text{=}\frac{\text{Mass of ethanol}}{\text{ Mass of}\text{ethanol}+\text{Mass of}\text{water }}\text{×100\%}\\ \\ \text{=}\left(0.127\right)\text{×100\% =}12.7\%.\end{array}$

The weight percent of ethanol is calculated as shown above. Hence, the weight percent obtained is $12.7\%.$

(c)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of ${\text{CS}}_{\text{2}}$ 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 $100\%$

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

Answer to Problem 58QRT

The mass fraction of ${\text{CS}}_{\text{2}}$ is $0.308$ and weight percent of ${\text{CS}}_{\text{2}}$ 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 is as follows: Mass of ${\text{CS}}_{\text{2}}$ is $\text{20}\text{.0 g}$ and ${\text{pure CHCl}}_3\text{ is 45}\text{.0 }\text{g}\text{.}$

Mass fraction of ${\text{CS}}_{\text{2}}$$\text{= }\frac{{\text{Mass of CS}}_{\text{2}}}{\text{Mass of}{\text{CS}}_{\text{2}}\text{+ Mass of}{\text{CHCl}}_3}$

  $\text{= }\frac{\text{20}\text{.0 g}}{\left(20.0\text{g}\text{+ 45}\text{.0g}\right)}\text{ = }0.308$.

Hence, the mass fraction of ${\text{CS}}_{\text{2}}$ is $0.308.$

• 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 is as follows: Mass of ${\text{CS}}_{\text{2}}$ is $\text{20}\text{.0 g}$ and ${\text{pure CHCl}}_3\text{ is 45}\text{.0 }\text{g}\text{.}$

  $\begin{array}{l}\underset{\_}{{\text{weight percent of CS}}_{\text{2}}\text{:}}\\ \\ {\text{weight \% of CS}}_{\text{2}}\text{=}\frac{{\text{Mass of CS}}_{\text{2}}}{\text{ Mass of}{\text{CS}}_{\text{2}}+\text{Mass of}{\text{CHCl}}_3}\text{×100\%}\\ \\ \text{=}\left(0.308\right)\text{×100\% =}30.8\%.\end{array}$

The weight percent of ${\text{CS}}_{\text{2}}$ is calculated as shown above. Hence, the weight percent obtained is $30.8\%.$

(d)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of benzene 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 $100\%$

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

Answer to Problem 58QRT

The mass fraction of benzene is $28.4\times 10^{-3}$ and weight percent of benzene is calculated as $2.8\%.$

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: Density of benzene is $\text{0}\text{.877 g/mL}$, volume is $\text{4}\text{.00 mL}$ and diethyl ether $\text{120}\text{.0}\text{g}\text{.}$

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

$\begin{array}{l}\text{=}\left(\text{0}\text{.877}\text{g/mL}\right)\left(\text{4}\text{.00}\text{mL}\right)\\ \\ \text{=}\text{3}\text{.51}\text{g}\text{.}\end{array}$

Mass fraction of benzene $\text{= }\frac{\text{Mass of benzene}}{\text{Mass of}\text{benzene + Mass of}\text{Diethyl ether }}.$

$\text{= }\frac{\text{3}\text{.51}\text{g}}{\left(\text{3}\text{.51}\text{g}\text{+ 120}\text{.0g}\right)}\text{ = }28.4\times {10}^{-3}$.

Hence, the mass fraction of benzene is $28.4\times 10^{-3}$

• The Weight percent:

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

  $\begin{array}{l}\underset{\_}{\text{weight percent of benzene:}}\\ \\ \text{weight \% of benzene}\text{=}\frac{\text{Mass of benzene}}{\text{ Mass of}\text{benzene}+\text{Mass of}\text{diethyl ether }}\text{×100\%}\\ \\ \text{=}\left(28.4\times {10}^{-3}\right)\text{×100\% =}2.8\%.\end{array}$

The weight percent of benzene is calculated as shown above. Hence, the weight percent obtained is $2.8\%.$