Chapter 13.6, Problem 13.8E

(a)

Interpretation Introduction

Interpretation: The mass fraction and weight percent of ${\text{NaHCO}}_{\text{3}}$ 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 13.8E

The mass fraction of $\left({\text{NaHCO}}_3\right)$ is $2.0\times {10}^{-4}$ and weight percent of $\left({\text{NaHCO}}_3\right)$ is calculated as $2.0\times 10^{-2}\%.$

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 data is as follows:

Mass of $\text{NaCl}$$\text{= 6}\text{.5 g}$ ; $\text{pure water = 1}\text{.0 L}\text{=}\text{1000}\text{g}$.

Mass of ${\text{NaHCO}}_3$$\text{= 0}\text{.20 g}$.

Mass of ${\text{CaCl}}_2$$\text{= 0}\text{.10 g}$.

Mass of $\text{KCl}$$\text{= 0}\text{.10 g}\text{.}$

Mass fraction of $\left({\text{NaHCO}}_3\right)$

  $\text{= }\frac{{\text{Mass of NaHCO}}_3}{\text{ Mass of}{\text{NaHCO}}_3\text{+ Mass of}\text{NaCl + Mass of}\text{water + Mass of}{\text{CaCl}}_2\text{+ Mass of KCl}}.$

  $\text{= }\frac{\text{0}\text{.20 g}}{\left(\text{1000}\text{g}\text{+}\text{6}\text{.5}\text{g}\text{+}\text{0}\text{.10}\text{g}\text{+}\text{0}\text{.20}\text{g}\text{+0}\text{.10g}\right)}\text{ = }1.986\times {10}^{-4}\approx 2.0\times {10}^{-4}$.

Hence, the mass fraction of $\left({\text{NaHCO}}_3\right)$ is $2.0\times {10}^{-4}.$

• The Weight percent:

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

  $\begin{array}{l}\underset{\_}{{\text{weight percent of NaHCO}}_3\text{:}}\\ \\ {\text{weight \% of NaHCO}}_3\\ \text{=}\frac{{\text{Mass of NaHCO}}_3}{\text{ Mass of}{\text{NaHCO}}_3\text{+ Mass of}\text{NaCl + Mass of}\text{water + Mass of}{\text{CaCl}}_2\text{+ Mass of KCl}}\text{×100\%}\\ \\ \text{=}\left(2.0\times {10}^{-4}\right)\text{×100\% =}2.0\times 10^{-2}\%.\end{array}$

The weight percent of $\left({\text{NaHCO}}_3\right)$ is calculated as shown above. Hence, the weight percent obtained is $2.0\times 10^{-2}\%.$

(b)

Interpretation Introduction

Interpretation:

The solute that has the lowest mass fraction in the solution has to be determined.

Concept introduction:

Refer to (a).

Answer to Problem 13.8E

The solutes that have the lowest mass fraction in the solution are both $\left(CaC{l}_2\right)$ and $\left(KCl\right).$

Explanation of Solution

Here, water is a solvent and by omitting the solvent, the mass fraction of remaining solute is calculated as shown below,

• The Mass fraction of $\left(NaHC{O}_3\right)$:

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

Given data is as follows:

Mass of $\text{NaCl}$$\text{= 6}\text{.5 g}$ ; $\text{pure water = 1}\text{.0 L}\text{=}\text{1000}\text{g}$.

Mass of ${\text{NaHCO}}_3$$\text{= 0}\text{.20 g}$.

Mass of ${\text{CaCl}}_2$$\text{= 0}\text{.10 g}$.

Mass of $\text{KCl}$$\text{= 0}\text{.10 g}\text{.}$

• Mass fraction of $\left(NaHC{O}_3\right)$

  $\text{= }\frac{{\text{Mass of NaHCO}}_3}{\text{ Mass of}{\text{NaHCO}}_3\text{+ Mass of}\text{NaCl + Mass of}\text{water + Mass of}{\text{CaCl}}_2\text{+ Mass of KCl}}$.

  $\text{= }\frac{\text{0}\text{.20 g}}{\left(\text{1000}\text{g}\text{+}\text{6}\text{.5}\text{g}\text{+}\text{0}\text{.10}\text{g}\text{+}\text{0}\text{.20}\text{g}\text{+0}\text{.10g}\right)}\text{ = }1.986\times {10}^{-4}\approx 2.0\times {10}^{-4}$.

Hence, the mass fraction of $\left({\text{NaHCO}}_3\right)$ is $2.0\times {10}^{-4}.$

• The Mass fraction of $\left(NaCl\right)$.

Mass fraction of $\left(NaCl\right)$

  $\text{= }\frac{\text{Mass of NaCl}}{\text{ Mass of}{\text{NaHCO}}_{\text{3}}\text{+ Mass of}\text{NaCl + Mass of}\text{water + Mass of}{\text{CaCl}}_{\text{2}}\text{+ Mass of KCl}}.$

  $\text{= }\frac{\text{6}\text{.5g}}{\left(\text{1000}\text{g}\text{+}\text{6}\text{.5}\text{g}\text{+}\text{0}\text{.10}\text{g}\text{+}\text{0}\text{.20}\text{g}\text{+0}\text{.10g}\right)}\text{ = }64.6\times {10}^{-4}$.

Hence, the mass fraction of $\left(NaCl\right)$ is $64.6\times {10}^{-4}.$

• The Mass fraction of $\left(CaC{l}_2\right)$

Mass fraction of $\left(CaC{l}_2\right)$

  $\text{= }\frac{\text{Mass of }\left(CaC{l}_2\right)}{\text{ Mass of}{\text{NaHCO}}_{\text{3}}\text{+ Mass of}\text{NaCl + Mass of}\text{water + Mass of}{\text{CaCl}}_{\text{2}}\text{+ Mass of KCl}}.$

  $\text{= }\frac{\text{0}\text{.10 g}}{\left(\text{1000}\text{g}\text{+}\text{6}\text{.5}\text{g}\text{+}\text{0}\text{.10}\text{g}\text{+}\text{0}\text{.20}\text{g}\text{+0}\text{.10g}\right)}\text{ = }9.9\times {10}^{-5}\approx 1.0\times {10}^{-4}$.

Hence, the mass fraction of $\left(CaC{l}_2\right)$ is $1.0\times {10}^{-4}$.

• The Mass fraction of $\left(KCl\right)$.

Mass fraction of $\left(KCl\right)$

  $\text{= }\frac{\text{Mass of }\left(KCl\right)}{\text{ Mass of}{\text{NaHCO}}_3\text{+ Mass of}\text{NaCl + Mass of}\text{water + Mass of}{\text{CaCl}}_2\text{+ Mass of KCl}}.$

  $\text{= }\frac{\text{0}\text{.10 g}}{\left(\text{1000}\text{g}\text{+}\text{6}\text{.5}\text{g}\text{+}\text{0}\text{.10}\text{g}\text{+}\text{0}\text{.20}\text{g}\text{+0}\text{.10g}\right)}\text{ = }9.9\times {10}^{-5}\approx 1.0\times {10}^{-4}$.

Hence, the mass fraction of $\left(KCl\right)$ is $1.0\times {10}^{-4}$.

From the above obtained values,

The solutes that have the lowest mass fraction in the solution are both $\left(CaC{l}_2\right)$ and $\left(KCl\right).$