Chapter 13, Problem 121QRT

Interpretation Introduction

Interpretation: The mass fraction, weight percent and $\text{ppm}$ of solute 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 121QRT

$\begin{array}{cccccc}\text{Compound}& \text{Mass}\text{of}\text{compound}& \text{Mass}\text{of}\text{water}& \text{Mass}\text{fraction}\text{of}\text{solute}& \text{Weight}\text{percent}\text{of}\text{solute}& \text{ppm}\text{of}\text{solute}\\ \text{Lye}& 75g& 125g& 0.375& 37.5\%& 37.5\times 10^4\\ \text{Glycerol}& 33g& 200g& 0.142& 14.2\%& 1.4\times 10^5\\ \text{Acetylene}& 0.0015g& 2\times 10^{2}g& 9\times 10^{-6}& 0.0009\%& 9ppm\end{array}$

Explanation of Solution

Lye$\left(NaOH\right)$:

• The weight percent:

  $\text{weight \% A = }\left(\text{Mass}\text{fraction}\right)\text{×100\%}$.

Given value of the mass fraction is $\mathrm{0.375.}$

Weight percent of $\text{NaOH}$:

  $\begin{array}{c}\text{=}\left(0.375\right)\text{×100\%}\\ \\ \text{=}37.5\%.\end{array}$

• The Mass of $NaOH$:

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

Given information as: Mass of $\text{NaOH}$ is unknown and pure water is $\text{125}\text{g}\text{.}$

  $\begin{array}{l}\text{weight \% of NaOH}\text{=}\frac{\text{Mass of NaOH}}{\text{ Mass of}\text{NaOH + Mass of}\text{water }}\text{×100\%}\\ \\ 37.5\%\text{=}\frac{\text{Mass of NaOH}}{\text{ Mass of}\text{NaOH + 125 g }}\text{×100\%}\\ \\ 0.375\text{=}\frac{\text{Mass of NaOH}}{\text{ Mass of}\text{NaOH + 125 g }}\\ \\ 0.375\left(\text{ Mass of}\text{NaOH + 125 g }\right)=\text{Mass of NaOH}\\ \\ 0.375\left(\text{Mass of}\text{NaOH}\right)-\text{Mass of NaOH}=-46.875\\ \\ \left[\left(0.375\right)-1\right]\left(\text{Mass of}\text{NaOH}\right)=-46.875\\ \\ \left(\text{Mass of}\text{NaOH}\right)=\frac{-46.875}{\left[\left(0.375\right)-1\right]}=75g.\end{array}$

The mass of Lye is $75g.$

The weight percent of Lye obtained is $37.5\%.$

As known, $\text{1}\text{\%}\text{=}\text{10,000}\text{ppm}$

Then, $37.5\%$ will be $37.5\text{×}\text{10,000}\text{ppm}\text{=}37.5\times 10^{4}ppm.$

Glycerol:

• 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 glycerol is $\text{33 g}$ and pure water is $\text{200}\text{g}\text{.}$.

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

  $\text{= }\frac{\text{33 g}}{\left(\text{33g}\text{+ 200g}\right)}\text{ = }0.142$.

Hence, the mass fraction of glycerol is $0.142.$

• The Weight percent:

  $\text{weight \% A = }\left(\text{Mass}\text{fraction}\right)\text{×100\%}$.

  $\begin{array}{l}\text{weight \% of glycerol}\text{=}\left(0.142\right)\text{×100\%}\\ \\ \text{ =}14.2\%.\end{array}$

The weight percent of glycerol obtained is $14.2\%.$

As known, $\text{1}\text{\%}\text{=}\text{10,000}\text{ppm}$

Then, $14.2\%.$ will be $\text{14}\text{.2}\text{×}\text{10,000}\text{ppm}\text{=}1.4\times 10^{5}ppm.$

Therefore, ppm of solute is $1.4\times 10^{5}ppm$.

Acetylene:

As known, $\text{1}\text{\%}\text{=}\text{10,000}\text{ppm}$.

Then, $\text{0}\text{.0009\%}$ will be $\left(\text{0}\text{.0009\%}\right)\text{×}\text{10,000}\text{ppm}\text{=}9ppm$.

• The mass of water using 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: weight percent of solute is $\text{0}\text{.0009}\text{\%}\text{.}$

  $\begin{array}{l}\text{weight \% of Acetylene}\text{=}\frac{\text{Mass of Acetylene}}{\text{ Mass of}\text{Acetylene + Mass of}\text{water }}\text{×100\%}\\ \\ 0.0009\%\text{=}\frac{0.0015\text{g}}{\text{ 0}\text{.0015}\text{g + Mass of}\text{water }}\text{×100\%}\\ \\ 0.000009\left(\text{ 0}\text{.0015}\text{g + Mass of}\text{water }\right)=0.00\text{15}\text{g}\\ \\ 0.000009\left(\text{Mass of}\text{water}\right)=0.0015g-1.35\times {10}^{-8}g=1.55\times {10}^{-3}g\\ \\ \left(\text{Mass of}\text{water}\right)=\frac{1.55\times {10}^{-3}g}{9\times {10}^{-6}}=\text{172}\text{.2}\text{g}\text{=}2\times 10^{2}g\left(\text{1}\text{sig}\text{fig}\right).\end{array}$

As the weight percent of acetylene is $0.0009\%$, then the Mass fraction of acetylene will be $\frac{0.0009}{100}=9\times 10^{-6}.$

Therefore, the completed table is given below:

$\begin{array}{cccccc}\text{Compound}& \text{Mass}\text{of}\text{compound}& \text{Mass}\text{of}\text{water}& \text{Mass}\text{fraction}\text{of}\text{solute}& \text{Weight}\text{percent}\text{of}\text{solute}& \text{ppm}\text{of}\text{solute}\\ \text{Lye}& 75g& 125g& 0.375& 37.5\%& 37.5\times 10^4\\ \text{Glycerol}& 33g& 200g& 0.142& 14.2\%& 1.4\times 10^5\\ \text{Acetylene}& 0.0015g& 2\times 10^{2}g& 9\times 10^{-6}& 0.0009\%& 9ppm\end{array}$