Chapter 13, Problem 122QRT

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 $100\%.$

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

Answer to Problem 122QRT

$\begin{array}{cccccc}\text{Compound}& \text{Mass}\text{of}\text{compound}& \text{Mass}\text{of}\text{water}& \text{Mass}\text{fraction}\text{of}\text{solute}& \text{Weight}\text{percent}\text{of}\text{solute}& \text{ppm}\text{of}\text{solute}\\ \text{Table}\text{salt}& 52g& 175g& 0.229& 22.9\%& 2.3\times 10^5\\ \text{Glucose}& 15g& 199g& 0.07& 7\%& 7\times {10}^4\\ \text{Methane}& 2.5\times 10^{-3}g& 100g& 2.5\times 10^{-5}& 0.0025\%& 25\end{array}$

Explanation of Solution

Table salt:

• 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 table salt is $\text{52 g}$ and pure water is $\text{175}\text{g}\text{.}$.

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

$\text{= }\frac{\text{52 g}}{\left(\text{52g}\text{+ 175g}\right)}\text{ = }0.229$.

Hence, the mass fraction of table salt is $0.229.$

• 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 table salt is $\text{52 g}$ and pure water is $\text{175}\text{g}\text{.}$.

  $\begin{array}{l}\underset{\_}{\text{weight percent of table salt:}}\\ \\ \text{weight \% of table salt}\text{=}\frac{\text{Mass of table salt}}{\text{ Mass of}\text{table salt + Mass of}\text{water }}\text{×100\%}\\ \\ \text{=}\left(0.229\right)\text{×100\% =}22.9\%.\end{array}$

The weight percent of table salt is calculated as shown above. Hence, the weight percent obtained is $22.9\%.$

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

Then, $22.9\%$ will be $\text{23×}\text{10,000}\text{ppm}\text{=}2.3\times 10^{5}ppm.$

Therefore, ppm of solute is $2.3\times 10^{5}ppm.$

Glucose:

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

Then, $\text{7}\text{×}{\text{10}}^{\text{4}}\text{ppm}$ will be

  $\begin{array}{l}\left(x\right)\text{×}\text{10,000}\text{ppm}\text{=}\text{7}\text{×}{\text{10}}^{\text{4}}\text{ppm}\\ \\ \left(x\right)\text{=}\frac{\text{7}\text{×}{\text{10}}^{\text{4}}\text{ppm}}{\text{10,000}\text{ppm}}=7\%\end{array}$

Therefore, weight percent of solute is $7\%.$

• 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 glucose is $\text{15 g}$ and $\text{pure water is ?}\text{g}$

  $\begin{array}{l}\underset{\_}{\text{weight percent of glucose:}}\\ \\ \text{weight \% of glucose}\text{=}\frac{\text{Mass of glucose}}{\text{ Mass of}\text{glucose + Mass of}\text{water }}\text{×100\%}\\ \\ 7\%\text{=}\frac{\text{15}\text{g}}{\text{ 15}\text{g + Mass of}\text{water }}\text{×100\%}\\ \\ 0.07\left(\text{ 15}\text{g + Mass of}\text{water }\right)=\text{15}\text{g}\\ \\ 0.07\left(\text{Mass of}\text{water}\right)=15g-1.05g=13.95g\\ \\ \left(\text{Mass of}\text{water}\right)=\frac{13.95g}{0.07}=199g.\end{array}$

The weight percent of glucose is $7\%.$

Mass fraction of glucose is $0.07.$

Mass of water is $199g.$

Methane:

• 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 methane is un-known and pure water is $\text{100}\text{g}\text{.}$

  $\begin{array}{l}\text{weight \% of Methane}\text{=}\frac{\text{Mass of methane}}{\text{ Mass of}\text{methane + Mass of}\text{water }}\text{×100\%}\\ \\ 0.0025\%\text{=}\frac{\text{Mass of methane}}{\text{ Mass of}\text{methane + 100 g }}\text{×100\%}\\ \\ 0.000025\text{=}\frac{\text{Mass of methane}}{\text{ Mass of}\text{methane + 100 g }}\\ \\ 0.000025\left(\text{ Mass of}\text{methane + 100 g }\right)=\text{Mass of methane}\\ \\ 0.000025\left(\text{Mass of}\text{methane}\right)-\text{Mass of methane}=-0.0025\\ \\ \left[\left(0.000025\right)-1\right]\left(\text{Mass of}\text{methane}\right)=-0.0025\\ \\ \left(\text{Mass of}\text{methane}\right)=\frac{-0.0025}{\left[\left(0.000025\right)-1\right]}=2.5\times 10^{-3}g.\end{array}$

The mass of methane is $2.5\times 10^{-3}g.$

• 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 methane is $\text{2}{\text{.5×10}}^{\text{-3}}\text{g}$ and pure water $\text{175}\text{g}\text{.}$

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

$\text{= }\frac{\text{0}\text{.0025}\text{g}}{\left(\text{0}\text{.0025}\text{g}\text{+ 100g}\right)}\text{ = }2.5\times 10^{-5}$

Hence, the mass fraction is $2.5\times 10^{-5}$

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

Then, $0.0025\%$ will be $0.0025\text{×}\text{10,000}\text{ppm}\text{=}25ppm.$

The ppm of solute is $25ppm.$

Therefore,

$\begin{array}{cccccc}\text{Compound}& \text{Mass}\text{of}\text{compound}& \text{Mass}\text{of}\text{water}& \text{Mass}\text{fraction}\text{of}\text{solute}& \text{Weight}\text{percent}\text{of}\text{solute}& \text{ppm}\text{of}\text{solute}\\ \text{Table}\text{salt}& 52g& 175g& 0.229& 22.9\%& 2.3\times 10^5\\ \text{Glucose}& 15g& 199g& 0.07& 7\%& 7\times {10}^4\\ \text{Methane}& 2.5\times 10^{-3}g& 100g& 2.5\times 10^{-5}& 0.0025\%& 25\end{array}$