Chapter 13, Problem 4PS

Interpretation Introduction

Interpretation: The mole fraction, molality, and weight percent of given compounds to be identified.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Molality: The amount of solute (mol) per kilogram of solvent.

$\text{Concentration (C, mol/kg) = molality of solute = }\frac{\text{Amount of solute (mol)}}{\text{Mass of solvent (Kg)}}$

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 4PS

$\begin{array}{cccc}Compound& Molality& Weightpercent& Molefraction\\ {\text{KNO}}_3& \underset{\_}{\text{1}\text{.1}}& 10.0& \underset{\_}{0.019}\\ \text{Ethanoic acid}& 0.0183& \underset{\_}{0.1}& \underset{\_}{3.29}\\ \text{Ethylene glycol}& \underset{\_}{16.1}& 18.0& \underset{\_}{0.224}\end{array}$

Explanation of Solution

• Calculate molality of ${\text{KNO}}_3$:

Given data:

$\text{Weight percent = 10}\text{.0}$

$\begin{array}{l}\text{Weight percent = 10}{\text{.0 indictaes 10g of KNO}}_3\text{ in}\text{100}\text{g of solution}\text{.}\\ {\text{Mass of Water (solvent) = mass of solution - mass of KNO}}_3\text{ = 100 g - 10 g = 90 g }\end{array}$

$\begin{array}{c}\text{No}\text{.}\text{of}\text{moles}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{no}{\text{.of moles (KNO}}_3\text{) = }\frac{\text{10 g}}{\text{ 101}\text{.10 g/mol}}\text{ = 0}\text{.099 mol}\end{array}$

Molality of a solute is defined as the amount of solute dissolves in kilogram of solvent.

$\text{Molality}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{kg}}$

$\begin{array}{l}{\text{n}}_{{\text{KNO}}_3}{\text{ ( moles of KNO}}_3\text{) = 0}\text{.099 mol}\\ \text{Amount of solvent = 0}\text{.090 Kg}\end{array}$

$\begin{array}{l}\underset{\_}{\text{Molality of solute:}}\\ \frac{\text{Amount of solute (mol)}}{\text{Mass of solvent (Kg)}}\text{= }\frac{\text{0}\text{.099 mol}}{\text{0}\text{.090 kg}}\text{ = 1}\text{.14 mol/Kg = 1}\text{.1 m}\end{array}$

The molality of ${\text{KNO}}_3$ is calculated as shown above. Hence, the calculated molality of ${\text{KNO}}_3$$\text{1}\text{.1 m}$

• Calculate mole fraction of ${\text{KNO}}_3$:

Known data:

$\text{Mass of water = 90 g}\text{.}$

$\text{No}\text{.}\text{of}\text{moles}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}$

$\text{Moles of water}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\text{=}\text{}\frac{\text{90g}}{\text{18}\text{}\text{g/mol}}\text{=}\text{5moles}$

$\begin{array}{l}{\text{n}}_{{\text{KNO}}_3}\text{= 0}\text{.099 mol}\\ {\text{n}}_{\text{Water}}\text{= 5 mol}\end{array}$

${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

${\text{ (χ}}_{{\text{KNO}}_3}\text{) = }\frac{{\text{n}}_{{\text{KNO}}_3}}{{\text{ n}}_{{\text{KNO}}_3}{\text{ + n}}_{\text{Water}}}\text{ = }\frac{\text{0}\text{.099mol}}{\text{0}\text{.099 mol + 5 mol}}\text{ = 0}\text{.019}$

Convert the mass of water into unit of moles and substitute in the respective mole fraction equation as shown above. Hence, the mole fraction of ${\text{KNO}}_3$ is $\text{0}\text{.019}$

• Calculate Weight percentage of ethanoic acid:

  Given data:

  $\text{Molality}\text{=}\text{0}\text{.0183m}$

$\text{0}\text{.0183mol}\text{of}\text{Acetic}\text{acid}\text{present}\text{in}\text{1Kg}\text{of}\text{water}\text{.}$

$\text{No}\text{.}\text{of}\text{moles}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}$

$\text{Mass of Acetic}\text{acid = }\left(\text{0}\text{.0183}\text{mol}\right)\left(\text{60}\text{.05g/mol}\right)\text{=}\text{1}\text{.10}\text{g}$

$\text{Mass of water = 1000 g}\text{.}$

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

$\begin{array}{l}\text{weight \% Acetic acid = }\frac{\text{Mass of Acetic acid}}{\text{ Mass of Acetic acid + Mass of water }}\text{×100\%}\\ \text{= }\frac{\text{1}\text{.10 g}}{\left(\text{1}\text{.10g}\right)\text{+}\left(\text{1000 g}\right)}\text{×100\% = 0}\text{.1\%}\end{array}$

The weight percent of ethanoic acid is calculated as shown above. Hence, the weight percent obtained is $\text{0}\text{.1\%}$

• Calculate mole fraction of ethanoic acid:

$\begin{array}{l}{\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}\\ \text{Known:}\\ \text{no}\text{.of moles }\left(\text{Acetic acid}\right)\text{ = 0}\text{.0183 mol }\\ \text{no}\text{.of moles (Water) = }\frac{\text{1000g}}{\text{18g/mol}}\text{=}\text{55}\text{.55mol}\\ {\text{n}}_{Aceticacid}\text{= 0}\text{.0183 mol}\\ {\text{n}}_{\text{Water}}\text{= 55}\text{.55 mol}\\ \text{Mole fraction :}\\ {\text{ (χ}}_{Aceticacid}\text{) = }\frac{{\text{n}}_{Aceticacid}}{{\text{ n}}_{Aceticacid}{\text{ + n}}_{\text{Water}}}\text{ = }\frac{\text{0}\text{.0183 mol}}{\text{0}\text{.0183 mol + 55}\text{.55 mol}}\text{ = 3}\text{.29}\end{array}$

Convert the mass of water into unit of moles and substitute in the respective mole fraction equation as shown above. Hence, the mole fraction of Acetic acid is $\text{3}\text{.29}$

• Calculate molality of ethylene glycol:

$\begin{array}{l}\text{Molality of solute = }\frac{\text{Amount of solute (mol)}}{\text{Mass of solvent (Kg)}}\\ \text{Given:}\\ \text{Weight percent = 18}\text{.0 indictaes 18g of }glycol\text{ in}\text{100}\text{g of solution}\text{.}\\ \text{Mass of Water (solvent) = mass of solution - mass of }glycol\text{ = 100 g - 18 g = 82 g }\\ \text{no}\text{.of moles (}glycol\text{) = }\frac{\text{82 g}}{\text{ 62}\text{.07 g/mol}}\text{ = 1}\text{.32 mol}\\ {\text{n}}_{glycol}\text{ ( moles of }glycol\text{) = 1}\text{.32 mol}\\ \text{Amount of solvent = 0}\text{.082 Kg}\\ \underset{\_}{\text{Molality of solute:}}\\ \frac{\text{Amount of solute (mol)}}{\text{Mass of solvent (Kg)}}\text{= }\frac{\text{1}\text{.32 mol}}{\text{0}\text{.082 kg}}\text{ = 16}\text{.1 mol/Kg = 16}\text{.1 m}\end{array}$

The molality of ethylene glycol is calculated as shown above. Hence, the calculated molality of ethylene glycol $\text{16}\text{.1 m}$

• Calculate mole fraction of ethylene glycol:

$\begin{array}{l}{\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}\\ \text{Mass of water = 82 g}\text{.}\\ \text{Moles of water}\text{= }\frac{\text{mass}}{\text{molar}\text{mass}}\text{ =}\text{}\frac{\text{82 g}}{\text{18}\text{}\text{g/mol}}\text{=}4.56\text{moles}\\ {\text{n}}_{glycol}\text{ ( moles of}glycol\text{) = 1}\text{.32 mol}\\ \text{no}\text{.of moles (Water) = 4}\text{.56 moles}\\ {\text{n}}_{glycol}\text{= 1}\text{.32 mol}\\ {\text{n}}_{\text{Water}}\text{= 4}\text{.56 mol}\\ \text{Mole fraction :}\\ {\text{ (χ}}_{glycol}\text{) = }\frac{{\text{n}}_{glycol}}{{\text{ n}}_{glycol}{\text{ + n}}_{\text{Water}}}\text{ = }\frac{\text{1}\text{.32 mol}}{\text{1}\text{.32 mol + 4}\text{.56 mol}}\text{ = 0}\text{.224}\end{array}$

Convert the mass of water into unit of moles and substitute in the respective mole fraction equation as shown above. Hence, the mole fraction of ethylene glycol is $\text{0}\text{.224}$

Conclusion

The mole fraction, molality, and weight percent of given compounds were identified