Chapter 13, Problem 3PS

Interpretation Introduction

Interpretation: The mole fraction, molality, and weight percent of given compounds has 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 3PS

$\begin{array}{cccc}Compound& Molality& Weightpercent& Molefraction\\ \text{NaI}& \text{0}\text{.15}& \underset{\_}{\text{2}\text{.2}}& \underset{\_}{\text{2}{\text{.7×10}}^{\text{-3}}}\\ \text{Ethanol}& \underset{\_}{\text{1}\text{.1}}& \text{5}\text{.0}& \text{0}\text{.020}\\ \text{Sucrose}& \text{0}\text{.15}& \underset{\_}{\text{4}\text{.9}}& \underset{\_}{\text{2}{\text{.7×10}}^{\text{-3}}}\end{array}$

Explanation of Solution

• Calculate mole fraction of $\text{NaI}$:

Given data:

$\text{Molality = 0}\text{.15 mol/ kg}$

$\text{ 0}\text{.15 moles of NaI is in 1Kg of water}\text{.}$

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

$\text{Mass of NaI = }\left(\text{moles of NaI}\right)\left(\text{molar mass of NaI}\right)\text{=}\left(\text{0}\text{.15 mol}\right)\left(\text{149}\text{.89g/mol}\right)\text{=}\text{22}\text{.48g}$

$\begin{array}{l}\text{Mass of water = 1000 g}\text{.}\\ \text{Moles of water}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\text{=}\text{}\frac{\text{1000g}}{\text{18}\text{}\text{g/mol}}\text{=}\text{55}\text{.55moles}\end{array}$

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

$\begin{array}{l}{\text{n}}_{\text{NaI}}\text{= 0}\text{.15 mol}\\ {\text{n}}_{\text{Water}}\text{= 55}\text{.55 mol}\\ \text{Mole fraction :}\\ {\text{ (χ}}_{NaI}\text{) = }\frac{{\text{n}}_{\text{NaI}}}{{\text{ n}}_{\text{NaI}}{\text{ + n}}_{\text{Water}}}\text{ = }\frac{\text{0}\text{.15mol}}{\text{0}\text{.15 mol + 55}\text{.55 mol}}\text{ = 0}\text{.0027}\text{=}\text{2}\text{.7}\times {10}^{-3}\end{array}$

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 $\text{NaI}$ is $\text{2}\text{.7}\times {10}^{-3}$

• Calculate weight percent of $\text{NaI}$:

  Known data:

  $\begin{array}{l}\text{Mass of NaI = }\text{22}\text{.48g}\\ \text{Mass of water = 1000 g}\text{.}\end{array}$

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

  $\begin{array}{l}\text{weight \% NaI = }\frac{\text{Mass of NaI}}{\text{ Mass of NaI + Mass of water }}\text{×100\%}\\ \text{= }\frac{\text{22}\text{.48 g}}{\left(\text{22}\text{.48 g}\right)\text{+}\left(\text{1000 g}\right)}\text{×100\% = 2}\text{.3\%}\end{array}$

The weight percent of $\text{NaI}$ is calculated as shown above. Hence, the weight percent obtained is $\text{2}\text{.3\%}$

• Calculate molality of Ethanol:

  Given data:

  $\text{Weight percent = 5}\text{.0}$

  $\begin{array}{l}\text{Weight percent = 5}\text{.0 indictaes 5g of ethanol in}\text{100}\text{g of solution}\text{.}\\ \text{Mass of Water (solvent) = mass of solution - mass of ethanol = 100 g - 5 g = 95 g}\end{array}$

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

  $\text{no}\text{.of moles (ethanol) = }\frac{\text{5 g}}{\text{ 46}\text{.07 g/mol}}\text{ = 0}\text{.1085 mol}$

  $\text{Amount of solvent = 0}\text{.095 Kg}$

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

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

  $\frac{\text{No}\text{.}\text{of}\text{moles (mol)}}{\text{Mass of solvent (Kg)}}\text{= }\frac{\text{0}\text{.1085 mol}}{\text{0}\text{.095 kg}}\text{ = 1}\text{.14 mol/Kg = 1}\text{.1 m}$

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

• Calculate mole fraction of ethanol:

$\begin{array}{l}{\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}\\ \text{Known:}\\ \text{no}\text{.of moles }\left(\text{Ethanol}\right)\text{ = 0}\text{.1085 mol }\\ \text{no}\text{.of moles (Water) = }\frac{\text{95g}}{\text{18g/mol}}\text{=}\text{5}\text{.3mol}\\ {\text{n}}_{\text{ethanol}}\text{= 0}\text{.1085 mol}\\ {\text{n}}_{\text{Water}}\text{= 5}\text{.3 mol}\\ \text{Mole fraction :}\\ {\text{ (χ}}_{ethanol}\text{) = }\frac{{\text{n}}_{\text{ethanol}}}{{\text{ n}}_{\text{ethanol}}{\text{ + n}}_{\text{Water}}}\text{ = }\frac{\text{0}\text{.1085mol}}{\text{0}\text{.1085 mol + 5}\text{.3 mol}}\text{ = 0}\text{.020}\end{array}$

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 ethanol is $\text{0}\text{.020}$

Sucrose:

• Calculate mole fraction:

  Given data:

  $\text{Molality = 0}\text{.15 mol/ kg }$

  $\text{ 0}\text{.15 moles of sucrose in 1Kg of water}\text{.}$

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

  $\text{Mass of sucrose = }\left(\text{moles of sucrose}\right)\left(\text{molar mass of sucrose}\right)\text{=}\left(\text{0}\text{.15 mol}\right)\left(\text{342}\text{.3g/mol}\right)\text{=}\text{51}\text{.345 g}$

  $\begin{array}{l}\text{Mass of water = 1000 g}\text{.}\\ \text{Moles of water}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\text{=}\text{}\frac{\text{1000g}}{\text{18}\text{}\text{g/mol}}\text{=}\text{55}\text{.55moles}\end{array}$

  $\begin{array}{l}{\text{n}}_{\text{sucrose}}\text{= 0}\text{.15 mol}\\ {\text{n}}_{\text{Water}}\text{= 55}\text{.55 mol}\\ \text{Mole fraction :}\\ {\text{ (χ}}_{sucrose}\text{) = }\frac{{\text{n}}_{\text{sucrose}}}{{\text{ n}}_{\text{sucrose}}{\text{ + n}}_{\text{Water}}}\text{ = }\frac{\text{0}\text{.15mol}}{\text{0}\text{.15 mol + 55}\text{.55 mol}}\text{ = 0}\text{.0027}\text{=}\text{2}\text{.7}\times {10}^{-3}\end{array}$

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 sucrose is $\text{2}\text{.7}\times {10}^{-3}$

• Calculate weight percent of Sucrose:

Known data:

$\begin{array}{l}\text{Mass of sucrose = }\text{51}\text{.345 g}\\ \text{Mass of water = 1000 g}\text{.}\end{array}$

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

$\begin{array}{l}\underset{\_}{\text{Mass percent of sucrose:}}\\ \text{weight \% sucrose = }\frac{\text{Mass of sucrose}}{\text{ Mass of sucrose + Mass of water }}\text{×100\%}\\ \text{= }\frac{\text{51}\text{.345 g}}{\left(\text{51}\text{.345 g}\right)\text{+}\left(\text{1000 g}\right)}\text{×100\% = 4}\text{.9\%}\end{array}$

The weight percent of sucrose is calculated as shown above. Hence, the weight percent obtained is $\text{4}\text{.9\%}$

Conclusion

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