Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Question
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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 ( nA +  nB +  n...).

Mole fraction of A (χA)= nA nA +  nB +  n...

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

Concentration (C, mol/kg) =  molality of solute = Amount of solute (mol)Mass of solvent (Kg)

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by 100%

weight % A = Mass of A Mass of A + Mass of B + Mass of C + ...×100%

Expert Solution & Answer
Check Mark

Answer to Problem 4PS

CompoundMolalityWeightpercentMolefractionKNO31.1_10.00.019_Ethanoic acid0.01830.1_3.29_Ethylene glycol16.1_18.00.224_

Explanation of Solution

  • Calculate molality of KNO3:

Given data:

Weight percent = 10.0

Weight percent = 10.0  indictaes 10g of KNO3 in100g of solution.Mass of Water (solvent) = mass of solution - mass of KNO3 = 100 g - 10 g = 90 g 

No.ofmoles=massmolarmassno.of moles (KNO3) = 10 g 101.10 g/mol = 0.099 mol

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

Molality=molesofsoluteMassofsolventinkg

nKNO3 ( moles of KNO3) = 0.099 molAmount of solvent = 0.090 Kg

Molality of solute:_Amount of solute (mol)Mass of solvent (Kg)0.099 mol0.090 kg = 1.14 mol/Kg = 1.1 m

The molality of KNO3 is calculated as shown above. Hence, the calculated molality of KNO31.1 m

  • Calculate mole fraction of KNO3:

Known data:

Mass of water = 90 g.

No.ofmoles=massmolarmass

Moles of water=massmolarmass=90g18g/mol=5moles

nKNO3= 0.099 molnWater= 5 mol

Mole fraction of A (χA)= nA nA +  nB +  n... 

 (χKNO3) = nKNO3 nKNO3 +  nWater  = 0.099mol0.099 mol + 5 mol = 0.019

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 KNO3 is 0.019

  • Calculate Weight percentage of ethanoic acid:

    Given data:

    Molality=0.0183m

0.0183molofAceticacidpresentin1Kgofwater.

No.ofmoles=massmolarmass

Mass of Aceticacid = (0.0183mol)(60.05g/mol)=1.10g

Mass of water = 1000 g.

weight % A = Mass of A Mass of A + Mass of B + Mass of C + ...×100%

weight % Acetic acid = Mass of Acetic acid Mass of Acetic acid + Mass of water ×100%1.10 g(1.10g)+(1000 g)×100% = 0.1%

The weight percent of ethanoic acid is calculated as shown above. Hence, the weight percent obtained is 0.1%

  • Calculate mole fraction of ethanoic acid:

Mole fraction of A (χA)= nA nA +  nB +  n...Known:no.of moles (Acetic acid) = 0.0183 mol no.of moles (Water) = 1000g18g/mol=55.55molnAceticacid= 0.0183 molnWater= 55.55 molMole fraction : (χAceticacid) = nAceticacid nAceticacid +  nWater  = 0.0183 mol0.0183 mol + 55.55 mol = 3.29

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 3.29

  • Calculate molality of ethylene glycol:

Molality of solute = Amount of solute (mol)Mass of solvent (Kg)Given:Weight percent = 18.0  indictaes 18g of glycol in100g of solution.Mass of Water (solvent) = mass of solution - mass of glycol = 100 g - 18 g = 82 g no.of moles (glycol) = 82 g 62.07 g/mol = 1.32 molnglycol ( moles of glycol) = 1.32 molAmount of solvent = 0.082 KgMolality of solute:_Amount of solute (mol)Mass of solvent (Kg)1.32 mol0.082 kg = 16.1 mol/Kg = 16.1 m

The molality of ethylene glycol is calculated as shown above. Hence, the calculated molality of ethylene glycol 16.1 m

  • Calculate mole fraction of ethylene glycol:

Mole fraction of A (χA)= nA nA +  nB +  n... Mass of water = 82 g.Moles of watermassmolarmass =82 g18g/mol=4.56molesnglycol ( moles ofglycol) = 1.32 molno.of moles (Water) = 4.56 molesnglycol= 1.32 molnWater= 4.56 molMole fraction : (χglycol) = nglycol nglycol +  nWater  = 1.32 mol1.32 mol + 4.56 mol = 0.224

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 0.224

Conclusion

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

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Chapter 13 Solutions

Chemistry & Chemical Reactivity

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