Chapter 13, Problem 4PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Fill in the blanks in the table. Aqueous solutions are assumed.

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%

Explanation
• Calculate molality of KNO3:

Given data:

WeightÂ percentÂ =Â 10.0

WeightÂ percentÂ =Â 10.0Â Â indictaesÂ 10gÂ ofÂ KNO3Â inâ€‰100â€‰gÂ ofÂ solution.MassÂ ofÂ WaterÂ (solvent)Â =Â massÂ ofÂ solutionÂ -Â massÂ ofÂ KNO3Â =Â 100Â gÂ -Â 10Â gÂ =Â 90Â gÂ

No.â€‰ofâ€‰molesâ€‰=â€‰massmolarâ€‰massno.ofÂ molesÂ (KNO3)Â =Â 10Â gÂ 101.10Â g/molÂ =Â 0.099Â mol

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

Molalityâ€‰=â€‰molesâ€‰ofâ€‰soluteMassâ€‰ofâ€‰solventâ€‰inâ€‰kg

nKNO3Â (Â molesÂ ofÂ KNO3)Â =Â 0.099Â molAmountÂ ofÂ solventÂ =Â 0.090Â Kg

MolalityÂ ofÂ solute:_AmountÂ ofÂ soluteÂ (mol)MassÂ ofÂ solventÂ (Kg)=Â â€‰0.099Â mol0.090Â kgÂ =Â 1.14Â mol/KgÂ =Â 1.1Â m

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

• Calculate mole fraction of KNO3:

Known data:

MassÂ ofÂ waterÂ =Â 90Â g.

No.â€‰ofâ€‰molesâ€‰=â€‰massmolarâ€‰mass

MolesÂ ofÂ waterâ€‰=â€‰massmolarâ€‰massâ€‰=â€‹â€‰90g18â€‹g/molâ€‰=â€‰5moles

nKNO3=Â 0.099Â molnWater=Â 5Â mol

MoleÂ fractionÂ ofÂ AÂ (Ï‡A)=Â â€‰nAÂ nAÂ +Â Â nBÂ +Â Â nCÂ +Â ...Â

Â (Ï‡KNO3)Â =Â â€‰nKNO3Â nKNO3Â +Â Â nWaterÂ Â =Â â€‰0.099mol0.099Â molÂ +Â 5Â molÂ =Â 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:

Molalityâ€‰=â€‰0.0183mâ€‰

0.0183molâ€‰ofâ€‰Aceticâ€‰acidâ€‰presentâ€‰inâ€‰1Kgâ€‰ofâ€‰water.

No.â€‰ofâ€‰molesâ€‰=â€‰massmolarâ€‰mass

MassÂ ofÂ Aceticâ€‰acidÂ =Â (â€‰0.0183â€‰mol)(60.05g/mol)â€‰=â€‰1.10â€‰g

MassÂ ofÂ waterÂ =Â 1000Â g.

weightÂ %Â AÂ =Â â€‰MassÂ ofÂ AÂ MassÂ ofÂ AÂ +Â MassÂ ofÂ BÂ +Â MassÂ ofÂ CÂ +Â ...Ã—100%

weightÂ %Â AceticÂ acidÂ =Â â€‰MassÂ ofÂ AceticÂ acidÂ MassÂ ofÂ AceticÂ acidÂ +Â MassÂ ofÂ waterÂ Ã—100%=Â â€‰1.10Â g(1.10g)+(1000Â g)Ã—100%Â =Â 0

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