Chapter 13, Problem 3PS

### 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 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 ( 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 mole fraction of NaI:

Given data:

MolalityÂ =Â 0.15Â mol/Â kg

Â 0.15Â molesÂ ofÂ NaIÂ isÂ inÂ 1KgÂ ofÂ water.

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

MassÂ ofÂ NaIÂ =Â (molesÂ ofÂ NaI)Â (molarÂ massÂ ofÂ NaI)=â€‰(0.15Â mol)(149.89g/mol)=â€‰22.48g

MassÂ ofÂ waterÂ =Â 1000Â g.MolesÂ ofÂ waterâ€‰=â€‰massmolarâ€‰massâ€‰=â€‹â€‰1000g18â€‹g/molâ€‰=â€‰55.55moles

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

nNaI=Â 0.15Â molnWater=Â 55.55Â molMoleÂ fractionÂ :Â (Ï‡NaI)Â =Â â€‰nNaIÂ nNaIÂ +Â Â nWaterÂ Â =Â â€‰0.15mol0.15Â molÂ +Â 55.55Â molÂ =Â 0.0027â€‰=â€‰2.7â€‰Ã—10âˆ’3

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 NaI is 2.7â€‰Ã—10âˆ’3

• Calculate weight percent of NaI:

Known data:

MassÂ ofÂ NaIÂ =Â â€‰22.48gMassÂ 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 NaIÂ =Â â€‰MassÂ ofÂ NaIÂ MassÂ ofÂ NaIÂ +Â MassÂ ofÂ waterÂ Ã—100%=Â â€‰22.48Â g(22.48Â g)+(1000Â g)Ã—100%Â =Â 2.3%

The weight percent of NaI is calculated as shown above. Hence, the weight percent obtained is 2.3%

• Calculate molality of Ethanol:

Given data:

WeightÂ percentÂ =Â 5.0

WeightÂ percentÂ =Â 5.0Â Â indictaesÂ 5gÂ ofÂ ethanolÂ inâ€‰100â€‰gÂ ofÂ solution.MassÂ ofÂ WaterÂ (solvent)Â =Â massÂ ofÂ solutionÂ -Â massÂ ofÂ ethanolÂ =Â 100Â gÂ -Â 5Â gÂ =Â 95Â g

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

no.ofÂ molesÂ (ethanol)Â =Â 5Â gÂ 46.07Â g/molÂ =Â 0.1085Â mol

AmountÂ ofÂ solventÂ =Â 0.095Â Kg

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

No.â€‰ofâ€‰molesÂ (mol)MassÂ ofÂ solventÂ (Kg)=Â â€‰0.1085Â mol0.095Â kgÂ =Â 1.14Â mol/KgÂ =Â 1

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started