Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 13, Problem 106SCQ

(a)

Interpretation Introduction

Interpretation: The molality, calculated freezing points and van’t Hoff factor (i) for the acids in the given table has to be determined

Concept introduction:

Freezing point depression: The freezing point of the solution varies with the solute concentration.

  ΔTfp=Kfpmsolute

  where,

  Kfp is the molal freezing point depression constant.

van’t Hoff factor, i: it is the ration between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

  van'tHofffactor,i=measuredΔTfpCalculatedΔTfp

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  Molality (m) =Numberofmolesofsolute1kgofsolvent

The number of moles of any substance can be determined using the equation

  Numberofmole=GivenmassofthesubstanceMolarmass

(a)

Expert Solution
Check Mark

Answer to Problem 106SCQ

Acid (1 mass %)Molality (mol/kg H2O)Tmeasured (C)Tcalculated (C)i
HNO30.1603m-0.56-0.29810C1.8785
CH3CO2H0.1682m-0.32-0.31280C1.023
H2SO40.1029m-0.42-0.19140C2.1943
H2C2O40.1122m-0.30-0.20860C1.4381
HCO2H0.2194m-0.42-0.4080C1.029
CCl3CO2H0.06182m-0.21-0.11500C1.826

Explanation of Solution

The number of moles of any substance can be determined using the equation

  Numberofmole=GivenmassofthesubstanceMolarmass

Molarity of any substance can be determined using the equation

,

  Molality (m) =Numberofmolesofsolute1kgofsolvent

Therefore,

For HNO3,

In 100g of solution, 1g of HNO3 is present (mass % = 1% ) and mass of water is 99g.

    Numberofmole=GivenmassofthesubstanceMolarmass=1g63.01g/mol=0.01587mol

  Molality (m) =NumberofmolesofsoluteSolventinkg=0.01587mol0.099kg=0.1603m

Depression in freezing point is calculated by using the equation

  ΔTfp= Kfp. msolute

Therefore,

For HNO3,

  ΔTfp= Kfp. msolute=(-1.860C/m)(0.1603m)=-0.29810C

For CH3CO2H,

In 100g of solution, 1g of CH3CO2H is present (mass % = 1%) and mass of water is 99g.

    Numberofmole=GivenmassofthesubstanceMolarmass=1g60.05g/mol=0.01665mol

  Molality (m) =NumberofmolesofsoluteSolventinkg=0.01665mol0.099kg=0.1682m

Depression in freezing point is calculated by using the equation

  ΔTfp= Kfp. msolute

For CH3CO2H,

  ΔTfp= Kfp. msolute=(-1.860C/m)(0.1682m)=-0.31280C

For H2SO4,

In 100g of solution, 1g of H2SO4 is present (mass % = 1%) and mass of water is 99g

    Numberofmole=GivenmassofthesubstanceMolarmass=1g98.08g/mol=0.01019mol

  Molality (m) =NumberofmolesofsoluteSolventinkg=0.01019mol0.099kg=0.1029m

Depression in freezing point is calculated by using the equation

  ΔTfp= Kfp. msolute

For H2SO4,

  ΔTfp= Kfp. msolute=(-1.860C/m)(0.1029m)=-0.19140C

For H2C2O4,

In 100g of solution, 1g of H2C2O4 is present (mass % = 1%) and mass of water is 99g

    Numberofmole=GivenmassofthesubstanceMolarmass=1g90.03g/mol=0.0111mol

Molality (m) =NumberofmolesofsoluteSolventinkg=0.0111mol0.099kg=0.1122m

Depression in freezing point is calculated by using the equation

  ΔTfp= Kfp. msolute

For H2C2O4,

ΔTfp= Kfp. msolute=(-1.860C/m)(0.1122m)=-0.20860C

For HCO2H,

In 100g of solution, 1g of HCO2H is present (mass % = 1%) and mass of water is 99g

    Numberofmole=GivenmassofthesubstanceMolarmass=1g46.03g/mol=0.02172mol

  Molality (m) =NumberofmolesofsoluteSolventinkg=0.02172mol0.099kg=0.2194m

Depression in freezing point is calculated by using the equation

  ΔTfp= Kfp. msolute

For HCO2H,

  ΔTfp= Kfp. msolute=(-1.860C/m)(0.2194m)=-0.4080C

For CCl3CO2H,

In 100g of solution, 1g of CCl3CO2H is present (mass % = 1%) and mass of water is 99g.

    Numberofmole=GivenmassofthesubstanceMolarmass=1g163.378g/mol=0.00612mol

  Molality (m) =NumberofmolesofsoluteSolventinkg=0.00612mol0.099kg=0.06182m

Depression in freezing point is calculated by using the equation

  ΔTfp= Kfp. msolute

For CCl3CO2H,

  ΔTfp= Kfp. msolute=(-1.860C/m)(0.06182m)=-0.11500C

Van’t Hoff factor can be find out using the equation,

  van'tHofffactor,i=measuredΔTfpCalculatedΔTfp

For HNO3,

  i=0.560C0.29810C=1.8785

For CH3CO2H,

  i=0.320C0.31280C=1.023

For H2SO4,

  i=0.420C0.19140C=2.1943

For H2C2O4,

  i=0.300C0.20860C=1.4381

For HCO2H,

  i=0.420C0.4080C=1.029

For CCl3CO2H,

  i=0.210C0.11500C=1.826

(b)

Interpretation Introduction

Interpretation: The relation of acidic strength with the i value has to be given.

Concept introduction:

van’t Hoff factor, i: it is the ration between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

  van'tHofffactor,i=measuredΔTfpCalculatedΔTfp

(b)

Expert Solution
Check Mark

Answer to Problem 106SCQ

The values of ‘i’ increases in the increasing order of strength of acids.

Explanation of Solution

The increasing order of acidic strength for the given acids is,

    CH3CO2H<HCO2H<H2C2O4<CCl3CO2H<HNO3<H2SO4

The increasing order of ‘i’ value for the given acids is,

  CH3CO2H<HCO2H<H2C2O4<CCl3CO2H<HNO3<H2SO4

Hence,

The values of ‘i’ increases in the increasing order of strength of acids.

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Chapter 13 Solutions

Chemistry & Chemical Reactivity

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