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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

The table below gives experimentally determined values for freezing points of 1.00% solutions (mass %) of a series of acids.

  1. (a) Calculate the molality of each solution, determine the calculated freezing points, and then calculate the values of the van’t Hoff factor i. Fill these values into the table.

    simages

  2. (b) Analyze the results, comparing the values of i for the various acids. How do these data relate to acid strengths? (The discussion of strong and weak acids in Section 3.6 will assist you to answer this question.)

(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

Explanation

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

(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

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