General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 16, Problem 16.30P

(a)

Interpretation Introduction

Interpretation:

The molality of potassium iodide in the aqueous solution has to be calculated.

Concept Introduction:

Raoult’s law:

It states that the equilibrium vapor pressure of the solvent over a solution is directly proportional to the mole fraction of the solvent in the solution.

The expression for the vapor pressure lowering is given below.

  ΔP=χ2P

Where, ΔP is vapor pressure lowering, χ2 is the mole fraction of solute and P is vapor pressure of the pure solvent.

(a)

Expert Solution
Check Mark

Answer to Problem 16.30P

The molality of KI in the solution is 3.58m.

Explanation of Solution

Given that, the vapor pressure lowering of an aqueous solution of potassium iodide is 2.00Torrat20C.

The vapor pressure of pure water at 20C is 17.54Torr.

Then,

  ΔP=χ2P=χ2×(17.54Torr)χ2=ΔP17.54Torr=2.00Torr17.54Torr=0.114

Suppose the molality of aqueous solution of KI is m.KI is a strong electrolyte and it dissociates completely into one mole of K+(aq) ion and one mole of I(aq) ion.  Therefore, the value of i for KI is two.  The colligative molality for aqueous solution of KI can be calculated as given below.

  mc=i×m=2×m=2mc.

The number of moles of KI can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolesofsolute=Molality×kgofsolventMolesofKI=2mcmolkg×1kg=2mmol.

The mole fraction of KI can be calculated as given below.

  χKI=nKInWater+nKI0.114=2mMassofwaterMolarmassofwater+2m0.114=2m1000g18.02g/mol+2m0.114=2m55.5+2m(0.114)×(55.5+2m)=2m2m=6.327+0.228mm=3.58.

Now, molality of aqueous solution of KI can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolality=3.58mol1kg=3.58m.

Therefore, the molality of KI in the solution is 3.58m.

(b)

Interpretation Introduction

Interpretation:

The molality of strontium chloride in the aqueous solution has to be calculated.

Concept Introduction:

Raoult’s law:

It states that the equilibrium vapor pressure of the solvent over a solution is directly proportional to the mole fraction of the solvent in the solution.

The expression for the vapor pressure lowering is given below.

  ΔP=χ2P

Where, ΔP is vapor pressure lowering, χ2 is the mole fraction of solute and P is vapor pressure of the pure solvent.

(b)

Expert Solution
Check Mark

Answer to Problem 16.30P

The molality of SrCl2 in the solution is 2.38m.

Explanation of Solution

Given that, the vapor pressure lowering of an aqueous solution of SrCl2 is 2.00Torrat20C.

The vapor pressure of pure water at 20C is 17.54Torr.

Then,

  ΔP=χ2P=χ2×(17.54Torr)χ2=ΔP17.54Torr=2.00Torr17.54Torr=0.114

Suppose the molality of aqueous solution of SrCl2 is m.SrCl2 is a strong electrolyte and it dissociates completely into one mole of Sr2+(aq) ion and two moles of Cl(aq) ions.  Therefore, the value of i for SrCl2 is three.  The colligative molality for aqueous solution of SrCl2 can be calculated as given below.

  mc=i×m=3×m=3mc.

The number of moles of SrCl2 can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolesofsolute=Molality×kgofsolventMolesofSrCl2=3mcmolkg×1kg=3mmol.

The mole fraction of SrCl2 can be calculated as given below.

  χSrCl2=nSrCl2nWater+nSrCl20.114=3mMassofwaterMolarmassofwater+3m0.114=3m1000g18.02g/mol+3m0.114=3m55.5+3m(0.114)×(55.5+3m)=3m3m=6.327+0.342mm=2.38.

Now, molality of aqueous solution of SrCl2 can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolality=2.38mol1kg=2.38m.

Therefore, the molality of SrCl2 in the solution is 2.38m.

(c)

Interpretation Introduction

Interpretation:

The molality of ammonium sulfate in the aqueous solution has to be calculated.

Concept Introduction:

Raoult’s law:

It states that the equilibrium vapor pressure of the solvent over a solution is directly proportional to the mole fraction of the solvent in the solution.

The expression for the vapor pressure lowering is given below.

  ΔP=χ2P

Where, ΔP is vapor pressure lowering, χ2 is the mole fraction of solute and P is vapor pressure of the pure solvent.

(c)

Expert Solution
Check Mark

Answer to Problem 16.30P

The molality of (NH4)2SO4 in the solution is 2.38m.

Explanation of Solution

Given that, the vapor pressure lowering of an aqueous solution of (NH4)2SO4 is 2.00Torrat20C.

The vapor pressure of pure water at 20C is 17.54Torr.

Then,

  ΔP=χ2P=χ2×(17.54Torr)χ2=ΔP17.54Torr=2.00Torr17.54Torr=0.114

Suppose the molality of aqueous solution of (NH4)2SO4 is m.(NH4)2SO4 is a strong electrolyte and it dissociates completely into two moles of NH4+(aq) ions and one mole of SO42(aq) ion.  Therefore, the value of i for (NH4)2SO4 is three.  The colligative molality for aqueous solution of (NH4)2SO4 can be calculated as given below.

  mc=i×m=3×m=3mc.

The number of moles of (NH4)2SO4 can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolesofsolute=Molality×kgofsolventMolesof(NH4)2SO4=3mcmolkg×1kg=3mmol.

The mole fraction of (NH4)2SO4 can be calculated as given below.

  χ(NH4)2SO4=n(NH4)2SO4nWater+n(NH4)2SO40.114=3mMassofwaterMolarmassofwater+3m0.114=3m1000g18.02g/mol+3m0.114=3m55.5+3m(0.114)×(55.5+3m)=3m3m=6.327+0.342mm=2.38.

Now, molality of aqueous solution of (NH4)2SO4 can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolality=2.38mol1kg=2.38m.

Therefore, the molality of (NH4)2SO4 in the solution is 2.38m.

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Chapter 16 Solutions

General Chemistry

Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35PCh. 16 - Prob. 16.36PCh. 16 - Prob. 16.37PCh. 16 - Prob. 16.38PCh. 16 - Prob. 16.39PCh. 16 - Prob. 16.40PCh. 16 - Prob. 16.41PCh. 16 - Prob. 16.42PCh. 16 - Prob. 16.43PCh. 16 - Prob. 16.44PCh. 16 - Prob. 16.45PCh. 16 - Prob. 16.46PCh. 16 - Prob. 16.47PCh. 16 - Prob. 16.48PCh. 16 - Prob. 16.49PCh. 16 - Prob. 16.50PCh. 16 - Prob. 16.51PCh. 16 - Prob. 16.52PCh. 16 - Prob. 16.53PCh. 16 - Prob. 16.54PCh. 16 - Prob. 16.55PCh. 16 - Prob. 16.56PCh. 16 - Prob. 16.57PCh. 16 - Prob. 16.58PCh. 16 - Prob. 16.59PCh. 16 - Prob. 16.60PCh. 16 - Prob. 16.61PCh. 16 - Prob. 16.62PCh. 16 - Prob. 16.63PCh. 16 - Prob. 16.64PCh. 16 - Prob. 16.65PCh. 16 - Prob. 16.66PCh. 16 - Prob. 16.67PCh. 16 - Prob. 16.68PCh. 16 - Prob. 16.69PCh. 16 - Prob. 16.70PCh. 16 - Prob. 16.71PCh. 16 - Prob. 16.72PCh. 16 - Prob. 16.73PCh. 16 - Prob. 16.74PCh. 16 - Prob. 16.75PCh. 16 - Prob. 16.76PCh. 16 - Prob. 16.77PCh. 16 - Prob. 16.78PCh. 16 - Prob. 16.79PCh. 16 - Prob. 16.80PCh. 16 - Prob. 16.81PCh. 16 - Prob. 16.82PCh. 16 - Prob. 16.83PCh. 16 - Prob. 16.84PCh. 16 - Prob. 16.85PCh. 16 - Prob. 16.86PCh. 16 - Prob. 16.87PCh. 16 - Prob. 16.88PCh. 16 - Prob. 16.89PCh. 16 - Prob. 16.90PCh. 16 - Prob. 16.91PCh. 16 - Prob. 16.93P
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