General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
Question
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Chapter 16, Problem 16.26P

(a)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of 1.50m ethanol at 25C 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 16.26P

The vapor pressure lowering for the aqueous solution of ethanol is 0.834mbar.

Explanation of Solution

Given that the molality of aqueous solution of ethanol is 1.5m.  Ethanol is a weak electrolyte.  Therefore, the value of i for ethanol is one.  The colligative molality for aqueous solution of ethanol can be calculated as given below.

  mc=i×m=1×1.5m=1.5mc.

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

  Molality=MolesofsolutekgofsolventMolesofsolute=Molality×kgofsolventMolesofethanol=1.5molkg×1kg=1.5mol.

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

  χethanol=nethanolnWater+nethanolχethanol=1.5MassofwaterMolarmassofwater+1.5χethanol=1.51000g18.02g/mol+1.5χethanol=0.0263

The vapor pressure of water at 25C is 31.7mbar.

The vapor pressure lowering can be calculated as given below.

  ΔP=χ2P=(0.0263)×(31.7mbar)=0.834mbar.

Therefore, the vapor pressure lowering for the aqueous solution of ethanol is 0.834mbar.

(b)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of 0.50mTl(Cl)3 at 25C has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 16.26P

The vapor pressure lowering for the aqueous solution of Tl(Cl)3 is 1.10mbar.

Explanation of Solution

Given that the molality of aqueous solution of thallium (III) chloride is 0.50m.Tl(Cl)3 is a strong electrolyte and it dissociates completely into one mole of Tl3+(aq) ion and three mole of Cl(aq) ion.  Therefore, the value of i for Tl(Cl)3 is four.  The colligative molality for aqueous solution of Tl(Cl)3 can be calculated as given below.

  mc=i×m=4×0.50m=2.00mc.

The number of moles of thallium chloride can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolesofsolute=Molality×kgofsolventMolesofTl(Cl)3=2.00molkg×1kg=2.00mol.

The mole fraction of thallium chloride can be calculated as given below.

  χTl(Cl)3=nTl(Cl)3nWater+nTl(Cl)3χTl(Cl)3=2.00MassofwaterMolarmassofwater+2.00χTl(Cl)3=2.001000g18.02g/mol+2.00χTl(Cl)3=0.035

The vapor pressure of water at 25C is 31.7mbar.

The vapor pressure lowering can be calculated as given below.

  ΔP=χ2P=(0.035)×(31.7mbar)=1.10mbar.

Therefore, the vapor pressure lowering for the aqueous solution of thallium chloride is 1.10mbar.

(c)

Interpretation Introduction

Interpretation:

The vapor pressure lowering of aqueous solution of 0.25mK2SO4 at 25C has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 16.26P

The vapor pressure lowering for the aqueous solution of K2SO4 is 0.423mbar.

Explanation of Solution

Given that the molality of aqueous solution of K2SO4 is 0.25m.K2SO4 is a strong electrolyte and it dissociates completely into two moles of K+(aq) ions and one mole of SO42(aq) ion.  Therefore, the value of i for K2SO4 is three.  The colligative molality for aqueous solution of K2SO4 can be calculated as given below.

  mc=i×m=3×0.25m=0.75mc.

The number of moles of K2SO4 can be calculated as given below.

  Molality=MolesofsolutekgofsolventMolesofsolute=Molality×kgofsolventMolesofK2SO4=0.75molkg×1kg=0.75mol

The mole fraction of K2SO4 can be calculated as given below.

  χK2SO4=nK2SO4nWater+nK2SO4χK2SO4=0.75MassofwaterMolarmassofwater+0.75χK2SO4=0.751000g18.02g/mol+0.75χK2SO4=0.0133

The vapor pressure of water at 25C is 31.7mbar.

The vapor pressure lowering can be calculated as given below.

  ΔP=χ2P=(0.0133)×(31.7mbar)=0.423mbar.

Therefore, the vapor pressure lowering for the aqueous solution of K2SO4 is 0.423mbar.

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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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