OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)
OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)
11th Edition
ISBN: 9781305673939
Author: Darrell Ebbing; Steven D. Gammon
Publisher: Cengage Learning US
Question
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Chapter 12, Problem 12.89QP
Interpretation Introduction

Interpretation:

An aqueous solution is 14 % NH4Cl by mass.  Given the density of the solution, molality, mole fraction and molarity of NH4Cl in solution has to be determined.

Concept Introduction:

Molality is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

Molality = number of moles of solutemass of solvent in kg

Molarity is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

Molarity = number of moles of solutevolume of solution in L

A solution is at least made up of two components.  Mole fraction of a component in the solution correlates to the ratio of number of moles of that component to the total number of moles.  It is expressed as,

Mole fraction = number of moles of a componenttotal number of moles in the solution

Mass percent is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

Mass percent  = mass of solutemass of solution × 100%

Expert Solution & Answer
Check Mark

Answer to Problem 12.89QP

Molality of 14% NH4Cl is determined as 3.04 m.

Mole fraction of NH4Cl in 14% NH4Cl is determined as 0.520.

Molarity of 14% NH4Cl is determined as 2.72 M.

Explanation of Solution

Assume that volume of solution is 1.000 L which is equivalent to 1.040 kg as density of the solution is 1.040 g/mL. Then mass of ammonium chloride used to prepare 14 % NH4Cl solution is calculated as,

Mass percent of NH4Cl = mass of NH4Clmass of solution × 100%

Rearranging the above expression,

mass of NH4Cl = Mass percent× mass of solution

Substitute the known values,

mass of NH4Cl  = 14% × 1.040 kg = 0.1456 kg

As we know,

Mass of solution  = mass of solute + mass of solvent                            = mass of NH4Cl + mass of water

Mass of the solute, NH4Cl is calculated above. Therefore,

1.040 kg of solution  = 0.1456 kg + mass of watermass of water        = 1.040 kg - 0.1456 kg = 0.8944 kg

Number of moles of ammonium chloride and water are calculated as,

no.of moles of NH4Cl mass of NH4Clmolar mass of NH4Cl = 145.6 g53.49 g/mol = 2.72 molNo.of moles of water mass of H2Omolar mass of H2O = 894.4 g18.02 g/mol = 49.65 mol

Total number of moles in solution, no.of moles of NH4Cl + no.of moles of H2O = 2.72 mol + 49.65 mol = 52.37 mol

Molality of the solution is calculated as,

Molality of NH4Cl number of moles of NH4Clmass of H2O in kg = 2.72 mol0.8944 kg = 3.043 m 3.04 m

Molarity of the solution is calculated as,

Molarity of NH4Cl number of moles NH4Clvolume of solution in L = 2.72 mol1.000 L = 2.72 M

Mole fraction of ammonium chloride is calculated as,

Mole fraction of NH4Cl number of moles NH4Cltotal number of moles in solution = 2.72 mol52.37 mol = 0.0519 = 0.0520

Conclusion

Given the density of aqueous solution of 14 % NH4Cl , the molality, mole fraction and molarity of NH4Cl in solution are determined by calculating the number of moles of NH4Cl in solution.

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Chapter 12 Solutions

OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)

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