   Chapter 15.6, Problem 2.1ACP

Chapter
Section
Textbook Problem

Freezing point depression is one means of determining the molar mass of a compound. The freezing point depression constant of benzene is −5.12 °C/m. a. When a 0.503 g sample of the white crystalline dimer is dissolved in 10.0 g benzene, the freezing point of benzene is decreased by 0542 °C. Verify that the molar mass of the dimer is 475 g/mol when determined by freezing point depression. Assume no dissociation of the dimer occurs. b. The correct molar mass of the dimer is 487 g/mol. Explain why the dissociation equilibrium causes the freezing point depression calculation to yield a lower molar mass for the dimer.

(a)

Interpretation Introduction

Interpretation: Molar mass of dimer has to be verified.

Concept Introduction: At equilibrium the concentration of the reaction and the product is equated to a constant K, where K is known as the equilibrium

If a reaction is as aA+bBcC+dD ,

The equilibrium constant K=[C]c[D]d[A]a[B]b

Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent.

Explanation

Concentration (molality) =0.5420C÷5.120C/m

= 0.1059moldimer/kg

Amount dimer = 0.1059moldimer/kg(0.0100Kgbenzene) =1

(b)

Interpretation Introduction

Interpretation: Dissociation equilibrium causes the freezing point depression to yield a lower molar mass has to be explained.

Concept Introduction: At equilibrium the concentration of the reaction and the product is equated to a constant K, where K is known as the equilibrium

If a reaction is as aA+bBcC+dD ,

The equilibrium constant K=[C]c[D]d[A]a[B]b

Mass is the amount of matter contained in it.

No:ofmoles=Mass×1molMolarmass

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