   Chapter 13, Problem 2PS

Chapter
Section
Textbook Problem

You dissolve 45.0 g of camphor, C10H16O, in 425 mL of ethanol, C2H5OH. Calculate the molality, mole fraction, and weight percent of camphor in this solution. (The density of ethanol is 0.785 g/mL.)

Interpretation Introduction

Interpretation: The mole fraction, molality, and weight percent of camphor has to be identified.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture ( nA +  nB +  n...).

Mole fraction of A (χA)= nA nA +  nB +  n...

Molality: The amount of solute (mol) per kilogram of solvent.

Concentration (C, mol/kg) =  molality of solute = Amount of solute (mol)Mass of solvent (Kg)

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by 100%

weight % A = Mass of A Mass of A + Mass of B + Mass of C + ...×100%

Explanation
• The Mole fraction:

Given data:

Mass ofcamphor = 45 gVolume of ethanol = 425 mlDensity of ethanol=0.785 g/ml

By using the density and volume, the mass of ethanol can be calculated.

Massofethanol=Volume×density=(425 ml)(0.785 g/ml)=333.6g

In-order to find out the number of moles, the equation given below can be used.

No.ofmoles=massmolarmass

no.of moles (camphor) = 45 g 152.23 g/mol = 0.29 molno.of moles (ethanol) = 333.6g 46.07 g/mol = 7.24 mol

Using the number of moles, mole fraction can be determined.

Mole fraction of A (χA)= nA nA +  nB +  n...ncamphor= 0.29 molnethanol= 7.24 molMole fraction : (χcamphor) = ncamphor ncamphor +  nethanol  = 0.29 mol0.29 mol + 7.24 mol = 0.038

Convert the mass of camphor and ethanol into unit of moles and substitute in the respective mole fraction equation as shown above. Hence, the mole fraction of camphor is 0

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