   Chapter 13, Problem 4PS

Chapter
Section
Textbook Problem

Fill in the blanks in the table. Aqueous solutions are assumed. Interpretation Introduction

Interpretation: The mole fraction, molality, and weight percent of given compounds 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
• Calculate molality of KNO3:

Given data:

Weight percent = 10.0

Weight percent = 10.0  indictaes 10g of KNO3 in100g of solution.Mass of Water (solvent) = mass of solution - mass of KNO3 = 100 g - 10 g = 90 g

No.ofmoles=massmolarmassno.of moles (KNO3) = 10 g 101.10 g/mol = 0.099 mol

Molality of a solute is defined as the amount of solute dissolves in kilogram of solvent.

Molality=molesofsoluteMassofsolventinkg

nKNO3 ( moles of KNO3) = 0.099 molAmount of solvent = 0.090 Kg

Molality of solute:_Amount of solute (mol)Mass of solvent (Kg)0.099 mol0.090 kg = 1.14 mol/Kg = 1.1 m

The molality of KNO3 is calculated as shown above. Hence, the calculated molality of KNO31.1 m

• Calculate mole fraction of KNO3:

Known data:

Mass of water = 90 g.

No.ofmoles=massmolarmass

Moles of water=massmolarmass=90g18g/mol=5moles

nKNO3= 0.099 molnWater= 5 mol

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

(χKNO3) = nKNO3 nKNO3 +  nWater  = 0.099mol0.099 mol + 5 mol = 0.019

Convert the mass of water into unit of moles and substitute in the respective mole fraction equation as shown above. Hence, the mole fraction of KNO3 is 0.019

• Calculate Weight percentage of ethanoic acid:

Given data:

Molality=0.0183m

0.0183molofAceticacidpresentin1Kgofwater.

No.ofmoles=massmolarmass

Mass of Aceticacid = (0.0183mol)(60.05g/mol)=1.10g

Mass of water = 1000 g.

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

weight % Acetic acid = Mass of Acetic acid Mass of Acetic acid + Mass of water ×100%1.10 g(1.10g)+(1000 g)×100% = 0

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