Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074



Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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 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%

  • Calculate mole fraction of NaI:

Given data:

Molality = 0.15 mol/ kg

 0.15 moles of NaI is in 1Kg of water.


Mass of NaI = (moles of NaI) (molar mass of NaI)=(0.15 mol)(149.89g/mol)=22.48g

Mass of water = 1000 g.Moles of water=massmolarmass=1000g18g/mol=55.55moles

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

nNaI= 0.15 molnWater= 55.55 molMole fraction : (χNaI) = nNaI nNaI +  nWater  = 0.15mol0.15 mol + 55.55 mol = 0.0027=2.7×103

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 NaI is 2.7×103

  • Calculate weight percent of NaI:

    Known data:

    Mass of NaI = 22.48gMass of water = 1000 g.

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

    weight % NaI = Mass of NaI Mass of NaI + Mass of water ×100%22.48 g(22.48 g)+(1000 g)×100% = 2.3%

The weight percent of NaI is calculated as shown above. Hence, the weight percent obtained is 2.3%

  • Calculate molality of Ethanol:

    Given data:

    Weight percent = 5.0

    Weight percent = 5.0  indictaes 5g of ethanol in100g of solution.Mass of Water (solvent) = mass of solution - mass of ethanol = 100 g - 5 g = 95 g


    no.of moles (ethanol) = 5 g 46.07 g/mol = 0.1085 mol

    Amount of solvent = 0.095 Kg

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


    No.ofmoles (mol)Mass of solvent (Kg)0.1085 mol0.095 kg = 1.14 mol/Kg = 1

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