MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 4, Problem 4.26P

(a)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in 130 mL of 0.45 M aluminium chloride is to be calculated.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.

The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows:

Moles of compound(mol)=[volume of solution(L)(molarityofsolution(mol)1L of solution)]

The expression to calculate the moles of ions is as follows:

moles ofion of compound(mol)=[(moles of compound(mol))(total moles of ion(mol)1mole of compound)]

The expression to calculate the number of ions is as follows:

numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)

(a)

Expert Solution
Check Mark

Answer to Problem 4.26P

The number of moles of is Al3+ and Cl is 0.058mol and 0.18mol respectively. The number of ions Al3+ and Cl is 3.5×1022 and 1.1×1023 respectively.

Explanation of Solution

The expression to calculate the moles of aluminium chloride is:

Moles of aluminium chloride(mol)=[volume of solution(mL)(1L1000mL)(molarityofaluminium chloride(mol)1L of solution)]

The relation between mL and L is:

1L=1000 mL

Substitute 130 mL for the volume of solution and 0.45mol/L for molarity of aluminium chloride in the above equation as follows:

Moles of aluminium chloride(mol)=130 mL(1L1000mL)(0.45mol1L of solution)=0.0585mol

One mole of aluminium chloride (AlCl3) on dissociation produces one mole of Al3+ ion and three moles of Cl ion.

AlCl3(s)Al3+(aq)+3Cl(aq)

The expression to calculate the amount of Al3+ ion in moles is as follows:

amountofAl3+(mol)=(moles of AlCl3(mol))(moles of Al3+ion(mol)1mole of AlCl3)

Substitute 0.0585mol for moles of AlCl3 and 1 mol for moles of Al3+ ion in the above equation as follows:

amountofAl3+(mol)=(0.0585mol)(1 mol1mole of AlCl3)=0.0585mol

The expression to calculate the number of Al3+ ions is:

numberof Al3+ions=(moles of Al3+(mol))(6.022×1023Al3+ions1mole of Al3+)

Substitute 0.0585mol for moles of Al3+ in the above equation as follows:

numberof Al3+ions=(0.0585mol)(6.022×1023 Al3+ions1mole of  Al3+)=3.522×10233.5×1023

The expression to calculate the amount of Cl ion in moles is as follows:

amountofCl(mol)=(moles of AlCl3(mol))(moles of Clion(mol)1mole of AlCl3)

Substitute 0.0585mol for moles of AlCl3 and 3 mol for moles of Cl ion in the above equation as follows:

amountofCl(mol)=(0.0585mol)(3 mol1mole of AlCl3)=0.18mol

The expression to calculate the number of Cl ions is:

numberof Clions=(moles of Cl(mol))(6.022×1023Clions1mole of Cl)

Substitute 0.0585mol for moles of Cl in the above equation as follows:

numberof  Clions=(0.18mol)(6.022×1023Clions1mole of Cl)=1.0568×10231.1×1023

Conclusion

The number of moles of is Al3+ and Cl is 0.058mol and 0.18mol respectively. The number of ions Al3+ and Cl is 3.5×1022 and 1.1×1023 respectively.

(b)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in 9.80 mL of a solution containing 2.59 g lithium sulphate per litre is to be calculated.

Concept introduction:

Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]

The expression to calculate the moles of ions is as follows:

moles ofion of compound(mol)=[(moles of compound(mol))(total moles of ion(mol)1mole of compound)]

The expression to calculate the number of ions is as follows:

numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)

(b)

Expert Solution
Check Mark

Answer to Problem 4.26P

The number of moles of is Li+ and SO42 is 4.62×104mol and 2.31×104mol respectively. The number of ions Li+ and SO42 is 2.78×1020 and 1.39×1020 respectively.

Explanation of Solution

The molecular mass of lithium sulphate (Li2SO4) is 109.94 g/mol.

The expression to calculate the moles of lithium sulphate (Li2SO4) per liter is:

Moles of Li2SO4(mol)=[volume of solution(mL)(1L1000mL)(given massof Li2SO4(g)1L)(1moleof Li2SO4(mol)molecular mass of Li2SO4(g))]

Substitute 9.80 mL for the volume of solution, 109.94 g/mol for the molecular mass of Li2SO4 and 2.59 g for given mass of Li2SO4 in the above equation as follows:

Moles of Li2SO4(mol)=[9.80 mL(1L1000mL)(2.59 g1L)(1moleof Li2SO4(mol)109.94 g)]=2.3087×104mol

One mole of lithium sulphate (Li2SO4) on dissociation produces two moles of Li+ ion and one mole of SO42 ion.

Li2SO4(s)2Li+(aq)+SO42(aq)

The expression to calculate the amount of Li+ ion in moles is as follows:

amountofLi+(mol)=(moles of Li2SO4(mol))(moles of Li+ion(mol)1mole of Li2SO4)

Substitute 2.3087×104mol for moles of Li2SO4 and 2 mol for moles of Li+ ion in the above equation as follows:

amountofLi+(mol)=(2.3087×104mol)(2 mol1mole of Li2SO4)=4.62×104mol

The expression to calculate the number of Li+ ions is:

numberof Li+ions=(moles of Li+(mol))(6.022×1023Li+ ions1mole of Li+)

Substitute 4.62×104mol for moles of Li+ in the above equation as follows:

numberof Li+ions=(4.62×104mol)(6.022×1023 Li+ ions1mole of  Li+)=2.7806×10202.78×1020

The expression to calculate the amount of SO42 ion in moles is as follows:

amountofSO42(mol)=(moles of Li2SO4(mol))(moles of SO42ion(mol)1mole of Li2SO4)

Substitute 2.3087×104mol for moles of Li2SO4 and 1 mol for moles of SO42 ion in the above equation as follows:

amountofSO42(mol)=(2.3087×104mol)(1 mol1mole of Li2SO4)=2.3087×104mol

The expression to calculate the number of SO42 ions is:

numberof SO42ions=(moles of SO42(mol))(6.022×1023SO42 ions1mole of SO42)

Substitute 2.3087×104mol for moles of SO42 in the above equation as follows:

numberof SO42ions=(2.3087×104mol)(6.022×1023 SO42 ions1mole of  SO42)=1.39030×10201.39×1020

Conclusion

The number of moles of is Li+ and SO42 is 4.62×104mol and 2.31×104mol respectively. The number of ions Li+ and SO42 is 2.78×1020 and 1.39×1020 respectively.

(c)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in 245 mL of a solution containing 3.68×1022 formula units of potassium bromide per liter is to be calculated.

Concept introduction:

A formula unit is used for the ionic compound to represent their empirical formula. The expression to calculate the moles of a compound when the volume of solution and formula unit of a compound is given is as follows:

moles of a compound(mol)=[(volume of solution(L))(given formula unit of compound(FU))(1 mole of compound6.022×1023FU)]

The expression to calculate the moles of ions is as follows:

moles ofion of compound(mol)=[(moles of compound(mol))(total moles of ion(mol)1mole of compound)]

The expression to calculate the number of ions is as follows:

numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)

(c)

Expert Solution
Check Mark

Answer to Problem 4.26P

The number of moles of is K+ and Br is 1.50×102mol and 1.50×102mol respectively. The number of ions K+ and Br is 9.02×1021 and 9.02×1021 respectively.

Explanation of Solution

One mole of potassium bromide (KBr) on dissociation produces one mole of K+ ion and one mole of Br ion.

KBr(s)K+(aq)+Br(aq)

The expression to calculate the moles of KBr is:

moles of KBr(mol)=[(volume of solution(mL))(1L1000mL)(given formula unit of KBr(FU))(1 mole of KBr6.022×1023FU)]

Substitute 3.68×1022 formula unit for given formula unit of KBr and 245 mL for the volume of solution in the above equation as follows:

moles of KBr(mol)=[(245 mL)(1L1000mL)(3.68×1022)(1 mole of KBr6.022×1023FU)]=0.01479mol

The expression to calculate the amount of K+ ion in moles is as follows:

amountofK+(mol)=(moles of KBr(mol))(moles of K+ion(mol)1mole of KBr)

Substitute 0.01479mol for moles of KBr and 1 mol for moles of K+ ion in the above equation as follows:

amountofK+(mol)=(0.01479mol)(1 mol1mole of KBr)=1.50×102mol

The expression to calculate the number of K+ ions is:

numberof K+ions=(moles of K+(mol))(6.022×1023K+ ions1mole of K+)

Substitute 1.50×102mol for moles of K+ in the above equation as follows:

numberof K+ions=(1.50×102mol)(6.022×1023 K+ ions1mole of  K+)=9.016×10219.02×1021

The expression to calculate the amount of Br ion in moles is as follows:

amountofBr(mol)=(moles of KBr(mol))(moles of Brion(mol)1mole of KBr)

Substitute 0.01479mol for moles of KBr and 1 mol for moles of Br ion in the above equation as follows:

amountofBr(mol)=(0.01479mol)(1 mol1mole of KBr)=1.50×102mol

The expression to calculate the number of Br ions is:

numberof Brions=(moles of Br(mol))(6.022×1023Br ions1mole of Br)

Substitute 1.50×102mol for moles of Br in the above equation as follows:

numberof Brions=(1.50×102mol)(6.022×1023 Br ions1mole of  Br)=9.016×10219.02×1021

Conclusion

The number of moles of is K+ and Br is 1.50×102mol and 1.50×102mol respectively. The number of ions K+ and Br is 9.02×1021 and 9.02×1021 respectively.

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Chapter 4 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 4.1 - A chemist dilutes 60.0 mL of 4.50 M potassium...Ch. 4.1 - Prob. 4.6BFPCh. 4.1 - Prob. 4.7AFPCh. 4.1 - Prob. 4.7BFPCh. 4.3 - Prob. 4.8AFPCh. 4.3 - Prob. 4.8BFPCh. 4.3 - Prob. 4.9AFPCh. 4.3 - Molecular views of the reactant solutions for a...Ch. 4.3 - It is desirable to remove calcium ion from hard...Ch. 4.3 - To lift fingerprints from a crime scene, a...Ch. 4.3 - Despite the toxicity of lead, many of its...Ch. 4.3 - Mercury and its compounds have uses from fillings...Ch. 4.4 - How many OH−(aq) ions are present in 451 mL of...Ch. 4.4 - Prob. 4.12BFPCh. 4.4 - Prob. 4.13AFPCh. 4.4 - Prob. 4.13BFPCh. 4.4 - Prob. 4.14AFPCh. 4.4 - Prob. 4.14BFPCh. 4.4 - Another active ingredient in some antacids is...Ch. 4.4 - Prob. 4.15BFPCh. 4.4 - What volume of 0.1292 M Ba(OH)2 would neutralize...Ch. 4.4 - Calculate the molarity of a solution of KOH if...Ch. 4.5 - Prob. 4.17AFPCh. 4.5 - Prob. 4.17BFPCh. 4.5 - Prob. 4.18AFPCh. 4.5 - Prob. 4.18BFPCh. 4.5 - Prob. 4.19AFPCh. 4.5 - Prob. 4.19BFPCh. 4.6 - Prob. 4.20AFPCh. 4.6 - Prob. 4.20BFPCh. 4 - Prob. 4.1PCh. 4 - What types of substances are most likely to be...Ch. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Which of the following scenes best represents how...Ch. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - A mathematical equation useful for dilution...Ch. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Does an aqueous solution of each of the following...Ch. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Calculate each of the following quantities: Mass...Ch. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Calculate each of the following quantities: Volume...Ch. 4 - Prob. 4.30PCh. 4 - Concentrated sulfuric acid (18.3 M) has a density...Ch. 4 - Prob. 4.32PCh. 4 - Muriatic acid, an industrial grade of concentrated...Ch. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Write two sets of equations (both molecular and...Ch. 4 - Why do some pairs of ions precipitate and others...Ch. 4 - Use Table 4.1 to determine which of the following...Ch. 4 - The beakers represent the aqueous reaction of...Ch. 4 - Complete the following precipitation reactions...Ch. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - If 25.0 mL of silver nitrate solution reacts with...Ch. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - With ions shown as spheres and solvent molecules...Ch. 4 - The precipitation reaction between 25.0 mL of a...Ch. 4 - A 1.50-g sample of an unknown alkali-metal...Ch. 4 - Prob. 4.54PCh. 4 - The mass percent of Cl− in a seawater sample is...Ch. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Write a general equation for a neutralization...Ch. 4 - Prob. 4.59PCh. 4 - (a) Name three common weak acids. 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