Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.27P

(a)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in 88 mL of 1.75 M magnesium 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 liter 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 4.27P

The number of moles of is Mg2+ and Cl ions is 0.15mol and 0.31mol respectively. The number of ions Mg2+ and Cl is 9.3×1022 and 1.9×1023 respectively.

Explanation of Solution

The expression to calculate the moles of magnesium chloride is:

Moles of magnesium chloride(mol)=[volume of solution(mL)(1L1000mL)(molarityofmagnesium chloride(mol)1L of solution)]

The relation between mL and L is:

1L=1000 mL

Substitute 88 mL for the volume of solution and 1.75 mol/L for molarity of magnesium chloride in the above equation as follows:

Moles of magnesium chloride(mol)=88 mL(1L1000mL)(1.75mol1L of solution)=0.154mol

One mole of magnesium chloride (MgCl2) on dissociation produces one mole of Mg2+ ion and two moles of Cl ion.

MgCl2(s)Mg2+(aq)+2Cl(aq)

The expression to calculate the amount of Mg2+ ion in moles is as follows:

amountofMg2+(mol)=(moles of MgCl2(mol))(moles of Mg2+ion(mol)1mole of MgCl2)

Substitute 0.154mol for moles of MgCl2 and 1 mol for moles of Mg2+ ion in the above equation as follows:

amountofMg2+(mol)=(0.154mol)(1 mol1mole of MgCl2)=0.154mol

The expression to calculate the number of Mg2+ ions is:

numberof Mg2+ions=(moles of Mg2+(mol))(6.022×1023Mg2+ions1mole of Mg2+)

Substitute 0.154mol for moles of Mg2+ in the above equation as follows:

numberof Mg2+ions=(0.154mol)(6.022×1023 Al3+ions1mole of  Mg2+)=9.2738×10229.3×1022

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

amountofCl(mol)=(moles of MgCl2(mol))(moles of Clion(mol)1mole of MgCl2)

Substitute 0.154mol for moles of MgCl2 and 2 mol for moles of Cl ion in the above equation as follows:

amountofCl(mol)=(0.154mol)(2 mol1mole of MgCl2)=0.31mol

The expression to calculate the number of Cl ions is:

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

Substitute 0.31mol for moles of Cl in the above equation as follows:

numberof  Clions=(0.31mol)(6.022×1023Clions1mole of Cl)=1.85477×10231.9×1023

Conclusion

The number of moles of is Mg2+ and Cl is 0.15mol and 0.31mol respectively. The number of ions Mg2+ and Cl is 9.3×1022 and 1.9×1023 respectively.

(b)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in 321 mL of a solution containing 0.22 g aluminium sulfate per liter 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 liter 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 4.27P

The number of moles of is Al3+ and SO42 is 4.1×104mol and 6.2×104mol respectively. The number of ions Al3+ and SO42 is 2.5×1020 and 3.7×1020 respectively.

Explanation of Solution

The molecular mass of aluminium sulfate (Al2(SO4)3) is 342.14 g/mol.

The expression to calculate the moles of aluminium sulfate (Al2(SO4)3) per liter is:

Moles of Al2(SO4)3(mol)=[volume of solution(mL)(1L1000mL)(given massof Al2(SO4)3(g)1L)(1moleof Al2(SO4)3(mol)molecular mass of Al2(SO4)3(g))]

Substitute 321 mL for the volume of solution, 342.14 g/mol for the molecular mass of Al2(SO4)3 and 0.22 g for given mass of Al2(SO4)3 in the above equation as follows:

Moles of Al2(SO4)3(mol)=[321 mL(1L1000mL)(0.22 g1L)(1moleof Al2(SO4)3(mol)109.94 g)]=2.06406×104mol

One mole of aluminium sulfate (Al2(SO4)3) on dissociation produces two moles of Al3+ ion and three moles of SO42 ion.

Al2(SO4)3(s)2Al3+(aq)+3SO42(aq)

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

amountofAl3+(mol)=(moles of Al2(SO4)3(mol))(moles of Al3+ion(mol)1mole of Al2(SO4)3)

Substitute 2.06406×104mol for moles of Al2(SO4)3 and 2 mol for moles of Al3+ ion in the above equation as follows:

amountofAl3+(mol)=(2.06406×104mol)(2 mol1mole of Al2(SO4)3)=4.1×104mol

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

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

Substitute 4.1×104mol for moles of Al3+ in the above equation as follows:

numberof Al3+ions=(4.1×104mol)(6.022×1023 Al3+ ions1mole of Al3+)=2.4680×10202.5×1020

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

amountofSO42(mol)=(moles of Al2(SO4)3(mol))(moles of SO42ion(mol)1mole of Al2(SO4)3)

Substitute 2.06406×104mol for moles of Al2(SO4)3 and 3 mol for moles of SO42 ion in the above equation as follows:

amountofSO42(mol)=(2.06406×104mol)(3 mol1mole of Al2(SO4)3)=6.2×104mol

The expression to calculate the number of SO42 ions is:

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

Substitute 6.2×104mol for moles of SO42 in the above equation as follows:

numberof SO42ions=(6.2×104mol)(6.022×1023 SO42 ions1mole of  SO42)=3.7289×10203.7×1020

Conclusion

The number of moles of is Al3+ and SO42 is 4.1×104mol and 6.2×104mol respectively. The number of ions Al3+ and SO42 is 2.5×1020 and 3.7×1020 respectively.

(c)

Interpretation Introduction

Interpretation:

The number of moles and number of ions of each type in 1.65 mL of a solution containing 8.83×1021 formula units of cesium nitrate 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 4.27P

The number of moles of is Cs+ and NO3 is 0.0242 mol and 0.0242 mol respectively. The number of ions Cs+ and NO3 is 1.46×1022 and 1.46×1022 respectively.

Explanation of Solution

One mole of cesium nitrate (CsNO3) on dissociation produces one mole of Cs+ ion and one mole of NO3 ion.

CsNO3(s)Cs+(aq)+NO3(aq)

The expression to calculate the moles of CsNO3 is:

moles of CsNO3(mol)=[(volume of solution(mL))(1L1000mL)(given formula unit of CsNO3(FU))(1 mole of CsNO36.022×1023FU)]

Substitute 8.83×1021 formula unit for given formula unit of CsNO3 and 1.65 mL for the volume of solution in the above equation as follows:

moles of CsNO3(mol)=[(1.65 mL)(1L1000mL)(8.83×1021)(1 mole of KBr6.022×1023FU)]=0.024194mol

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

amountofCs+(mol)=(moles of CsNO3(mol))(moles of Cs+ion(mol)1mole of CsNO3)

Substitute 0.024194mol for moles of CsNO3 and 1 mol for moles of Cs+ ion in the above equation as follows:

amountofCs+(mol)=(0.024194mol)(1 mol1mole of CsNO3)=0.0242mol

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

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

Substitute 0.0242mol for moles of Cs+ in the above equation as follows:

numberof Cs+ions=(0.0242mol)(6.022×1023 Cs+ ions1mole of  Cs+)=1.456×10221.46×1022

The expression to calculate the amount of NO3 ion in moles is as follows:

amountofNO3(mol)=(moles of CsNO3(mol))(moles of NO3ion(mol)1mole of CsNO3)

Substitute 0.024194mol for moles of CsNO3 and 1 mol for moles of NO3 ion in the above equation as follows:

amountofNO3(mol)=(0.024194mol)(1 mol1mole of CsNO3)=0.0242mol

The expression to calculate the number of NO3 ions is:

numberof NO3ions=(moles of NO3(mol))(6.022×1023NO3 ions1mole of NO3)

Substitute 0.0242mol for moles of NO3 in the above equation as follows:

numberof NO3ions=(0.0242mol)(6.022×1023 NO3 ions1mole of  NO3)=1.4569×10221.46×1022

Conclusion

The number of moles of is Cs+ and NO3 is 0.0242 mol and 0.0242 mol respectively. The number of ions Cs+ and NO3 is 1.46×1022 and 1.46×1022 respectively.

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Chapter 4 Solutions

Chemistry: The Molecular Nature of Matter and Change

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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