EBK INTRODUCTORY CHEMISTRY
EBK INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 8220100480485
Author: DECOSTE
Publisher: CENGAGE L
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Chapter 8, Problem 39QAP

Calculate the number of molecules present in each of the following samples.

l type="a">

  • 4.75 mmol of phosphine, PH 3
  • i>4.75 g of phosphine, PH 3

    i> 1.25 × 10 2 g of lead(II) acetate, Pb ( CH 3 CO 2 ) 2

    i>1.25 X 10 2 moles of lead(II) acetate, Pb ( CH 3 CO 2 ) 2

    i>a sample of benzene, C 6 H 6

    1. , which contains a total of 5.40 moles of carbon

    Expert Solution
    Check Mark
    Interpretation Introduction

    (a)

    Interpretation:

    The number of molecules of substance should be calculated.

    Concept Introduction:

    Number of moles is related to mass and molar mass as follows:

    n=mM

    Here, m is mass and M is molar mass.

    According to Avogadro’s law, 1 mole of a substance contains 6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbol NA.

    Thus, number of molecules can be calculated from number of moles using the following conversion factor:

    (6.023×1023 molecules1 mol).

    Answer to Problem 39QAP

    2.86×1021 molecules.

    Explanation of Solution

    Number of moles of phosphine PH3 is 4.75 mmol.

    First convert mmol into mol as follows:

    (103 mol1 mmol)

    Thus, number of moles will be:

    n=4.75 mmol(103 mol1 mmol)=4.75×103 mol

    Since, 1 mole of phosphine contains 6.023×1023 molecules thus, 4.75×103 mol of phosphine will have following molecules:

    4.75×103 mol(6.023×1023 molecules1 mol)=2.86×1021 molecules

    Therefore, number of molecules in phosphine is 2.86×1021 molecules.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (b)

    Interpretation:

    The number of molecules of substance should be calculated.

    Concept Introduction:

    Number of moles is related to mass and molar mass as follows:

    n=mM

    Here, m is mass and M is molar mass.

    According to Avogadro’s law, 1 mole of a substance contains 6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbol NA.

    Thus, number of molecules can be calculated from number of moles using the following conversion factor:

    (6.023×1023 molecules1 mol).

    Answer to Problem 39QAP

    8.432×1022 molecules.

    Explanation of Solution

    Mass of posphine is 4.75 g.

    Calculate number of moles as follows:

    n=mM

    Molar mass of phosphine is 34 g/mol, putting the values,

    n=4.75 g34 g/mol=0.14 mol

    Since, 1 mole of phosphine contains 6.023×1023 molecules thus, 0.14 mol of phosphine will have following molecules:

    0.14 mol(6.023×1023 molecules1 mol)=8.432×1022 molecules

    Therefore, number of molecules in phosphine is 8.432×1022 molecules.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (c)

    Interpretation:

    The number of molecules of substance should be calculated.

    Concept Introduction:

    Number of moles is related to mass and molar mass as follows:

    n=mM

    Here, m is mass and M is molar mass.

    According to Avogadro’s law, 1 mole of a substance contains 6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbol NA.

    Thus, number of molecules can be calculated from number of moles using the following conversion factor:

    (6.023×1023 molecules1 mol).

    Answer to Problem 39QAP

    2.32×1019 molecules.

    Explanation of Solution

    Mass of lead (II) acetate is 1.25×102 g.

    Calculate number of moles as follows:

    n=mM

    Molar mass of lead (II) acetate is 325.3 g/mol, putting the values,

    n=1.25×102 g324.3 g/mol=3.85×105 mol

    Since, 1 mole of lead (II) acetate contains 6.023×1023 molecules thus, 3.85×105 mol of lead (II) acetate will have following molecules:

    3.85×105 mol(6.023×1023 molecules1 mol)=2.32×1019 molecules

    Therefore, number of molecules in lead (II) acetate is 2.32×1019 molecules.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (d)

    Interpretation:

    The number of molecules of substance should be calculated.

    Concept Introduction:

    Number of moles is related to mass and molar mass as follows:

    n=mM

    Here, m is mass and M is molar mass.

    According to Avogadro’s law, 1 mole of a substance contains 6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbol NA.

    Thus, number of molecules can be calculated from number of moles using the following conversion factor:

    (6.023×1023 molecules1 mol).

    Answer to Problem 39QAP

    7.53×1021 molecules.

    Explanation of Solution

    Number of moles of lead (II) acetate is 1.25×102 mol.

    Since, 1 mole of lead (II) acetate contains 6.023×1023 molecules thus, 1.25×102 mol of lead (II) acetate will have following molecules:

    1.25×102 mol(6.023×1023 molecules1 mol)=7.53×1021 molecules

    Therefore, number of molecules in lead (II) acetate is 7.53×1021 molecules.

    Expert Solution
    Check Mark
    Interpretation Introduction

    (e)

    Interpretation:

    The number of molecules of substance should be calculated.

    Concept Introduction:

    Number of moles is related to mass and molar mass as follows:

    n=mM

    Here, m is mass and M is molar mass.

    According to Avogadro’s law, 1 mole of a substance contains 6.023×1023 atoms. This is known as Avogadro’s number and denoted by symbol NA.

    Thus, number of molecules can be calculated from number of moles using the following conversion factor:

    (6.023×1023 molecules1 mol).

    Answer to Problem 39QAP

    5.42×1023 molecules.

    Explanation of Solution

    Total number of moles of carbon is 5.40 mol.

    From the formula of benzene C6H6, 1 mol of benzene contains 6 mol of carbon thus, 1 mol of carbon will be present in 1/6 moles of benzene and 5.40 mol will be present in 16×5.40 mol=0.9 mol

    Since, 1 mole of benzene contains 6.023×1023 molecules thus, 0.9 mol of benzene will have following molecules:

    0.9 mol(6.023×1023 molecules1 mol)=5.42×1023 molecules

    Therefore, number of molecules in benzene is 5.42×1023 molecules.

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    Chapter 8 Solutions

    EBK INTRODUCTORY CHEMISTRY

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