Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 13, Problem 13.126P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the compound is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

  ΔTf=ikfm        (1)

Here,

ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

(a)

Expert Solution
Check Mark

Answer to Problem 13.126P

89.9g/mol is the molar mass of the compound.

Explanation of Solution

The formula to calculate the change in freezing point is as follows:

  ΔTf=Tf(solvent)Tf(solution)        (2)

Substitute 0°C for Tf(solvent) and 0.201°C for Tf(solution) in equation (2).

  ΔTf=0°C(0.201°C)=0.201°C

The solute is a nonvolatile non-electrolyte so its van’t Hoff factor is 1.

Rearrange equation (1) to calculate the molarity of the solution as follows:

  m=ΔTfikf        (3)

Substitute 1 for i, 0.201°C for ΔTf, 1.86°C/m for kb in equation (3).

  m=0.201°C(1)(1.86°C/m)=0.1080645m

The density of the solution is calculated as follows:

  Density of solution(ρ)=Mass (M)of solutionVolume (V)of solution        (4)

Rearrange equation (4) to calculate the mass of the solution as follows:

  Mass of solution=(Density of solution)(Volume of solution)        (5)

Substitute 25mL for the volume of solution and 1g/mL for the density of the solution in equation (5) to calculate the mass of water.

  Mass of water=(25mL)(1g1mL)(1kg103g)=0.0250kg

The formula to calculate the molality of the solution is as follows:

  Molality=amount(mol)ofsolutemass(kg)ofsolvent        (6)

Rearrange equation (6) to calculate the moles of solute as follows:

  Amountofsolute=(Molality)(massofsolvent)        (7)

Substitute 0.1080645m for the molality and 0.0250kg for the mass of solvent in equation (7) to calculate the amount of solute.

  Amount of solute=(0.1080645m)(0.0250kg)=0.0027016mol

The formula to calculate the number of moles is as follows:

  Number of moles=Given massMolar mass        (8)

Rearrange equation (8) to calculate the molar mass as follows:

  Molar mass=Given massNumber of moles        (9)

Substitute 0.243g for the given mass and 0.0027016mol for the number of moles in equation (9).

  Molar mass=0.243g0.0027016mol=89.946698g/mol89.9g/mol.

(b)

Interpretation Introduction

Interpretation:

The empirical and the molecular formula of the compound are to be determined.

Concept introduction:

An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 13.126P

The empirical and molecular formula of the compound is C2H5O and C4H10O2.

Explanation of Solution

Consider the mass of the compound to be 100g. The mass of carbon is 53.31g, the mass of hydrogen is 11.18g.

The formula to calculate the mass of the compound is as follows:

  Mass of compound=Mass of C+Mass of H+Mass of O        (10)

Rearrange equation (10) to calculate the mass of oxygen as follows:

  Mass of O=Mass of compound(Mass of C+Mass of H)        (11)

Substitute 100g for the mass of the compound, 53.31g for the mass of carbon and 11.18g for the mass of hydrogen in equation (11).

  Mass of O=100 g(53.31 g+11.18 g)=35.51g

The formula to calculate the moles of the compound is as follows:

  Moles of compound=Given massMolar mass        (12)

Substitute 53.31g for the given mass and 12.01 g/mol for the molar mass in equation (12) to calculate the moles of carbon.

  Moles of C=(53.31 g)(1mol12.01 g)=4.43880mol

Substitute 11.18g for the given mass and 1.008 g/mol for the molar mass in equation (12) to calculate the moles of hydrogen.

  Moles of H=(11.18 g)(1mol1.008 g)=11.09127mol

Substitute 35.51g for the given mass and 16 g/mol for the molar mass in equation (12) to calculate the moles of oxygen.

  Moles of O=(35.51 g)(1mol16 g)=2.219375mol

Write the amount of carbon, hydrogen, and oxygen as subscripts of their symbols to obtain a preliminary formula as follows:

  C4.43880H11.09127O2.219375

The smallest subscript is 2.219375. Therefore, divide each subscript by 2.219375 as follows:

  C4.438802.219375H11.091272.219375O2.2193752.219375C2H5O

The subscripts are in the whole number. Hence, the empirical formula of the compound is C2H5O.

The expression to calculate the empirical formula mass of C2H5O is as follows:

  Empirical formula mass of C2H5O=(2)(M of C)+(5)(M of H)+(1)(M of O)        (13)

Substitute 12.01 g/mol for M of C, 1.008 g/mol for M of H, and 16.00 g/mol for M of O in equation (13) as follows:

  Empirical formula mass of C2H5O=(2)(12.01 g/mol)+(5)(1.008 g/mol)+(1)(16.00 g/mol)=45.06 g/mol.

The molar mass of the compound is 89.9g/mol that is double its empirical mass. So the molecular formula of the compound is C4H10O2.

(c)

Interpretation Introduction

Interpretation:

The Lewis structures for the compound that forms hydrogen bonds and one that does not form hydrogen bonds are to be drawn.

Concept introduction:

Lewis structure is basically a simplified representation of the structure of any molecule or atom. Lewis structure shows the bonding between the atoms and the lone pairs of electrons as dot.

The steps to draw the Lewis structure of any molecule are as follows:

1. Write the letter symbol for each element present in the compound and draw the skeletal structure. Generally, the least electronegative element is considered as the central element.

2. Count the total number of valence electrons in the molecule. In case of charged molecules subtract the positive charge from the total number of valence electrons and add the negative charge to the total number of valence electrons.

3. Assign two electrons between two atoms and join them via a single bond. Place the remaining valence electrons as lone pairs such that octet of each element is achieved. Use multiple bonds to complete the octet.

(c)

Expert Solution
Check Mark

Answer to Problem 13.126P

The Lewis structure of the compound that forms hydrogen bonds is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 13, Problem 13.126P , additional homework tip  1

The Lewis structure of the compound that does not form hydrogen bonds is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 13, Problem 13.126P , additional homework tip  2

Explanation of Solution

The compound will form hydrogen bonds with the electronegative element is present at the end positions. But no hydrogen bonding occurs when the electronegative element is placed in between the other elements.

So the Lewis structure of the compound that forms hydrogen bonds is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 13, Problem 13.126P , additional homework tip  3

The Lewis structure of the compound that does not form hydrogen bonds is as follows:

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 13, Problem 13.126P , additional homework tip  4

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Chapter 13 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 13.5 - Prob. 13.6AFPCh. 13.5 - Prob. 13.6BFPCh. 13.6 - Calculate the vapor pressure lowering of a...Ch. 13.6 - Prob. 13.7BFPCh. 13.6 - Prob. 13.8AFPCh. 13.6 - Prob. 13.8BFPCh. 13.6 - Prob. 13.9AFPCh. 13.6 - Prob. 13.9BFPCh. 13.6 - A solution is made by dissolving 31.2 g of...Ch. 13.6 - Prob. 13.10BFPCh. 13.7 - Prob. B13.1PCh. 13.7 - Prob. B13.2PCh. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Which would you expect to be more effective as a...Ch. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10PCh. 13 - Prob. 13.11PCh. 13 - What is the strongest type of intermolecular force...Ch. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - What is the relationship between solvation and...Ch. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Use the following data to calculate the combined...Ch. 13 - Use the following data to calculate the combined...Ch. 13 - State whether the entropy of the system increases...Ch. 13 - Prob. 13.39PCh. 13 - Prob. 13.40PCh. 13 - Prob. 13.41PCh. 13 - Prob. 13.42PCh. 13 - Prob. 13.43PCh. 13 - Prob. 13.44PCh. 13 - For a saturated aqueous solution of each of the...Ch. 13 - Prob. 13.46PCh. 13 - Prob. 13.47PCh. 13 - Prob. 13.48PCh. 13 - Prob. 13.49PCh. 13 - Prob. 13.50PCh. 13 - Prob. 13.51PCh. 13 - Prob. 13.52PCh. 13 - Prob. 13.53PCh. 13 - Prob. 13.54PCh. 13 - Prob. 13.55PCh. 13 - Calculate the molarity of each aqueous...Ch. 13 - Calculate the molarity of each aqueous...Ch. 13 - Prob. 13.58PCh. 13 - Calculate the molarity of each aqueous...Ch. 13 - How would you prepare the following aqueous...Ch. 13 - Prob. 13.61PCh. 13 - Prob. 13.62PCh. 13 - Prob. 13.63PCh. 13 - Prob. 13.64PCh. 13 - Prob. 13.65PCh. 13 - Prob. 13.66PCh. 13 - Prob. 13.67PCh. 13 - Prob. 13.68PCh. 13 - Prob. 13.69PCh. 13 - Prob. 13.70PCh. 13 - Prob. 13.71PCh. 13 - Prob. 13.72PCh. 13 - Prob. 13.73PCh. 13 - Prob. 13.74PCh. 13 - Prob. 13.75PCh. 13 - Prob. 13.76PCh. 13 - Prob. 13.77PCh. 13 - Prob. 13.78PCh. 13 - Prob. 13.79PCh. 13 - Prob. 13.80PCh. 13 - Prob. 13.81PCh. 13 - What are the most important differences between...Ch. 13 - Prob. 13.83PCh. 13 - Prob. 13.84PCh. 13 - Prob. 13.85PCh. 13 - Prob. 13.86PCh. 13 - Prob. 13.87PCh. 13 - Prob. 13.88PCh. 13 - Classify each substance as a strong electrolyte,...Ch. 13 - Prob. 13.90PCh. 13 - Prob. 13.91PCh. 13 - Which solution has the lower freezing point? 11.0...Ch. 13 - Prob. 13.93PCh. 13 - Prob. 13.94PCh. 13 - Prob. 13.95PCh. 13 - Prob. 13.96PCh. 13 - Prob. 13.97PCh. 13 - Prob. 13.98PCh. 13 - Prob. 13.99PCh. 13 - The boiling point of ethanol (C2H5OH) is 78.5°C....Ch. 13 - Prob. 13.101PCh. 13 - Prob. 13.102PCh. 13 - Prob. 13.103PCh. 13 - Prob. 13.104PCh. 13 - Prob. 13.105PCh. 13 - Prob. 13.106PCh. 13 - Prob. 13.107PCh. 13 - Prob. 13.108PCh. 13 - Prob. 13.109PCh. 13 - In a study designed to prepare new...Ch. 13 - The U.S. Food and Drug Administration lists...Ch. 13 - Prob. 13.112PCh. 13 - Prob. 13.113PCh. 13 - Prob. 13.114PCh. 13 - In a movie theater, you can see the beam of...Ch. 13 - Prob. 13.116PCh. 13 - Prob. 13.117PCh. 13 - Prob. 13.118PCh. 13 - Prob. 13.119PCh. 13 - Gold occurs in seawater at an average...Ch. 13 - Prob. 13.121PCh. 13 - Prob. 13.122PCh. 13 - Prob. 13.123PCh. 13 - Prob. 13.124PCh. 13 - Pyridine (right) is an essential portion of many...Ch. 13 - Prob. 13.126PCh. 13 - Prob. 13.127PCh. 13 - Prob. 13.128PCh. 13 - Prob. 13.129PCh. 13 - Prob. 13.130PCh. 13 - Prob. 13.131PCh. 13 - Prob. 13.132PCh. 13 - Prob. 13.133PCh. 13 - Prob. 13.134PCh. 13 - Prob. 13.135PCh. 13 - Prob. 13.136PCh. 13 - Prob. 13.137PCh. 13 - Prob. 13.138PCh. 13 - Prob. 13.139PCh. 13 - Prob. 13.140PCh. 13 - Prob. 13.141PCh. 13 - The release of volatile organic compounds into the...Ch. 13 - Although other solvents are available,...Ch. 13 - Prob. 13.144PCh. 13 - Prob. 13.145PCh. 13 - Prob. 13.146PCh. 13 - Prob. 13.147PCh. 13 - Prob. 13.148PCh. 13 - Prob. 13.149PCh. 13 - Suppose coal-fired power plants used water in...Ch. 13 - Urea is a white crystalline solid used as a...Ch. 13 - Prob. 13.152PCh. 13 - Prob. 13.153PCh. 13 - Prob. 13.154PCh. 13 - Prob. 13.155PCh. 13 - Prob. 13.156PCh. 13 - Prob. 13.157PCh. 13 - Prob. 13.158PCh. 13 - Prob. 13.159PCh. 13 - Prob. 13.160PCh. 13 - Figure 12.11 shows the phase changes of pure...Ch. 13 - KNO3, KClO3, KCl, and NaCl are recrystallized as...Ch. 13 - Prob. 13.163PCh. 13 - Prob. 13.164PCh. 13 - Prob. 13.165P
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