Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 12, Problem 12.138P

(a)

Interpretation Introduction

Interpretation:

The unit-cell edge length of the densest diamond is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

(a)

Expert Solution
Check Mark

Answer to Problem 12.138P

The unit-cell edge length of the densest diamond is 3.57×108cm.

Explanation of Solution

The formula to calculate the volume of one mole of carbon atom with a higher density is as follows:

Volume of one mole of C atom=(molar mass of CDensity of C) (1)

Substitute the value of 12.01g/mol for molar mass of C atom and 3.52 g/cm3 for the density of C in the equation (1).

Volume of one mole of C atom=(12.01g/mol3.52 g/cm3)=3.411931 cm3/mol

Diamond has a face-centered cubic unit cell. Face-centered cubic unit cell, 8 atoms are present at the corners of the cell and 6 atoms at the face of the cell. The contribution of an atom present at the corner is 18 and the contribution of an atom at the face of a cell is 12. Therefore the number of an atom in the face-centered cubic unit cell is calculated as follows:

Number of atoms=(18atom per cell)(8atoms)+(12atom per cell)(6atom)=1atom+3atoms=4atoms

There are 4 atoms at the tetrahedral voids and therefore, the diamond has 8 carbon atoms in the lattice.

The formula to calculate the volume of the unit cell is as follows:

Volume of unit cell=[(Volume of one mole ofC atom)(1molC6.022×1023Catoms)(8C atom1unit cell)] (2)

Substitute the value of 3.411931 cm3/mol for volume of one mole of C atom in the equation (2).

Volume of unit cell=[(3.411931 cm3/mol)(1molC6.022×1023Catoms)(8C atom1unit cell)]=4.56326228×1023cm3

The formula to calculate edge length of the cube is as follows:

Edge length of the cube=Volume of unit cell3 (3)

Substitute the value of 4.56326228×1023cm3 for volume of the unit cell in the equation (3).

Edge length of the cube=4.56326228×1023cm33=3.5654679×108cm3.57×108cm

Conclusion

The unit-cell edge length of the densest diamond is 3.57×108cm.

(b)

Interpretation Introduction

The number of C atoms present diamond unit cell with the lowest density is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is 18, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is 12 and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is 1. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

(b)

Expert Solution
Check Mark

Answer to Problem 12.138P

The number of C atoms present diamond unit cell with the lowest density is 6.84Catoms.

Explanation of Solution

The formula to calculate the volume of one mole of carbon atom with lower density is as follows:

Volume of one mole of C atom=(molar mass of CDensity of C) (4)

Substitute the value of 12.01g/mol for molar mass of C atom and 3.01 g/cm3 for the density of C in the equation (4).

Volume of one mole of C atom=(12.01g/mol3.01 g/cm3)=3.9900 cm3/mol

The formula to calculate the volume of the unit cell is as follows:

Volume of unit cell=(A)3 (5)

Substitute 3.3058Ao for A in the equation (5).

Volume of unit cell=(3.5654679×108cm)3=12.712561×1024cm3

The formula to calculate the number of C atoms with lower density is as follows:

Number of C atom=[(Volume of unit cellVolume of one mole of C atom)(6.022×1023Catoms1molC)] (6)

Substitute 12.712561×1024cm3 for the volume of unit cell and 3.9900 cm3/mol for the volume of one mole of carbon atom in the equation (6).

Number of C atom=[(12.712561×1024cm33.9900 cm3/mol)(6.022×1023Catoms1molC)]=6.8409Catoms6.84Catoms

Conclusion

The number of C atoms present diamond unit cell with the lowest density is 6.84Catoms.

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Chapter 12 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 12.6 - For each of the following crystalline solids,...Ch. 12.6 - Prob. 12.6BFPCh. 12.6 - Prob. 12.7AFPCh. 12.6 - Iron crystallizes in a body-centered cubic...Ch. 12.6 - Prob. 12.8AFPCh. 12.6 - Prob. 12.8BFPCh. 12.6 - Prob. B12.1PCh. 12.6 - Prob. B12.2PCh. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Name the phase change in each of these events: (a)...Ch. 12 - Prob. 12.9PCh. 12 - Many heat-sensitive and oxygen-sensitive solids,...Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Prob. 12.14PCh. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - From the data below, calculate the total heat (in...Ch. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23PCh. 12 - Prob. 12.24PCh. 12 - Prob. 12.25PCh. 12 - Prob. 12.26PCh. 12 - Prob. 12.27PCh. 12 - Prob. 12.28PCh. 12 - Prob. 12.29PCh. 12 - Prob. 12.30PCh. 12 - Use Figure 12.10 to answer the following: Carbon...Ch. 12 - Prob. 12.32PCh. 12 - Prob. 12.33PCh. 12 - Prob. 12.34PCh. 12 - Prob. 12.35PCh. 12 - Prob. 12.36PCh. 12 - Distinguish between polarizability and polarity....Ch. 12 - Prob. 12.38PCh. 12 - Prob. 12.39PCh. 12 - Prob. 12.40PCh. 12 - Prob. 12.41PCh. 12 - Prob. 12.42PCh. 12 - Prob. 12.43PCh. 12 - Prob. 12.44PCh. 12 - Prob. 12.45PCh. 12 - Prob. 12.46PCh. 12 - Prob. 12.47PCh. 12 - Prob. 12.48PCh. 12 - Prob. 12.49PCh. 12 - Which liquid in each pair has the lower vapor...Ch. 12 - Which substance has the lower boiling point?...Ch. 12 - Which substance has the higher boiling point?...Ch. 12 - Prob. 12.53PCh. 12 - Prob. 12.54PCh. 12 - Prob. 12.55PCh. 12 - Prob. 12.56PCh. 12 - Why does the antifreeze ingredient ethylene glycol...Ch. 12 - Prob. 12.58PCh. 12 - Prob. 12.59PCh. 12 - Why does an aqueous solution of ethanol (CH3CH2OH)...Ch. 12 - Prob. 12.61PCh. 12 - Prob. 12.62PCh. 12 - Prob. 12.63PCh. 12 - Prob. 12.64PCh. 12 - Prob. 12.65PCh. 12 - Prob. 12.66PCh. 12 - Prob. 12.67PCh. 12 - Prob. 12.68PCh. 12 - Prob. 12.69PCh. 12 - Prob. 12.70PCh. 12 - Prob. 12.71PCh. 12 - Prob. 12.72PCh. 12 - Prob. 12.73PCh. 12 - Prob. 12.74PCh. 12 - Prob. 12.75PCh. 12 - Prob. 12.76PCh. 12 - Prob. 12.77PCh. 12 - Prob. 12.78PCh. 12 - Prob. 12.79PCh. 12 - Prob. 12.80PCh. 12 - Prob. 12.81PCh. 12 - Prob. 12.82PCh. 12 - Prob. 12.83PCh. 12 - Prob. 12.84PCh. 12 - Besides the type of unit cell, what information is...Ch. 12 - What type of unit cell does each metal use in its...Ch. 12 - What is the number of atoms per unit cell for each...Ch. 12 - Calcium crystallizes in a cubic closest packed...Ch. 12 - Chromium adopts the body-centered cubic unit cell...Ch. 12 - Prob. 12.90PCh. 12 - Prob. 12.91PCh. 12 - Prob. 12.92PCh. 12 - Prob. 12.93PCh. 12 - Prob. 12.94PCh. 12 - Prob. 12.95PCh. 12 - Prob. 12.96PCh. 12 - Prob. 12.97PCh. 12 - Prob. 12.98PCh. 12 - Prob. 12.99PCh. 12 - Prob. 12.100PCh. 12 - Prob. 12.101PCh. 12 - Prob. 12.102PCh. 12 - Prob. 12.103PCh. 12 - Polonium, the Period 6 member of Group 6A(16), is...Ch. 12 - Prob. 12.105PCh. 12 - Prob. 12.106PCh. 12 - Prob. 12.107PCh. 12 - Prob. 12.108PCh. 12 - Prob. 12.109PCh. 12 - Prob. 12.110PCh. 12 - Prob. 12.111PCh. 12 - Prob. 12.112PCh. 12 - Prob. 12.113PCh. 12 - Prob. 12.114PCh. 12 - Prob. 12.115PCh. 12 - Prob. 12.116PCh. 12 - Prob. 12.117PCh. 12 - Prob. 12.118PCh. 12 - Prob. 12.119PCh. 12 - Prob. 12.120PCh. 12 - Prob. 12.121PCh. 12 - Prob. 12.122PCh. 12 - Prob. 12.123PCh. 12 - Prob. 12.124PCh. 12 - Prob. 12.125PCh. 12 - Prob. 12.126PCh. 12 - Bismuth is used to calibrate instruments employed...Ch. 12 - Prob. 12.128PCh. 12 - Prob. 12.129PCh. 12 - Prob. 12.130PCh. 12 - Prob. 12.131PCh. 12 - Prob. 12.132PCh. 12 - Prob. 12.133PCh. 12 - Prob. 12.134PCh. 12 - Prob. 12.135PCh. 12 - Prob. 12.136PCh. 12 - Prob. 12.137PCh. 12 - Prob. 12.138PCh. 12 - Prob. 12.139PCh. 12 - Prob. 12.140PCh. 12 - Prob. 12.141PCh. 12 - Prob. 12.142PCh. 12 - Prob. 12.143PCh. 12 - Prob. 12.144PCh. 12 - Prob. 12.145PCh. 12 - The crystal structure of sodium is based on the...Ch. 12 - Prob. 12.147PCh. 12 - One way of purifying gaseous H2 is to pass it...
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