Copy and complete table 3 . Element Density / g cm - 3 Length of cubic crystal edge/ pm Radius of atom from X - ray diffraction data/ pm Crystal type iron 125 BCC sodium 0 . 968 429 platinum 21 . 09 FCC

The density of the unit cell, d = zM a 3 N A . Where, z = No . of atoms ; M = atomic mass ; a = edge length ; N A = Avogadro ' s number . The relationship between the edge length and the radius of SC ; BCC ; FCC are given as, SC ⇒ a = 2 r . BCC ⇒ 3 a = 4 r . FCC ⇒ 2 a = 4 r . For iron, The edge length, 3 a = 4 r ⇒ a = 4 r 3 = 4 × 125 1 . 732 = 500 1 . 732 = 288 . 68 pm . The density, d = zM a 3 N A = 2 × 56 g mol - 1 2 . 8868 × 10 - 8 cm 3 × 6 . 022 × 10 23 mol - 1 . d = 112 g mol - 1 2 . 405 × 10 - 23 cm 3 × 6 . 022 × 10 23 mol - 1 . d = 112 g 14 . 48 cm 3 = 7 . 73 g cm - 3 . For sodium, d = zM a 3 N A ⇒ z = d × a 3 × N A M = 0 . 968 g cm - 3 × 4 . 29 × 10 - 8 cm 3 × 6 . 022 × 10 23 mol - 1 23 g mol - 1 . z = 0 . 968 × 7 . 89 × 10 - 23 × 6 . 022 × 10 23 23 = 45 . 993 23 = 2 . The number of atom is two. Therefore, the crystal structure is body-centred cubic unit cell. The radius will be, 3 a = 4 r ⇒ r = 3 a 4 = 1 . 732 × 429 4 = 185 . 757 pm . For platinum, d = zM a 3 N A ⇒ a 3 = z × M d × N A . a 3 = 4 × 195 g mol - 1 21 . 09 g cm - 3 × 6 . 022 × 10 23 mol - 1 a 3 = 780 × 10 - 23 cm 3 127 = 6 . 14 × 10 - 23 cm 3 . a 3 = 61 . 4 × 10 - 24 cm 3 ⇒ a = 61 . 4 × 10 - 24 cm 3 3 a = 3 . 94 × 10 - 8 cm = 394 pm . The radius for face-centred cubic unit cell, 2 a = 4 r ⇒ r = 2 a 4 = 1 . 414 × 394 4 = 139 . 27 pm . The completed table 3 is given as, Element Density/ g cm - 3 Length of cubic crystal edge/ pm Radius of atom from X - ray diffraction data/ pm Crystal type iron 7 . 73 288 . 68 125 BCC sodium 0 . 968 429 185 . 75 BCC platinum 21 . 09 394 139 . 27 FCC

E M B I B E

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 2

EASY

Diploma

IMPORTANT

Earn 100

Outline why is $PCB$ considered “dioxin like”, but not a dioxin.

Important Questions on Materials

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 4

LDPE

has a specific gravity of

0.92 g {cm}^{- 3}

and

HDPE

has a density of

0.95 g {cm}^{- 3}

. Suggest a reason why flotation is not a good method of separating these two plastics.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 5

Many plastic materials are disposed of by combustion. State two disadvantages of disposing of polyvinyl chloride in this way.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 6

Identify two difficulties and two advantages in recycling plastics.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Quick question

Nickel has a density of

8.91 g {cm}^{- 3}

and

X -

ray diffraction shows that the unit cell edge length is

0.3524 nm

. Determine the packing structure for

Ni

.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 4

Draw structures representing a face centred cubic and body centred cubic unit cell.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 6

Sketch a graph of resistance versus temperature for a conductor and a superconductor.

MEDIUM

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 22 - Materials>Questions>Q 8

Copy and complete table $3$ .

Element	Density / $g {cm}^{- 3}$	Length of cubic crystal edge/ $pm$	Radius of atom from $X -$ ray diffraction data/ $pm$	Crystal type
iron			$125$	$BCC$
sodium	$0.968$	$429$
platinum	$21.09$			$FCC$