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Date November 2017 Marks available 1 Reference code 17N.3.hl.TZ0.8
Level HL Paper 3 Time zone TZ0
Command term Calculate Question number 8 Adapted from N/A

Question

Metals have various crystal structures. Cobalt forms a face-centred cubic (FCC) lattice. Two representations of FCC are shown.

Calculate the total number of cobalt atoms within its unit cell.

[1]
a.

The atomic radius, r, of cobalt is 1.18 × 10–8 cm. Determine the edge length, in cm, of the unit cell, a, using the second diagram.

[1]
b.i.

Determine a value for the density of cobalt, in g cm–3, using data from sections 2 and 6 of the data booklet and your answers from (a) and (b) (i).

If you did not obtain an answer to (b) (i), use 3.00 × 10–8 cm but this is not the correct answer.

[2]
b.ii.

Markscheme

« 8 × 1 8 + 6 × 1 2 = » 4

a.

face diagonal  = 2 a = 4 r

« a = ( 4 × 1.18 × 10 8 cm ) 2 = » 3.34 x 10–8 «cm»

b.i.

mass of 4 atoms = 4 × 58.93 g m o l 1 6.02 × 10 23 m o l 1 = 3.916 × 10 22 «g»

«density  = 3.916 × 10 22 g ( 3.34 × 10 8 c m ) 3 = » 10.5 «g cm–3»

Answer using 3.00 x 10–8 cm:

mass of 4 atoms =  4 × 58.93 g m o l 1 6.02 × 10 23 m o l 1 = 3.916 × 10 22  «g»

«density  = 3.916 × 10 22 g ( 3.00 × 10 8 c m ) 3 = » 14.5 «g cm–3»

Award [2] for correct final answer.

b.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.

Syllabus sections

Options » A: Materials » A.8 Superconducting metals and X-ray crystallography (HL only)
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