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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]

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.

« $8 \times \frac{1}{8} + 6 \times \frac{1}{2} =$ » 4

face diagonal $= \sqrt {2a} = 4r$

« $a = \frac{{\left( {4 \times 1.18 \times {{10}^{ - 8}}\,{\text{cm}}} \right)}}{{\sqrt 2 }} =$ » 3.34 x 10^–8 «cm»

b.i.

mass of 4 atoms = $\frac{{4 \times 58.93\,g\,mo{l^{ - 1}}}}{{6.02 \times {{10}^{23}}\,mo{l^{ - 1}}}} = 3.916 \times {10^{ - 22}}$ «g»

«density $= \frac{{3.916 \times {{10}^{ - 22}}\,g}}{{{{\left( {3.34 \times {{10}^{ - 8}}\,cm} \right)}^3}}} =$ » 10.5 «g cm^–3»

Answer using 3.00 x 10^–8 cm:

mass of 4 atoms = $\frac{{4 \times 58.93\,g\,mo{l^{ - 1}}}}{{6.02 \times {{10}^{23}}\,mo{l^{ - 1}}}} = 3.916 \times {10^{ - 22}}$ «g»

«density $= \frac{{3.916 \times {{10}^{ - 22}}\,g}}{{{{\left( {3.00 \times {{10}^{ - 8}}\,cm} \right)}^3}}} =$ » 14.5 «g cm^–3»

Award [2] for correct final answer.

b.ii.

[N/A]

b.i.

[N/A]

b.ii.