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

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.

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.

«\(8 \times \frac{1}{8} + 6 \times \frac{1}{2} = \)» 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»

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.