Answer
$ρ = 6.80 g/cm^{3}$
Work Step by Step
Required:
theoretical density of NiO given that it has the rock salt crystal structure.
Solution:
Using Equation 12.1:
$ρ = \frac{n^{'} (A_{Ni} +A_{O})}{V_{C}N_{A}}$
For rock salt crystal structure, $n^{'}$ = 4 formula units per unit cell; using the ionic radii from Table 12.3:
$V_{C} = a^{3} = (2r_{Ni} + 2r_{O})^{3} = [ 2 (0.069 nm) + 2(0.140 nm)]^{3} = 0.0730 \frac{nm^{3}}{unit cell} = 7.30 \times 10^{-23} \frac{cm^{3}}{unit cell}$
Solving for ρ:
$ρ = \frac{(4 formula units/unit cell) (58.69 g/mol + 16.00 g/mol)}{(7.30 \times 10^{-23} \frac{cm^{3}}{unit cell})(6.022 \times 10^{23} formula units/mol)} = 6.80 g/cm^{3}$
