‹ Back to the paper
Solve the following.
An element has atomic weight 93 g mol⁻¹ and density 11·5 g cm⁻³. If the edge length of its unit…
An element has atomic weight $93\text{ g mol}^{-1}$ and density $11\cdot5\text{ g cm}^{-3}$. If the edge length of its unit cell is $300\text{ pm}$, identify the type of unit cell.
($N_A = 6\cdot023 \times 10^{23}\text{ mol}^{-1}$)
Answer
Answer
Formula used: Density of a unit cell: rho = z*M/(a**3*NA) with M = 93 g mol^-1, a = 300 pm, NA = 6.023e23 mol^-1, rho = 11.5 g cm^-3, so z = 2.01
Number of atoms per unit cell: $\rho = \dfrac{zM}{a^3 N_A}$, so $z = \dfrac{\rho a^3 N_A}{M}$
$z = \dfrac{11\cdot5 \times (300\times10^{-10}\text{ cm})^3 \times 6\cdot023\times10^{23}}{93} = 2\cdot01 \approx 2$
Final answer: 2 atoms per unit cell (body centred cubic)