Question

Chromium crystallises in BCC structure if its Atomic diameter is 245pm.Find the density and atomic mass of Chromium 52 amu. Avogadro number: 6.022*10²³

Answer 1

The formula for the density of a crystal lattice can be written as :

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{Z}\mathrm{M}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{where}\mathrm{Z}=\mathrm{number}\mathrm{of}\mathrm{atom}\mathrm{per}\mathrm{unit}\mathrm{cell} \\ \mathrm{a}=\mathrm{edge}\mathrm{length} \\ \mathrm{M}\mathrm{is}\mathrm{the}\mathrm{molar}\mathrm{mass} \end{gathered}$$

For BCC crystal lattice, Z = 2
Atomic mass for Cr = 52 amu, therefore M = 52 g/ mol
Na = 6.022 $\times$10²³ mol^-1

​For Cr, 2 r = 245 pm
We known, for BCC

$$\begin{gathered} \sqrt{3}\mathrm{a}=4\mathrm{r} \\ \mathrm{a}=\frac{4\mathrm{r}}{\sqrt{3}} \\ 2\mathrm{r}=245\mathrm{pm} \\ \mathrm{Thus}, \\ \mathrm{a}=\frac{2\times 245}{1.73} \\ \mathrm{a}=282.90\mathrm{pm} \end{gathered}$$

a can also be written as 282.90 x 10^-10 cm.
On substituting all the above values in the formula of density, we get

$$\begin{gathered} d=\frac{2\times 52}{(282.90\times {10}^{-10}{)}^3\times 6.022\times {10}^{23}} \\ d=7.627g/c{m}^3 \end{gathered}$$

Hence, density of Cr metal is 7.627 g/cm³