Question

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium.

Answer 1

Answer:

$\rho=1.54\ g/cm^3$

Explanation:

The expression for density is:

$\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}$

$N_a=6.023\times 10^{23}\ {mol}^{-1}$

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

$\rho$ is the density

Radius = 197 pm = $1.97\times 10^{-8}\ cm$

Also, for fcc, $Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm$

Thus,  

$\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3$

$\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3$

$\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3$

$\rho=\frac{160.312}{104.19483}\ g/cm^3$

$\rho=1.54\ g/cm^3$