Question

Calcium has a cubic closest packed structure (fcc) as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium. (1 pm = 10-12 m, 100 cm = 1 m)

Answer 1

Answer:

$1.536 g/cm^3$ is the density of solid calcium.

Explanation:

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

We have:

r = 197 pm = $197\times 10^{-12} m = 197\times 10^{-12}\times 100 cm$

$r=197\times 10^{-10} m$

$a=r\times 2\sqrt{2}$

$a=197\times 10^{-10} m\times 2\sqrt{2}=557.200\times 10^{-10} cm$

M = 40 g/mol

Z = 4

On substituting all the given values , we will get the value of 'a'.

$\rho =\frac{4\times 40 g/mol}{6.022\times 10^{23} mol^{-1}\times (557.200\times 10^{-10} cm)^{3}}$

$\rho =1.536 g/cm^3$

$1.536 g/cm^3$ is the density of solid calcium.