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

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

Step-by-step explanation:

Formula used :

`\rho=(Z* M)/(N_(A)* 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* 10^(-12) m = 197* 10^(-12)* 100 cm`

`r=197* 10^(-10) m`

`a=r* 2√(2)`

`a=197* 10^(-10) m* 2√(2)=557.200* 10^(-10) cm`

M = 40 g/mol

Z = 4

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

`\rho =(4* 40 g/mol)/(6.022* 10^(23) mol^(-1)* (557.200* 10^(-10) cm)^(3))`

`\rho =1.536 g/cm^3`

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