Nickel crystallises in a cubic close-packed structure. In this structure, the distance between two nearest neighbours is 250pm. If the atomic mass of nickel is 58.7amu, then calculate the density of nickel crystal.

Explanation

In ccp structure of atoms, the unit cell is face-centred cubic (fcc) unit cell.
In fcc unit cell, the distance between two nearest neighbours = `1/sqrt2xx` edge length of the unit cell (a)
`therefore" "a=sqrt2xx250`pm = 353.55pm = `3.53xx10^(-8)`cm
We know, `rho = (ZxxM)/(Nxxa^(3))`
Here Z = 4, M = `58.7g*mol^(-1)anda=3.53xx10^(-8)cm`
`rho = (4xx58.7)/(6.022xx10^(23)xx(3.53xx10^(-8))^(3))=8.862g*cm^(-3)`
`therefore" "`Density of nickel crystal = `8.862g*cm^(-3)`