Chromium metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be `287`pm. Calculate the atomic radius. What woulds be the density of chromium in `g cm^(-3)`?

Answer

`124.27"pm";7.3 "g cm"^(-3)`

Explanation

Step I. Calculation of the radius of the atom in the lattice.
For b.c.c. lattice, `r=(sqrt3a)/(4)=(sqrt3)/(4)xx("287pm")=124.27"pm"`
Step II. Calculation of density of chromium (Cr)
Density of unit cell `(rho)=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))`
No. of atoms per unit cell (Z) = 2
Atomic mass of chromium (M) = `52.0"g mol"^(-1)`
Edge length of unit cell (a) = 287 pm.
Avogadro's no. `(N_(0))=6.022xx10^(23)"mol"^(-1)`
`"Density"(rho)=(2xx("52.0g mol"^(-1)))/((287)^(3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)xx"cm"^(3)))=7.30"g cm"^(-3)`