Question

A element crystallises in a bcc structure. The edge length of its unit cell is 288 pm. If the density of the crystal is 7.3 g `cm^(-3)` , what is the atomic mass of the element ?

Answer 1

We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))orM=(rhoxxa^(3)xxN_(0)xx10^(-30))/(Z)`
According to available data,
Edge length (a) = 288 pm = 288
No. of atoms per unit cell (Z) = 2
Density of the element =`7.3"g cm"^(-3)`
`therefore"Atomic mass of element (M)"=((7.3"g cm"^(-3))xx(288)^(3)xx(6.022xx10^(23)"mol"^(-1)))/(2)xx(10^(-30)"cm"^(3))`
`=52.51"g mol"^(-1)=52.51 u`.