An element with atomic mass 93 g `mol^(-1)` has density 11.5 g `cm^(-3)`. If the edge length of its unit cell is 300 pm, identify the type of unit cell.

Answer

bcc type

Explanation

`Z=(rhoxxa^(3)xxNo)/(M)`
Density of unit cell = 11.5 g `cm^(-3)` , molar mass of the element = 93 g `mol^(-1)`
Edge length of the unit cell = 300 pm , Avogadro's No.`(N_(0))=6.022xx10^(23)`
`Z=((11.5gcm^(-3))xx(3xx10^(-8)cm)^(3)xx(6.02xx10^(23)mol^(-1)))/((93 g mol^(-1)))=2.01=~~"2(b.c.c.)"`