An element with molar mass `2.7xx10^(-2)` kg `"mol"^(-1)` forms a cubic unit cell with edge length 405 pm. If its density is `2.7xx10^(3)` `kgm^(-3)`, what is the nature of the cubic unit cell ?

Explanation

`Z=(d xx a^3xxNA)/M`
`M = 2. 7 xx 10^(-2) kg mol^(-1)`
`a = 405 pm = 405 xx 10^(-12) m`
`=4.05xx10^(-10)m`
`=4.05xx10^(-10)m`
`d=2.7xx10^(3)kgm^(-3)`
`N_A=6.022xx10^(23)mol^(-1)`
Substituting these values in expression (i), we get
`Z=((2.7xx10^3kgm^(-1))(4.05xx10^(-10)m)^3(6.022xx0.23mol^(-1)))/(2.7xx10^(-2)kgmol^(-1))=3.99=4`
`therefore` 4 atoms of the clement present per unit cell Hence the cubic unit cell must be face centered.