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

`a=405"pm"=405xx10^(-12)"m"`
`=4.05xx10^(-10)"m"`
`M=2.7xx10^(-2)"kg mol"^(-1)`,
`N_(A)=6.022xx10^(23)"mol"^(-1)`
`d=2.7xx10^(23)"kg m"^(-3),z=?`
`d=(zxxM)/(a^(3)xxN_(A))`
`Z=(d xxa^(3)xxN_(A))/(M)`
`(2xx7xx10^(3)"kgm"^(-3))xx4xx0.5xx^(-10)m)^(3)`
`=(xx(6.022xx10^(23)"mol"^(-1)))/((2.7xx10^(23)"kg mol"^(-1)))`
`=3.99~~4` (nearest whole number)
Thus, there are 4 atoms of the element per unit cell. Therefore, element forms f.c.c. (for c.c.p.) lattice i.e., cubic unit cell is face centred.