An element 'X' with an atomic mass of 60g/ mol has density of `6.23"g cm"^(-1)`. If the edge length of its unit cell is 400 pm, identity the type of cell. Also calculate the radius of an atom of this element.

Explanation

`M=60"g mol"^(-1)d=6.23"g cm"^(-3)`
`a=400"pm"=4xx10^(-8)"cm"`
`Z=?,N_(A)=6.002xx10^(23)"mol"^(-1)`
`d=(ZxxM)/(a^(3)xxN_(A))`
`z=(d xxa^(3)xxN_(A))/(M)`
`(6.23"g cm"^(-3)xx(4xx10^(-8))^(3)"cm"^(3)xx6.022xx10^(23)"mol"^(-1))/(60"g mol"^(-1))`
`=(6.23xx64xx6.022)/(600)=4.00`
`therefore` Type of Unit Cell = f.c.c.
For f.c.c. lattice
`4r=sqrt(2)a`
`r=sqrt(2)/(4)a=(1.414xx400)/(4)"pm"`
= 141.4 pm