An element having atomic radius 144 pm crystallises in a ccp structure. The density of the element is `10.6g*cm^(-3)`. Identify the element.

Explanation

Since the element crystallises in a ccp structure, the unit cell of its crystal structure belongs to fcc system.
We know, `rho=(ZxxM)/(Nxxa^(3))`
For an fcc unit cell, Z = 4 and a = `2sqrt2xxr`
Given: `rho=10.6g*cm^(-3)` and r = 144pm = `1.44xx10^(-8)cm`
`thereforea=2sqrt2xx1.44xx10^(-8)=4.07xx10^(-8)cm`
`therefore" "M=(rhoxxNxxa^(3))/Z`
`=(10.6xx(6.022xx10^(23))xx(4.07xx10^(-8))^(3))/4`
`=107.6g*mol^(-1)`
The calculated atomic mass is equal to the atomic mass of silver. Hence, the element is silver.