Silver crystallisaes in fcc lattice. If the edge length of th cell is `4xx10^(-8)` cm and density is 11.2 g `"cm"^(-3)`, calculate the atomic mass of silver.

Explanation

`M=108"g mol"^(-1),d=10.5"g cm"^(-3)`
`a=409"pm"=4.09xx10^(-8)"cm"`
`N_(A)=6.022xx10^(23)"mol"^(-1)`
`d=(ZxxM)/(a^(3)xxN_(A))`
`Z=(d xxa^(3)xxN_(A))/(4)`
`(10xx5"g cm"^(-3)xx(4.09xx10^(-8))^(3)"cm"^(3)xx6xx0.22xx10^(23)"mol"^(-1))/(108"g mol"^(-1))`
`=(10.5xx68.42xx6.022)/(1080)=4`
Type of crystal lattice = f.c.c.
For f.c.c. lattice
`4r=sqrt(2)a`
`r=sqrt(2)/(4)a=(1.414xx409"pm")/(4)`
`=144.6"pm"`