Silver crystallizes in fcc lattic. If the edge length of the cell is `4.07 xx 10^(-8) cm` and density is `10.5 g cm^(-3)`. Calculate the atomic mass of silver.
`d=(ZM)/(a^(3) xx Na)`
`rArr” “M = (d xx a^(3) xx Na)/Z = (10.5 xx(4.07)^(3)xx10^(-24)xx6.02 xx 10^(23))/4 = 106.5391 g//mol`.
Edge length `(a)=4.077xx10^(-8)cm`
`:. ` Volume of unit cell `(a^(3))=(4.077xx10^(-8))^(3)`
`=67.77xx10^(-24)cm^(3)`
In a `f.c.c.` uinit , there are four atoms per unit cell `(i.e., n=4)`
Atomic mass of `Ag`
`=(“Density” xxAv. no .xx”Volume of unit cell”)/(z)`
`=(10.5xx6.023xx10^(23)xx67.77xx10^(-24))/(4)`
`=107.15`