Silver crystallises in fcc lattice. If edge length of the cell is 4.07xx`10^(-8)`
cm and density is `10.g cm^(-3)`, calculate the atomic mass of silver.

Explanation

`a = 4.077 xx 10^(8) cm`
`rho = 10.5 g cm^(-3)`
`Z_(eff) = 4, N_(A) = 6.023 xx 10^(23)`
`rho = (Z xx Mw)/(a^(3) xx N_(A))`
or `Mw = (rho xx a^(3) xx N_(A))/(Z_(eff))`
` = (10.5 xx (4.-77 xx 10^(-8))^(3) xx 6.023 xx 10^(23))/(4)`
` = (10.5 xx 67.77 xx 6.023)/(40) = 107.8u`