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.

Explanation

Density, `rho = (Z xx M)/(a^3 xx N_A)`
or `" " M = (rho xx a^3 xx N_A)/(Z)`
`Z = 4` (fcc lattice) , rho = 10.5 g cm^(-3), N_A = 6.022 xx 10^(23)`
`(a = 4.077 xx 10^(-8) cm)`
`:. M = (10.5 xx (4.077 xx 10^(-8))^(3) xx 6.022 xx 10^(23))/(4)`
`= 107.12 g mol^(-1)`