Question

Silver crystallizes in FCC lattice . If edge of the cell is `4.07 xx 10^(-8)` and density is `10.5 g. cm^(3)` . Calculate the atomic mass of silver .

Answer 1

Given data
`d = 10.5 g//cm^(3)`
`a = 4.07 xx 10^(-8) cm`
Z = 4 atoms
`N_(A) = 6.023 xx 10^(23)`
Formula M `= (d xx a^(3) xx N_(A))/(Z)`
`= (10. 5 xx (4.07 xx 10^(-8))^(3) xx 6.023 xx 10^(23))/(4)`
`= (10.5 xx 67.767 xx 10^(-24) xx 6.023 xx 10^(23))/(4)`
= `107.09` gm/mole