An element crystallises as face-centred cubic lattice with density as `5.20 g//cm^(3)` and edge length of the side of unit cell as 300pm. Calculate mass of the element which contains `3.01 xx 10^(24)` atoms

Explanation

`Z = 4` in fcc lattice `M = ?`, `d = 5.20 g//cm^(3)`
`a = 300` pm `= 3 xx 10^(-8)` cm
`:. a^(3) = 27 xx 10^(-24) cm^(3)`
`N_(a) = 6.02 xx 10^(23)`
`:. M = (dN_(a)a^(3))/(Z) = (5.20 g //cm^(3) xx 6.02 xx 10^(23) mol^(-1) xx 27 xx 10^(-24) cm^(3))/(4) = 21.13 g mol^(-1)`
Thus, `6.02 xx 10^(23)` atoms have `= 21.13` g
`:. 3.01 xx 10^(24)` atoms have `= (21.13)/(6.02 xx 10^(23)) xx 3.01 xx 10^(24) g = 105.65 g`