An element 'X' (At. Mass = 40 g `mol^(-1)`) having f.c.c structure has unit cell edge length of 400 pm . Calculate the density of 'X' and the number of unit cells in 4 g of 'X' .
`(N_(A) = 6.022 xx 10^(23) mol^(-1))`.

Explanation

Density `(rho) = (Z xx M)/(a^(3) xx N_(a)) " " a to ` edge length
For `" " ` f.c.c(z)= 4 `" " Z to` number of atoms
a = 400 pm = `400 xx 10^(-10) cm " " M to` molar mass
M = 40 g/mol
`rho = (4 xx 40)/((400 xx 10^(-10))^(3) xx 6.022 xx 10^(23)) = (cancel4 xx cancel40^(10))/(cancel4 xx cancel4 xx 4 xx 10^(-1) xx 6.022)`
`rho = 4.15 g//cm^(3)`
1 mole of 'X' = `6.022 xx 10^(23)` atoms of X
40 g of 'X' = `6.022 xx 10^(23)` atoms of X
4 g of 'X' = `(6.022 xx 10^(23))/(40) xx 4 = 6.022 xx 10^(22)` atoms of X.
f.c.c. structure contains 4 atoms in one unit cell .
`therefore` Number of unit cell in `6.022 xx 10^(22)` atoms = `(1)/(4) xx 6.022 xx 10^(22) = 1.51 xx 10^(22)` unit cells.