An element crystallises in the fcc crystal lattice and has a density of 10 g `cm^(-3)` with unit cell edge length of 100 pm . Calculate number of atom

Correct Answer - Number of atoms in 1 gram element `=4xx 10^(23)` atoms
Given : Density of crystal = d= 10 `g cm^(-3)`
Edge length =a= 100 pm `=1xx 10^(-8) cm`
Avogadro number `= N_(A) = 6.022 xx 10^(23) mol^(-1)`
Number of atom = s present in 1 g crystal = ?
Let N be the atomic mass of the element .
Mass of one atom `= (M )/(6.022 xx 10^(23))`
Fcc type unit cell contains 4 atoms.
`:. ` Mass of unit cell = Mass of 4 atoms `= 4 xx (M)/(6.022 xx 10^(23))`
Volume of unit cell `=a^(3) = (1 xx 10^(-8))^(3) =1 xx 10^(-24) cm^(3)`
Density of the crystal =d= `("Mass of unit cell" )/("volume of unit cell" )`
`10= 4 xx M //6.022 xx (10^(23))/(1xx 10^(-24))`
`= (4 xx M)/(6.022 xx 10^(23) xx 1 xx 10^(-24)) = (4 M)/(0.6022)`
`:. M = (10 xx 0.6022)/(4) =1.5 g mol^(-4)`
Now 1 gram atom = 1 mole of element =1.5g
`:.` 1.5 g element contains 6.022 `xx 10^(23)` atoms
`:.` 1 g element will contains .` (6.022 xx 10^(23))/(1.5) =4 xx 10^(23) ` atoms