An atom crystallizes in fcc crystal lattice and has a density of `10 g cm^(-3)` with unit cell edge leght of 100 pm . Calculate the number of atoms present in 1 g of crystal.

Explanation

Given density = 10 g `cm^(-3)`
Edge length = 100 pm
` = 100 xx 10^(-10) cm`
` = 10^(-8)` cm
` =4 ` ( since it is fcc)
M = ?
` N = 6.023xx10^(-23)`
` M = (rho xx a^(3) xx N)/(Z)`
` = (10 xx (10^(-8))^(3) xx 6.023 xx 10^(23))/(4)`
` = (10 xx 10^(-24) xx 6.023xx 10^(23))/(4)`
`=(6.023)/(4)`
` = 1.505`
The molecular weight comes to be `1.505 g mol^(-1)` 1.505 g of the atom ` =6.023 xx 10^(23)` atoms
1g of the atm ` =(6.023xx 10^(23))/(1.55=05)`
` = 4 xx 10^(23)` atm