An atom crystallizes in fcc crystal lattice and has a density of`10gcm^(-3)` with unit cell edge length of100pm. Calculate the number of atoms present in 1g of crystal.

Explanation

`d = Z xx M//a^(3) xx N_(A)`
Where a is edge of unit cell.
`N_(A)= `Avogadro no `( 6.022 xx 10^(23))`
M = Molar mass
Z = No. of atom per unit cell
For fcc Z = 4
`m = d xx a^(3) xx NA//Z`
No of atom = (given mass `//` molar mass `// m ) xx NA `
Putting the value of M in above equation we get
No. of moles = given mass` xx z xx N/d xx a^(3) xx NA `
No. of atom = Given mass `xx z //d xx a^(3)`
putting the values we get
No. of atoms `= 100g xx(4)/( 10) g cm^(3) xx 100 xx10^(10) cm^(3)`
`=( 400)/( 100) xx 10^(24)`
`= 4 xx 10^(24) ` atoms.