An element crystallized in f.c.c lattice …

Q u e s t i o n : An element crystallizes in a fcc lattice with cell edge of 400 pm.  The density of the element is 7 g cm^-3. How many atoms are Present in 280 g of the element ?

A n s w e r :

Volume of the unit cell a³ = (400 pm)³ 

Step I. Calculation of mass of an atom of the elements
Let the mass of an atom of the element = m gram
Mass of the unit cell = 4×m=(4m) 
Edge length (a)= 400 pm = 400pm ×10⁻¹⁰ 
Volume of the unit cell =(400×10⁻¹⁰cm)³=64×10⁻²⁴cm³
Density of unit cell = 7 g cm⁻³ 
Density of unit cell =Mass of unit cellvolume of unit cell=Mass of unit cellvolume of unit cell
(7 g cm⁻³)=(4m)g/ (64×10⁻²⁴cm³) 
m=(7gcm−3)×(64×10−24cm3)/ 4=112×10⁻²⁴g 
Step II. Calculation of the number of atoms in 280 g of the element.
No. of atoms=Mass of the elementMass of one atom of the elementNo. of atoms=Mass of the elementMass of one atom of the element
=(280g)/(112×10⁻²⁴g)=2.5×10²⁴atoms