Question

An element has
a FCC stoucture with
edge length 167 pm. The density of the unit cell is 5.8 gcm.how many atoms will be present in 68g of the unit cell?​

Answer 1

Answer:

Correct option is

B

2.8×10−23cm3

Density ρ=N0×a3Z×M

ρ=5.96 g/cm3= density

Z=2 atoms= numeber of atoms per unit cell.

M=50 g/mol= atomic weight of metal

N0=6.023×1023 atoms/mol=Avogadro's number

a= edge length

Density ρ=N0×a3Z×M

Density 5.96 g/cm3=6.023×1023 atoms/mol×(a)32 atoms×50 g/mol

V=a3=2.8×10−23 cm3

V is the volume of unit cell