An element has a bcc structure with unit cell edge length is 288 pm. The density is 1 g/cm³.How many unit cells are present in 200 g of the element?
Answers
Answered by
0
Answer:
Volume of unit cell =(288 pm)
3
=(288×10
−10
cm)
3
=2.389×10
−23
cm
3
Volume of 208 g of the element =
Density
Mass
=
7.2
208
=28.89 cm
3
Number of unit cells =
Volume of a unit cell
Total Volume
=
2.389×10
−23
28.89
=12.09×10
23
For a BCC structure, number of atoms per unit cell =2
∴ Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells
=2×12.09×10
23
=24.18×10
23
=2.418×10
24
Similar questions