Question

Q. 16) A metallic element has BCC lattice Edge length
of unit cell is 2.88 x 10-8cm. The density is 7.20 g
cm", calculate
a) Volume of unit cell b) Mass of unit cell
c) Number of atoms in 100g of metal.​

Answer 1

The volume of unit cell = 2.39 x 10^-23, Mass of unit cell = 51.77 g mol−1

Explanation:

density (d) = z(no. of atoms in a unit cell) M(molar mass) / NAa3(volume of a unit cell)

M = dNAa^3.z = 7.2 g.cm^−3×6.022×10^23 mol−1 × (2.88 × 10^−8 cm)^3/2 M=51.77 g mol−1

Therefore, 51.77 g of this element have 6.022 x 10^23 atoms.

Then, 156 g of this element will have = 6.022×10^23 / 51.77 × 156

= 1.81 × 10^24 atoms

Volume of 156 g of element = mass / density = 156 g / 7.2 g cm-3 = 21.66 cm^3

Volume of a unit cell = a^3 = (2.88 x 10^-8 cm)^3 = 2.39 x 10^-23

Number of unit cell present in 156 g of element = (volume of 156 g of element) / (Volume of a unit cell) = 21.66 cm^3​ / 2.39 x 10-23 = 9.06 x 1023 unit cells

Also learn more

Calculate the number of unit cells in 8.1 gram of aluminium if it crystallizes in a fcc structure. ?

https://brainly.in/question/2093224

Answer 2

9.6 × 10²³ cell units

hope this will help u