Question

an element crystalizes in bcc lattice with cell edge of 500 pm. the density of the elements is 7.5 g cm-3 .how many atom are present in 300 g of the element​

Answer 1

Answer:

here's the answer, hope it might help ya!

Explanation:

The edge length, a = 500pm = 500×10^−12m = 5×10^−8cm

The volume of the unit cell = a^3 = (5×10^−8cm)3 = 1.25×10^−22cm^3

For b.c.c unit cell, the number of atoms per unit cell Z = 2

Volume occupied by each atom = 1.25×10^−22cm^3 / 2 ​

= 6.25×10^−23cm^3

The density d = 7.5g / cm^3

Mass of sample = 300g

Volume of sample = mass / density ​= 300 / 7.5 ​= 40cm^3

The number of atoms present in 300 g of the element =40cm^3 /

6.25×10^−23cm^3  ​= 6.4×10^23  atoms.