Question

An element of atomic mass 98.5 gram per mole of cards in fcc structure if its unit cell is length is 500 cm and its density is 5.22 gram per centimetre cube calculate the value of the avogadro constant

Answer 1

Correction:

Unit cell length = 500 pm

Given:

M = 98.5 gm

a = 500 pm = 5 × 10⁻⁸ cm

d = 5.22 gm/cm³

The lattice is FCC.

To Find:

The value of Avogadro's Constant.

Calculation:

- As the given structure is FCC, it means that z = 4

- Volume of 1 unit cell = a³

⇒ V = (5× 10⁻⁸)³ = 125 × 10⁻²⁴ cm³

- Now applying the formula:

V × NA × d = z × M

⇒ NA = (z × M)/(V × d)

⇒ NA = (4 × 98.5)/(125 × 10⁻²⁴ × 5.22)

⇒ NA = 394 × 10²⁴/ 652.5

⇒ NA = 6.038 × 10²³

- So, the value of Avogadro's Constant is 6.038 × 10²³.