Question

An element crystallizes in foc lattice with a cell edge of 300 pm. The density of the
element is 10.8 g cm. Calculate the number of atoms in 108 g of the element.

Answer 1

Answer:

1.48 × 10²⁴ atoms

Explanation:

Given,

FCC lattice                   cell edge a=300 pm           density d=10.8 g/cm³

Volume of 108 g of element= $\frac{mass}{density}$= $\frac{108}{10.8}$ = 10 cm³

Volume of unit cell= a³ = (300 pm)³ = 27 × 10⁶ pm³ = 2.7 × 10⁻²³ cm³

Number of unit cell in that volume = $\frac{volume of total element}{volume of one unit cell}$=10 / 2.7 × 10⁻²³ cm³

                                                                                         =10²⁴ / 2.7

Each FCC lattice have 4 atoms.

∴         one unit cell = 4 atoms

∴ 10²⁴ / 2.7 unit cell = ?

∴ (10²⁴ / 2.7) × 4

= 1.48 × 10²⁴ atoms