Question

Calculate the number of unit cells present in 1 g of gold which has a face-centered cubic lattice.

Options

• 3.06 × 10²⁵

• 30.56 × 10²⁰

• 7.64 × 10²⁵

• 7.64 × 10²⁰

Answer 1

7.64 × 10²⁰

Explanation:

1 mole of gold = 197 g = 6.02 × 10²³

Hence, the number of atoms present in 1 g of gold = `(6.02 xx 10^23)/197`

As face-centered cubic (FCC) unit cells contain 4 atoms,

Therefore, in 1 g of gold number of unit cells present = `(6.02 xx 10^23)/(197 xx 4)` = 7.64 × 10²⁰