Question

• Nickel has FCC crystal structure, Determine its density if the atomic weight is 58.71 and the atomic radius is 1.25Aº

Answer 1

Answer:

20.6×10⁷ g/m³

Explanation:

We know that

$\boxed{\sf Density (\rho)=\dfrac{ZM}{a^3N_A}}$

For FCC , Z = 4

Na = 6.02 × 10²³

M = 58.71

a = 1.25 A° = 1.25 × ${\sf 10^{-10}}$ m

➞ 4 × 58.71/(1.25×10^-10)³×6.02×10²³

➞234.84/1.9 ×10^-30×6.02×10²³

➞ 234.84/11.4 × 10^−7

➞20.6×10⁷

Density of Nickel is 20.6×10⁷ g/m³