Question

An element A (atomic mass 60) has simple cubic lattice of edge 100 pm. The density of crystal (N₀ = 6 × 10²³) is ______.

पर्याय

• 600 g cm⁻³

• 1 × 10⁴ g cm⁻³

• 6 × 10⁻² g cm⁻³

• 1 × 10² g cm⁻³

Answer 1

An element A (atomic mass 60) has simple cubic lattice of edge 100 pm. The density of crystal (N₀ = 6 × 10²³) is 1 × 10² g cm⁻³.

Explanation:

Given: Convert the edge length to centimeters.

a = 100 pm

= 100 × 10⁻¹² m

= 100 × 10⁻¹⁰ cm

= 1 × 10⁻⁸ cm

Volume of the unit cell (V) = a³

= (1 × 10⁻⁸)³

= 1 × 10⁻²⁴ cm³

Formula: ρ = `(Z xx M)/(V xx N_0)`

= `(1 xx 60  g//mol)/((1 xx 10^-24  cm^3) xx (6 xx 10^23  mol^-1))` 

= `60/(6 xx 10^-1)  "g/cm"^3`

= `60/0.6  "g/cm"^3`

= 100 g/cm³

= 1 × 10² g/cm³

= 1 × 10² g cm⁻³