Question

An element of density 8.0 g/cm^3 forms an FCC lattice with unit cell edge of 300 pm. Calculate the number of atoms present in 0.5kg of the element.

(a) 95 x 10^23 atoms

(b) 93.59 x 10^23 atoms

(c) 92.59 x 10^23 atoms

(d) 91.38 x 10^23 atoms

I got this question in my homework.

The above asked question is from Solid State topic in division Solid State of Chemistry – Class 12

Answer 1

Right choice is (c) 92.59 x 10^23 atoms

The explanation: Given,

Density (ρ) = 8.0 g/cm^3

For FCC structure, Z = 4

Avogadro’s number (N0) = 6.02 x 10^23

Edge length of the unit cell (a) = 300 x 10^-10 cm

The density of the element (ρ) = (Z x M)/ (a^3 x N0)

 Therefore, the Molar Mass (M) = (ρ x a^3 x N0)/(Z)

    = (8.0 x 6.02 x 10^23 x 27.0 x 10^-24) /4

    = 32.508 g.

Therefore, 32.508 g of the element contains 6.02 x 10^23 atoms.

500 g of the element would contain = (6.02 x 10^23 x 500)/ 32.508 = 92.59 x 10^23 atoms.