An atomic substance A of molar mass 12 g mol⁻¹ has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is ____________. (Nearest integer)
Given the density of A is 3.0 g mL⁻¹ and N_A=6.02 × 10²³ mol⁻¹

Solution

Chemistry ma’am · 3 steps

Find the mass of one unit cell from density and volume, then ask how many atoms make up that mass.

1. Volume in cm³. Density is in g mL⁻¹, which is the same as g cm⁻³, so work in cm. a = 300 pm = 3 × 10⁻⁸ cm, so a³ = 27 × 10⁻²⁴ = 2.7 × 10⁻²³ cm³.
2. Mass of unit cell. Mass of one unit cell = ρ × a³ = 3.0 × 2.7 × 10⁻²³ = 8.1 × 10⁻²³ g. Each atom of A has mass 12/(6.02 × 10²³) = 1.99 × 10⁻²³ g.

   Z = (ρ a³ N_A)/M = (3.0 × 2.7 × 10⁻²³ × 6.02 × 10²³)/12 = 48.76/12 ≈ 4.06

3. So. One unit cell holds about 4.06 atoms, so the nearest integer is 4. This matches a face-centred cubic (fcc) lattice.

The idea to keep. For any cubic unit cell, ρ = (Z M)/(a³ N_A). Convert the edge length to cm before cubing, so that a³ is in cm³ to match the density in g cm⁻³.

Answer from the NTA final answer key. Solution by PrepWiser.