An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0 g cm⁻³. The number of atoms present in 180 g of the element is ____________ × 10²³. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

You do not need the molar mass here: find the mass of one unit cell, then count cells in 180 g.

1. Volume of one cell. Convert the edge to centimetres: a = 300 pm = 3 × 10⁻⁸ cm. So a³ = 27 × 10⁻²⁴ = 2.7 × 10⁻²³ cm³.
2. Mass of one cell. Mass equals density times volume: 6.0 × 2.7 × 10⁻²³ = 1.62 × 10⁻²² g. A bcc cell holds 8 × 1/8 + 1 = 2 atoms, so each atom weighs 8.1 × 10⁻²³ g.

   N = (180 g)/(8.1 × 10⁻²³ g) = 2.22 × 10²⁴

3. So. N = 22.2 × 10²³ atoms, which rounds to 22.

The common slip: 11. Forgetting that a bcc cell holds 2 atoms counts unit cells instead of atoms and gives 11.1 × 10²³, so 11 instead of 22.

The idea to keep. Density times a³ gives the mass of one unit cell directly. Divide by Z to get the mass of one atom, and you can count atoms in any sample without N_A.

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