An element X has a body centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm⁻³. The number of atoms present in 300 g of the element X is _______________.
Given : Avogadro constant, N A = 6.0 × 10²³ mol⁻¹.

Solution

Chemistry ma’am · 4 steps

Find the mass of one atom from the unit cell, then count how many of them make 300 g.

1. Volume of the cell. Convert the edge to centimetres first: a = 200 pm = 2 × 10⁻⁸ cm. So a³ = 8 × 10⁻²⁴ cm³.
2. Mass of one atom. One unit cell weighs 5 × 8 × 10⁻²⁴ = 4 × 10⁻²³ g. A bcc cell holds Z = 2 atoms, so one atom weighs 2 × 10⁻²³ g.

   N = (300 g)/(2 × 10⁻²³ g) = 1.5 × 10²⁵

3. Express in NA. (1.5 × 10²⁵)/(6.0 × 10²³) = 25. As a check, the molar mass is 2 × 10⁻²³ × 6.0 × 10²³ = 12 g mol⁻¹, and 300 / 12 = 25 mol.
4. So. 300 g of X contains 25 N_A atoms, which is option (d).

The idea to keep. Density times a³ gives the mass of one unit cell. Divide by Z to get the mass of one atom, and always put the edge length in cm before you cube it.

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