An element with molar mass 2.7 × 10⁻² kg mol⁻¹ forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10³ kg m⁻³, the radius of the element is approximately ______ × 10⁻¹² m (to the nearest integer).

Solution

Chemistry ma’am · 4 steps

Use the density to find Z first, because Z tells you which cubic cell it is and so which radius formula fits.

1. Density formula. Density, molar mass and edge length are linked through Z, the number of atoms per unit cell. Rearrange to make Z the subject.

   Z = (ρ a³ N_A)/M

2. Put in numbers. a = 405 pm = 4.05 × 10⁻¹⁰ m, so a³ = 6.643 × 10⁻²⁹ m³. Then Z = (2.7 × 10³ × 6.643 × 10⁻²⁹ × 6.022 × 10²³)/(2.7 × 10⁻²) = 0.108/0.027 ≈ 4.
3. Pick the cell. Z = 4 means a face-centred cubic (fcc) cell. In fcc the atoms touch along the face diagonal, so 4r = √2 a, which gives r = a/2√2.
4. So. r = 405/(2 × 1.414) = 405/2.828 = 143.2 pm, which rounds to 143.

The common slip: 175. Using the bcc relation r = √3/4a instead of the fcc one gives about 175, not 143. Find Z before choosing the formula.

The idea to keep. When the cell type is not given, density hands it to you: find Z first. Then remember where atoms touch: edge for simple cubic, body diagonal for bcc, face diagonal for fcc.

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