Metal M crystallizes into a fcc lattice with the edge length of 4.0 × 10⁻⁸ cm. The atomic mass of the metal is ____________ g / mol. (Nearest integer)
( Use : N_A=6.02 × 10²³ mol⁻¹, density of metal, M=9.03 g cm⁻³ )

Solution

Chemistry ma’am · 4 steps

Density links one unit cell's mass to its volume, and in fcc that cell holds 4 atoms.

1. Atoms per cell. In an fcc cell, 8 corner atoms each give 1/8 and 6 face atoms each give 1/2. So Z = 1 + 3 = 4 atoms per unit cell.

   d = (Z × M)/(a³ × N_A) ⇒ M = (d × a³ × N_A)/Z

2. Cell volume. a³ = (4.0 × 10⁻⁸ cm)³ = 64 × 10⁻²⁴ cm³. The length is already in cm, which matches the density unit.
3. Put in numbers. M = (9.03 × 64 × 10⁻²⁴ × 6.02 × 10²³)/4 = 347.9/4 = 86.98 g mol⁻¹.
4. So. The atomic mass of M is about 86.98, which rounds to 87 g mol⁻¹.

The common slip: 348. Forgetting to divide by Z = 4 for fcc gives 348, the mass of all four atoms in the cell, not one.

The idea to keep. For any density question, first fix Z (1 for simple cubic, 2 for bcc, 4 for fcc), and check that the edge length and density use the same length unit before you multiply.

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