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A metal has an fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm⁻³. The molar mass of the metal is (Nₐ Avogadro’s constant = 6.02 × 10²³ mol⁻¹):

1. 30 g mol⁻¹
2. 27 g mol⁻¹
3. 20 g mol⁻¹
4. 40 g mol⁻¹

Correct answer: 27 g mol⁻¹

Solution

For an fcc lattice, the number of atoms per unit cell is 4. Using the density formula, \( \text{Density} = \frac{Z \cdot M}{a^3 \cdot N_A} \), where \( Z = 4 \), \( a = 404 \times 10^{-10} \) cm, and \( N_A = 6.02 \times 10^{23} \). Solving for \( M \), we get \( M = 27 \, \text{g mol}^{-1} \).

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