Problem 10 - Entrance Test

1. **Calculate the volume of the unit cell (a³):** a = 288 pm = 288 × 10⁻¹⁰ cm = 2.88 × 10⁻⁸ cm a³ = (2.88 × 10⁻⁸ cm)³ = 23.887872 × 10⁻²⁴ cm³ 2. **Calculate the molar mass (M) of the element using the density formula:** Density (ρ) = (Z × M) / (a³ × NA) For a BCC structure, Z = 2. M = (ρ × a³ × NA) / Z M = (7.2 g/cm³ × 23.887872 × 10⁻²⁴ cm³ × 6.022 × 10²³ mol⁻¹) / 2 M = (7.2 × 23.887872 × 6.022 × 10⁻¹) / 2 M = (1034.92) / 2 = 517.46 g/mol Let's re-calculate M precisely: M = (7.2 × 23.887872 × 0.6022) / 2 = (103.492) / 2 = 51.746 g/mol. Oh, I made an error in decimal place for 10^-1. 10^-24 * 10^23 = 10^-1. M = (7.2 * 23.887872 * 0.6022) / 2 = 103.492 / 2 = 51.746 g/mol. 3. **Calculate the number of moles in 208 g of the element:** Number of moles (n) = Given mass / Molar mass n = 208 g / 51.746 g/mol ≈ 4.019 moles 4. **Calculate the number of atoms:** Number of atoms = Number of moles × Avogadro's number Number of atoms = 4.019 mol × 6.022 × 10²³ atoms/mol Number of atoms = 24.197 × 10²³ atoms = 2.4197 × 10²⁴ atoms. This value is very close to 2.40 × 10²⁴ atoms (option B). Let's check the calculation of M again. ρ = 7.2 g/cm³ a = 2.88 × 10⁻⁸ cm a³ = (2.88)³ × 10⁻²⁴ cm³ = 23.887872 × 10⁻²⁴ cm³ NA = 6.022 × 10²³ mol⁻¹ Z = 2 M = (7.2 × 23.887872 × 10⁻²⁴ × 6.022 × 10²³) / 2 M = (7.2 × 23.887872 × 6.022 × 10⁻¹) / 2 M = (7.2 × 23.887872 × 0.6022) / 2 M = 103.49206 / 2 = 51.74603 g/mol. Number of atoms = (208 g / 51.74603 g/mol) × 6.022 × 10²³ mol⁻¹ Number of atoms = 4.0197 × 6.022 × 10²³ = 24.197 × 10²³ = 2.4197 × 10²⁴ atoms. Rounding to two significant figures for the options, 2.40 × 10²⁴ atoms is the best match.