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A metal M having atomic mass 31.25 crystallizes in cubic close packing (CCP/FCC) and shows Schottky defects. The edge length of the unit cell is 500 pm and the density of the metal is 1.6075 g/mL. Calculate the number of moles of metal atoms missing per litre of the crystal. [Given: 1 amu = 1.67 * 10⁻²⁴ g]

1. 0.5 mol
2. 1 mol
3. 2 mol
4. 4 mol

Correct answer: 2 mol

Solution

Ideal density (no defects, Z=4): rho_ideal = 4*31.25 / (5.988*10²³ * (500*10⁻¹⁰)³) = 125 / 74.85 = 1.6700 g/mL. Fraction of missing sites = (1.6700 - 1.6075) / 1.6700 = 0.03743. Missing atoms per unit cell = 4 * 0.03743 = 0.1497. Volume of unit cell = (500 pm)³ = 1.25*10⁻²² mL. Cells per litre = 10³ / 1.25*10⁻²² = 8*10²⁴. Total missing atoms per litre = 0.1497 * 8*10²⁴ = 1.198*10²⁴. Moles missing = 1.198*10²⁴ / 5.988*10²³ = 2.0 mol.

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