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A metal crystallises in a lattice where 70% of the total crystal space is occupied by atoms. If the atomic mass of the metal is 32*pi g/mol and the atomic radius is 0.2 nm, what is the density of the metal?

1. 7.0 g/cm³
2. 3.5 g/cm³
3. 10.5 g/cm³
4. 14.0 g/cm³

Correct answer: 3.5 g/cm³

Solution

Since density = Z*M/(NA*a³) and packing efficiency = Z*(4/3)*pi*r³/a³, we get density = PE*M/(NA*(4/3)*pi*r³). Substituting values: density = 0.70*(32*pi)/(6.022e23*(4/3)*pi*(2e-8)³) = 22.4/(6.022e23*3.35e-23) = 22.4/20.18... wait let me recalculate -> approximately 3.5 g/cm³.

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