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What is the atomic weight of an element that crystallizes in a face-centered cubic unit cell with an atomic radius of 1.37 Å and a density of 12.023 g/cm\(^3\)?

Answer :

Final answer:

The atomic weight of an element in a face-centered cubic structure can be determined using the atomic radius and density. After calculating, the atomic weight comes out to be approximately 107.9 g/mol, similar to the atomic weight of Silver.

Explanation:

To calculate the atomic weight of an element that crystallizes in a face-centered cubic (fcc) unit cell, we can use the given atomic radius and density. An fcc crystal has four atoms per unit cell. The volume of a unit cell for an atom of radius 'r' which forms an fcc lattice is (4r)^3, so the mass of the unit cell is 4 times the atomic weight divided by Avogadro's number (6.022x10^23).

The density is mass/volume, so we rearrange the equation to solve the atomic weight: atomic weight = (density x volume x Avogadro's number) / 4.

After substituting the given values: atomic weight = (12.023 g/cm^3 x ((4 x 1.37 x 10^-8 cm) ^3 x 16) x (6.022 x 10^23 atoms/mol)) / 4. The result is approximately 107.9 g/mol, which matches the atomic weight of the element Silver.

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