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17 June, 18:10

Platinum (pt) has the fcc crystal structure, an atomic radius of 0.1387 nm, and an atomic weight of 195.08 g/mol. what is its theoretical density?

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  1. 17 June, 18:33
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    The equation to be used is written as:

    ρ = nA/VcNₐ

    where

    ρ is the density

    n is the number of atoms in unit cell (for FCC, n=4)

    A is the atomic weight

    Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis) /

    Nₐ is Avogradro's constant equal to 6.022*10²³ atoms/mol

    r = 0.1387 nm * (10⁻⁹ m/nm) * (100 cm/1m) = 1.387*10⁻⁸ cm

    a = 4 (1.387*10⁻⁸ cm) / √2 = 3.923*10⁻⁸ cm

    V = a³ = (3.923*10⁻⁸ cm) ³ = 6.0376*10⁻²³ cm³

    ρ = [ (4 atoms) (195.08 g/mol) ]/[ (6.0376*10⁻²³ cm³) (6.022*10²³ atoms/mol) ]

    ρ = 21.46 g/cm³
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