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Question

# Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold =197 amu.

A
19.4g/cm3, 143.9 pm
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B
38.8g/cm3, 143.9 pm
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C
19.4g/cm3, 287.8 pm
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D
None of the above
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Solution

## The correct option is A 19.4g/cm3, 143.9 pmFor an FCC crystal, atoms are in contact along the diagonal of the unit cell. 4r=√2a, where a is the edge length of unit cell and r is the radius of atom.Substituting values in the above expression, we get r=√2×4074=143.9 pmNumber of atoms per unit cell (FCC) =4Mass of 1 unit cell =197×4=788amu=1.66×10−27×788kg=1.30808×10−24kgVolume of 1 unit cell =(407×10−12)3m3=6.741×10−29m3 =2.824×10−28m3Density=1.30808×10−246.741×10−29=19404.83kg/m3=19.4g/cm3

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