Question

Silver crystallises in fcc lattice. If edge length of the cell is ${{4}}{{.07 \times 1}}{{{0}}^{{{ - 8}}}}{{cm}}$ , and density is ${{10}}{{.5\,g}}\,{{c}}{{{m}}^{{{ - 3}}}}$
Calculate the atomic mass of silver

Answer 1

Hint: Face-centered cubic lattice (fcc or cubic-F), like all lattices, has lattice points at the eight corners of the unit cell plus additional points at the centers of each face of the unit cell. It has unit cell vectors \[{{a = b = c}}\] and interaxial angles \[{{\alpha = \beta = \gamma = 90^\circ }}\] . The number of atoms in a unit cell is four \[{{8 \times 1/8 + 6 \times 1/2 = 4}}\] There are 26 metals that have the fcc lattice

Complete step by step solution:
${{Density = }}\dfrac{{{{ZM}}}}{{{{V}}{{{N}}_{{0}}}}}$ , where ‘Z’ is the number of atoms in a unit cell
\[{{Z = 4}}\] for fcc lattice
${{10}}{{.5 = }}\dfrac{{{{4 \times M}}}}{{{{{{(4}}{{.077 \times 1}}{{{0}}^{{{ - 8}}}}{{)}}}^{{3}}}\times{{6}}{{.023 \times 1}}{{{0}}^{{{23}}}}}}$
Therefore, \[{{M = 107}}{{.1\,g}}\]

The mass of silver = 107.1 g

Additional Information:
The face-centered cubic system is closely related to the hexagonal close packed (hcp) system, where two systems differ only in the relative placements of their hexagonal layers. The [111] plane of a face-centered cubic system is a hexagonal grid.

Note:
The coordination number of each atom in face centered cubic structure is 6: each cation is coordinated to 6 anions at the vertices of an octahedron, and similarly, each anion is coordinated to 6 cations at the vertices of an octahedron
In fcc structure, The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.