Silver crystallisaes in fcc lattice. If the edge length of th cell is `4xx10^(-8)` cm and density is 11.2 g `"cm"^(-3)`, calculate the atomic mass of silver.

Explanation

Density, `d=(ZxxM)/(a^(3)xxN_(A))`
where Z = Number of atoms per unit cell
M = Molar mass,
a= Edge length of cube
`N_(A)` Avogado.s number `(6.022xx10^(23)"mol"^(-1))`
`thereforeM=(dxxa^(3)xxN_(A))/(Z)`
Here `d=10.5"g cm"^(-3),a=4.077xx10^(-8)"cm"`
z = (for f.c.c. lattice), M = ?
`thereforeM=((11.2g"cm"^(-3))xx(4xx10^(-8)"cm")^(3)xx(6.022xx10^(23)"mol"^(-1)))/(4)`
`therefore` Atomic mass of silver = 107.9 u