Question

Silver crystallises in face centred cubic structure. The edge length of unit cell is found to be 408.7 pm. Calculate density of silver ( Atomic mass of $Ag$ = 108 g/$mo{l}^{-1}).$

Answer 1

Given:

The edge length of the unit cell, $a=408.7pm$ $=408.7\times {10}^{-10}cm$

Atomic mass of silver, M= 108 g ${mol}^{-1}$

$N_A=6.022\times {10}^{23}$ ${mol}^{-1}$

The number of atoms in the unit cell of a FCC structure, $n=4$

Density of silver,

$=\frac{n\times M}{N\times a^3}$

$=\frac{4\times 108g{mol}^{-1}}{6.022\times {10}^{23}{mol}^{-1}\times (408.7\times {10}^{-10}cm{)}^3}$

Density of silver $=10.51g/{cm}^3$