Question

An element 'X' (At. mass $=40$g $mo{l}^{-1}$) having f.c.c structure, has unit cell edge length of $400$ pm. Calculate the density of 'X' and the number of unit cells in $4$g of 'X'. $(N_A=6.022\times {10}^{23}mo{l}^{-1})$.

Answer 1

$\text{edge length}=\text{400 pm}=\text{400}\times {10}^{-10}\text{cm}=\text{4}\times {10}^{-8}\text{cm}$
$\text{Density}=\frac{\text{Atomic mass}\times \text{Number of atoms in 1 unit cell}}{\text{Avogadro's number}\times (\text{edge length}{)}^3}$
$\text{Density}=\frac{40gmo{l}^{-1}\times \text{4}}{6.022\times {10}^{23}mo{l}^{-1}\times (\text{4}\times {10}^{-8}\text{cm}{)}^3}$

$\text{Density}={\text{4.15 g/cm}}^3$
Volume of 4 g of 'X' $=\frac{\text{4 g}}{{\text{4.15 g/cm}}^3}={\text{0.964 cm}}^3$
Volume of one unit cell $=(\text{edge length}{)}^3=(\text{4}\times {10}^{-8}\text{cm}{)}^3=6.4\times {10}^{-23}{\text{cm}}^3$
The number of unit cells in 4g of 'X' $=\frac{{\text{0.964 cm}}^3}{6.4\times {10}^{-23}{\text{cm}}^3}=1.5\times {10}^{22}$