Question

An element 'X' (At. mass = 40 g $mo{l}^{-1}$) having FCC 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

m = 40 g $mo{l}^{-1}$

Z = 4 (fcc)

a = 400 pm = $400\times {10}^{-10}cm$

D = $\frac{ZM}{a^3{N}_0}$

= $\frac{4\times 40}{(400\times {10}^{10}{)}^3\times 6.022\times {10}^{23}}$

= $\frac{160}{64\times {10}^6\times {10}^{-30}\times 6.022\times {10}^{23}}$

= $\frac{160}{64\times 6.022\times {10}^{-1}}=4.15gc{m}^{-3}$

No. of unit cells =$\frac{W.No.}{Z.M}$

= $\frac{4\times 6.022\times {10}^{23}}{4\times 40}$

= $1.505\times {10}^{22}$