Question

An element crystallizes in a body centred cubic lattice. The edge length of the unit cell is $200\, pm$ and the density of the element is $5.0 \,gcm ^{-3}$. Calculate the number of atoms in $100 \,g$ of this element.

• A. $2.5 \times 10^{23}$
• B. $2.5 \times 10^{24}$
• C. $5.0 \times 10^{23}$
• D. $5.0 \times 10^{24}$

Answer 1

Answer: (D)

$\rho_{\text {unit cell }}(g / c c)=\frac{Z \times M . \text{Mass}(g / m o l)}{N_{A}\left(m o l^{-1}\right) \times(a p m)^{3} \times 10^{-30}}$
$Z_{\text {B.C.C }}=2$
or $5=\frac{2 \times \text { M.Mass }}{N_{A} \times(200)^{3} \times 10^{-30}}$
or M.Mass $=\frac{5 \times 8 \times N_{A} \times 10^{-24}}{2}=2 N_{A} \times 10^{-23}$
Now, no. of moles in $100\, g=\frac{100}{2 N_{A} \times 10^{-23}}$
$\therefore $ no. of atoms $=\frac{100}{2 N_{A} \times 10^{-23}} \times N_{A}$
$=50 \times 10^{23}=5.0 \times 10^{24}$