Question

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

${\rho }_{\text{unit cell }}(g/cc)=\frac{Z\times M.\text{Mass}(g/mol)}{N_A\left(mo{l}^{-1}\right)\times (apm{)}^3\times 10^{-30}}$
$Z_{\text{B.C.C }}=2$
or $5=\frac{2\times \text{ M.Mass }}{N_A\times (200{)}^3\times 10^{-30}}$
or M.Mass $=\frac{5\times 8\times N_A\times 10^{-24}}{2}=2N_A\times 10^{-23}$
Now, no. of moles in $100g=\frac{100}{2N_A\times 10^{-23}}$
$\therefore$ no. of atoms $=\frac{100}{2N_A\times 10^{-23}}\times N_A$
$=50\times 10^{23}=5.0\times 10^{24}$