Question

The number of atoms in 2.4 g of a body centred cubic crystal with edge length 200 pm is ($density=10gc{m}^{-3},N_A=6\times {10}^{23}atoms/mol$)

• A. $6\times {10}^{22}$
• B. $6\times {10}^{20}$
• C. $6\times {10}^{23}$
• D. $6\times {10}^{19}$

Answer 1

The correct option is A $6\times {10}^{22}$

Volume of unit cell = $(200pm{)}^3$
$\Rightarrow (200\times {10}^{-10}cm{)}^3$
$\Rightarrow 8\times {10}^{-24}c{m}^3$
Volume of 2.4 g of the element =$\frac{Mass}{Density}$
$\Rightarrow \frac{2.4}{10}\Rightarrow 0.24c{m}^3$
Number of unit cells =$\frac{Totalvolume}{Volumeofaunitcell}$
$\Rightarrow \frac{0.24c{m}^3}{8\times {10}^{-24}c{m}^3}$
$\Rightarrow 0.03\times {10}^{24}$
For a bcc structure, number of atom per unit cell = 2
Number of atoms present in 2.4g = Number of atoms per unit cell × Number of unit cells
$\Rightarrow 2\times 0.03\times {10}^{24}$
$\Rightarrow 6\times {10}^{22}$