Question

An element has a body-centred cubic (bcc) structure with a cell edge of 28 pm. The density of the element is $7.2g/{cm}^3$. How many atoms are present in 208 g of the element?

Answer 1

Theoretical density of crystal is $\rho =\frac{nM}{N_A{a}^3}g/c{m}^3$

$\Rightarrow M=\frac{\rho .N_A.a^3}{n}$....(i)

Now, molar mass $\left(M\right)=\frac{Givenmass\left(m\right)}{No.ofatoms\left(N_o\right)}$....(ii)

(i) and (ii)

$\Rightarrow \frac{m}{N_o}=\frac{\rho .N_A.a^3}{n}$

$\Rightarrow N_o=\frac{m.n}{\rho .N_A.a^3}$

For BCC, $n=2;\rho =7.2g/c{m}^3;m=200g;a=28\times {10}^{-10}cm$

$\Rightarrow N_o=\frac{208\times 2}{7.2\times 6.022\times {10}^{23}\times (28{)}^3\times {10}^{-30}}=4370atoms$