Question

An element crystalline in body centered cubic lattice has an edge of 500 pm. If its density is 4 g $c{m}^{-3}$, the atomic mass of the element (in g $mo{l}^{-1}$) is:

(consider $N_A=6\times {10}^{23}$)

• A. 100
• B. 250
• C. 125
• D. 150

Answer 1

Answer: (D)

150

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

Given that for bcc, $n=2,a=500\times {10}^{-10}cm,N_o=6\times {10}^{23}\rho =4g/c{m}^3$

$\therefore M=\frac{\rho .N_o.a^3}{n}$

$=\frac{4\times 6\times {10}^{23}\times (500{)}^3\times {10}^{-30}}{2}$

$\Rightarrow M=150g$