Question

An element crystallising in body centred cubic lattice has an edge length of $500pm$. If its density is $4g{cm}^{-3}$, the atomic mass of the element (in $g{mol}^{-1}$) is: (consider $N_A=6\times {10}^{23}$)

• A. $100$
• B. $250$
• C. $125$
• D. $150$
• E. $50$

Answer 1

The correct option is D $150$

Relation used,
Density $\left(\rho \right)=\frac{Z\times M}{a^3\times N_A}$
where, $Z=$ contribution factor
$M=$ molar mass of particle
$a=$ edge length
$\rho =$ density of crystal

Given, $N_A=$ Avogadro's number
$=6\times {10}^{23}$
$\rho =4g/{cm}^3,a=500pm=500\times {10}^{-10}cm$
$Z=2$ (crystal has bcc structure)

$\therefore M=\frac{\rho \times a^3\times N_A}{Z}$

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

$M=\frac{4\times {\left(5\right)}^3\times 6\times {10}^{-1}}{2}$

$=\frac{4\times 125\times 6\times {10}^{-1}}{2}$

$M=\frac{300}{2}=150g{mol}^{-1}$

$\therefore$ The atomic mass of the element (in $g{mol}^{-1}$) is $150$.

Hence, Option "D" is the correct answer.