Question

An element crystallising in body centred cubic lattice has an edge length of $500\,pm$. If its density is $4\,g\,cm^{-3}$, 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. $150$

Answer 1

The Correct Option is D

Relation used,

Density $(\rho) =\frac{Z \times M}{a^{3} \times N_{A}}$

where, $Z =$ contribution factor

$M =$ molar mass of particle

$a =$ edge length

$\rho =$ density of crystal

Given, $N_{A} =$ Avogadro's number

$=6 \times 10^{23}$
$\rho =4 g / cm ^{3}$
$Z =2$ (crystal has bcc structure)

$\therefore M=\frac{\rho \times a^{3} \times N_{A}}{Z}$
$=\frac{4 \times\left(500 \times 10^{-10}\right)^{3} \times 6 \times 10^{23}}{2}$
$M =\frac{4 \times(5)^{3} \times 6 \times 10^{-1}}{2}$
$=\frac{4 \times 125 \times 6 \times 10^{-1}}{2}$
$M =\frac{300}{2}=150\, g\, mol ^{-1}$
$\therefore$ The atomic mass of the element (in $g\, mol ^{-1}$ ) is 150 .