Answer:
Option A
Explanation:
Given ,
edge length
(a)= 500 pm= 500 x 10-12 m= 5 x 10-8 cm
Density (d) =7.5 g cm-3
Mass (m) = 300 g
For bcc unit , Z=2
the volume of unit cell (a3 )
$=(5\times 10^{-8}cm)^{3}=1.25 \times 10^{-22}cm^{3}$
Volume occupied by each atom
$\frac{1.25\times 10^{-22}cm^{-3}}{2}=6.25\times 10^{-23}cm^{-3}$
Volume of sample = $\frac{mass}{density}=\frac{300g}{7.5 gcm^{-3}}=40 cm^{3}$
The number of atoms present in 300g of the element
= $\frac{40 cm^{3}}{6.25\times 10^{-23}cm^{3}}=6.4\times 10^{23}atoms$