Discussion Forum



1)

 An element crystallises in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g cm-3. How many atoms are present in 300g of metal?


A) $6.4\times 10 ^{23}$ atoms

B) $12.8\times 10 ^{23}$ atoms

C) $3.2\times 10 ^{23}$ atoms

D) $1.6\times 10 ^{23}$ atoms

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

   = $\frac{40 cm^{3}}{6.25\times 10^{-23}cm^{3}}=6.4\times 10^{23}atoms$