An element crystallizes as body centred cubic lattice. Its density is 7.12 g cm^–3 and the length of the side of the unit cell is 2.88 $\stackrel{\mathrm{o}}{\mathrm{A}}$ . Calculate the number of atoms present in 288 g of an element is 3.386×10^4×x then find the x.

Density=7.12 gcm⁻³

 NA​=6.022×10²³

 z = 2 (B.C.C lattice)

 a = side length = 2.88A0=2.88×10⁻⁸cm

 Putting the values of all this in above equation,

$7.12=\frac{2\times \mathrm{M}}{6.022\mathrm{x}{10}^{23}\mathrm{x}{(2.88\mathrm{x}{10}^{-8})}^3}$

After solving, we get 

 M = 51.2 g

 Number of mole (n) = $\frac{ \mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$

 n=$\frac{288}{51.2}$=5.63

 Number of atoms = n ×NA​

 Number of atoms = 5.63×6.02×10²³

 Number of atoms = 3.39×10²⁴

 Hence, number of atoms present is 288 g of the element = 3.39×10^4x6

Then x=6