**An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element?**

Given: Types of unit cell = bcc, for bcc unit cell n = 2

Edge length (a) = 288 pm = 2.88 x 10^{-8 }cm^{3}

Mass of element (x) =200g

Density of element (r) = 7.2 gm/cm^{3}

Number of unit cell in x g of element = \(\frac{x}{ρa^3}\)

= \(\frac{200}{7.2×(2.88×10^{-8})^3}\) = **1.16 x 10 ^{24}**

Number of atoms in x g of element = \(\frac{nx}{ρa^3}\)

= \(\frac{2x}{ρa^3}\)= 2 × 1.16 x 10^{24 }**= 2.32 x 10 ^{24}**

**Ans. **Number of unit cell in x g of element** = 1.16 x 10 ^{24}**

Number of atoms in x g of element** = 2.32 x 10 ^{24}**