An element ‘Y’ has a body-centred cubic (bcc) structure with a cell edge of 144 pm. The density of the element is 3.6 \(g/cm^3\). How many atoms are present in 700g of the element?

\(13.42 × 10^{25}\)atoms

The correct answer is option 4. \(13.42 × 10^{25}\)atoms.

Given,

For, bcc geometry, the number of atoms per unit cell \((z) = 2\)

Edge length, \((a) = 144 pm = 144 × 10^{-10} cm\)

Density, \((\rho ) = 3.6 g/cm^3\)

We know that,

\(\text{Density, }(\rho ) = \frac{Z × M}{N_A × a^3}\)

\(⇒ 3.6 = \frac{2 × M}{(6.02 × 10^{23})(144 × 10^{-10})^3}\)

\(⇒ 3.6 = \frac{2 × M}{1.7405}\)

\(⇒ 2 × M = 6.28228\)

\(⇒ M = 3.1411\)

\(∴ \text{ The number of moles = }\frac{\text{mass}}{\text{molar mass}}\)

\(⇒ \text{ The number of moles = } \frac{700}{3.1411} = 222.849\)

\(∴ \text{ The number of atoms = }222.849 × 6.022 × 10^{23}\)

\(⇒\text{ The number of atoms  }≈ 13.42 × 10^{25}\)