Question

An element has a body-centred cubic (bcc) structure with a cell edge of

288 pm. The density of the element is 7.2 g/cm³. How many atoms are present in 208 g of the element?​

Answer 1

Answer :

Volume of the unit cell = (288 pm)³

Volume of the unit cell $\sf = (288×10^{-12} m)³$

Volume of the unit cell $\sf= (288×10^{-10} cm)³$

Volume of the unit cell $\sf= 2.39×10^{-23} cm³$

Volume of 208 g of the element

= $\sf \frac{mass}{density} = \frac{208 \: \: g}{7.2 \: \: g \: \: cm^{-3}} = 28.88 \: \: cm³$

Number of unit cells in this volume

= $\sf \frac{28.88 \: \: cm³}{2.39 × 10^{-23} \: \: cm³ / unit \: \: cell } = 12.08 × 10²³ unit cell$

Since each bcc cubic unit cell contains 2 atoms, therefore, the total number

of atoms in 208 g = 2 (atoms/unit cell) × 12.08 × 10²³ unit cells

= 24.16×10²³ atoms

I hope it helps you ❤️✔️

Answer 2

Answer:

Volume of the unit cell = (288 pm)³

Volume of the unit cell \sf = (288×10^{-12} m)³ =(288×10

−12

m)³

Volume of the unit cell \sf= (288×10^{-10} cm)³ =(288×10

−10

cm)³

Volume of the unit cell \sf= 2.39×10^{-23} cm³ =2.39×10

−23

cm³

Volume of 208 g of the element

= \begin{gathered} \sf \frac{mass}{density} = \frac{208 \: \: g}{7.2 \: \: g \: \: cm^{-3}} = 28.88 \: \: cm³ \\ \end{gathered}

density

mass

=

7.2gcm

−3

208g

=28.88cm³

Number of unit cells in this volume

= \begin{gathered} \sf \frac{28.88 \: \: cm³}{2.39 × 10^{-23} \: \: cm³ / unit \: \: cell } = 12.08 × 10²³ unit cell \\ \end{gathered}

2.39×10

−23

cm³/unitcell

28.88cm³

=12.08×10²³unitcell

Since each bcc cubic unit cell contains 2 atoms, therefore, the total number

of atoms in 208 g = 2 (atoms/unit cell) × 12.08 × 10²³ unit cells

= 24.16×10²³ atoms

I hope it helps you