Question

An element has a BCC structure with a cell edge of 288pm.The density of the element is 7.2g/cm3.How many atoms are present in 208g of the element?

Answer 1

Volume of the unit cell = (288 * 10^-12)^3

= 2.39 * 10^-23 cm^3

Volume of 208 g of element can be calculated as follows:

Mass / Density = 208 / 7.2

= 28.88 cm^3

Therefore number of unit cells in this volume:

28.88 cm^3 / 2.9 × 10^-23 cm^3 / unit cell

= 12.03 * 10^23 unit cells 

We know that each bcc unit cell contains 2 atoms, therefore the total number of atoms in 208 g:

2 * 12.03 * 10^23 atoms

= 24.16 * 10^23 atoms




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Answer 2

Hello mate here is your answer.

An element has a body centeredcubic structure with a cell edge of 288 pm. The density of the elements is 7.2 gcm-3. Calculate the number of atoms present is 208 g of the element.

Hope it helps you.