an element has bcc with a cell edge of 288pm the density of the element is 7.2 gm/cm^3. how many atoms are present in 208gm of the element?

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:

= 28.88 cm^{3}

Therefore number of unit cells in this volume:

= 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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