an element has bcc with a cell edge of 288pm the density of the element is 7 2 gm/cm^3 - Chemistry - The Solid State

Question

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?

Anita Dahiya · Accepted Answer

Volume of the unit cell = (288 * 10-12)3 = 2 39 * 10-23 cm3 Volume of 208 g of element can be calculated as follows: $\frac{M a s s}{D e n s i t y} = \frac{208}{7 2}$ = 28 88 cm3 Therefore number of unit cells in this volume: $\frac{28 88 {c m}^{3}}{2 39 \times 10^{- 23} {c m}^{3} / u n i t c e l l}$ = 12 03 * 1023 unit cells We know that each bcc unit cell contains 2 atoms, therefore the total number of atoms in 208 g: 2 * 12 03 * 1023 atoms = 24 16 * 1023 atoms

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?