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

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Answer 2

“The total number of atoms” are present in 208 g of element is $\bold{=2.42 \times 10^{24}}$.

The formula for Density $d=\frac{Z M}{N A} \times a 3$

Rearranging the formula  

$M=\frac{d \times N A \times a 3}{Z}$

Z=2,  Because the body centered cubic cell contains 2 atoms, then the 208 grams of element contains 2 atoms per unit cell.

On Substituting values, we get  

$M=\frac{\left(\frac{7.2 g}{c m 3} \times\left(6.022 \times 10^{23}\right) \times\left(6.022 \times 10^{23}\right)\right)}{2}$

$=51.8 \mathrm\ {gmol}^{-1}$.  

51.8 g of element contains $6.022 \times 10^{23}$

208g of this element contains=?

$=\frac{\left(6.022 \times 10^{23} \times 208\right)}{51.8}$

$=2.42 \times 10^{24}\ atoms$