An element forms a body centered cubic (bcc) lattice with edge length of 300 pm, If the density of the element is $7.2 g c m^{- 3}$ ,the number of atoms present in $324 g$ of it approximately is

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$6.66 \times 10^{24}$

$6.66 \times 10^{23}$

$3.33 \times 10^{24}$

$3.33 \times 10^{23}$

answer is C.

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Detailed Solution

Volume of bcc unit cell = 300 pm

$= {(300 \times 10^{- 10} c m)}^{3} = 27 \times 10^{- 24} c m^{3}$

Volume of 324 g of the element = $\frac{M a s s}{D e n s i t y}$

$= \frac{324 g}{7.2 g c m^{- 3}} = 45 c m^{3}$

For a bcc structure, number of atoms per unit cell = 2

So, number of atoms = $\frac{T o t a l v o l u m e}{v o l u m e o f a u n i t c e l l}$

$= \frac{2 \times 45}{27} \times 10^{24} = 333 \times 10^{24}$

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