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A metallic element exists as cubic lattice. Edge length of the unit cell is $2 A^{0}$ . If the density
of the metal is $6.25 g / {cm}^{3}$ then the number of unit cells present in 100g of the metal is $x \times 10^{24}$ The value of x is:

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answer is 2.

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Detailed Solution

Volume of metal taken $= \frac{m}{d} = \frac{100}{6.25} = 16 {cm}^{3}$
Volume of each unit cell $= {(2 \times 10^{- 8} cm)}^{3} = 8 \times 10^{- 24} {cm}^{3}$
No. of unit cells $= \frac{16}{8 \times 10^{- 24}} = 2 \times 10^{24}$

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