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A metallic element exists as cubic lattice. Each edge of the unit cell is Å. The density of the metaSolid State Chemistry Question

Question

A metallic element exists as cubic lattice. Each edge of the unit cell is $4.0$ Å. The density of the metal is $6.25 \text{ g/cm}^3$. How many unit cells will be present in 100 g of the metal?

Answer: B

💡 Solution & Explanation

Volume of one unit cell = $a^3 = (4.0 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3$. Mass of one unit cell = volume $\times$ density = $64 \times 10^{-24} \times 6.25 = 400 \times 10^{-24} \text{ g} = 4 \times 10^{-22} \text{ g}$. Number of unit cells in 100 g = $100 / (4 \times 10^{-22}) = 25 \times 10^{22} = 2.5 \times 10^{23}$. Therefore, correct answer is B.

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