An atom crystallizes in fcc crystal lattice and has a density of $10\ \text{g cm}^{-3}$ with unit cell edge length of 100pm. Calculate the number of atoms present in 1 g of crystal.

An atom crystallizes in fcc crystal lattice and has a density of 10g cm⁻³ with unit cell edge length of 100pm.

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10\ \text{g cm}^{-3}

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Step 1. Convert the edge length: $a = 100$ pm $= 1\times10^{-8}$ cm, so $a^3 = (1\times10^{-8})^3 = 1\times10^{-24}\ \text{cm}^3$ .

Step 2. Mass of one unit cell $= \rho \times a^3 = 10 \times 1\times10^{-24} = 1\times10^{-23}$ g.

Step 3. An fcc unit cell contains 4 atoms, so mass of ONE atom $= \dfrac{1\times10^{-23}}{4} = 2.5\times10^{-24}$ g. …