What is the number of atoms present per unit cell of aluminium having edge length $4 \mathrm{~A}^{\circ}$ ?…

What is the number of atoms present per unit cell of aluminium having edge length $4 \mathrm{~A}^{\circ}$ ? (If density of $\mathrm{Al}=2 \cdot 7 \mathrm{~g} \mathrm{~cm}^{-3}$, At. mass of $\mathrm{Al}=27$ )
  1. 8
  2. 1
  3. 2
  4. 4

Solution

$\mathrm{a}=4 Å=4 \times 10^{-8} \mathrm{~cm}, \rho=2.7 \mathrm{~g} \mathrm{~cm}^{-3}$ $\mathrm{M}=27, \mathrm{n}=?$ $\mathrm{n}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}}=\frac{2.7 \times\left(4 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{27}=3.85 \approx 4$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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