Calculate the number of atoms in 5 gram metal that crystallises to form simple cubic unit cell structure…

Calculate the number of atoms in 5 gram metal that crystallises to form simple cubic unit cell structure having edge length $336 \mathrm{pm}$. (Density of metal $=9.4 \mathrm{~g} \mathrm{~cm}^{-3}$ )
  1. $1.4 \times 10^{22}$
  2. $1.8 \times 10^{22}$
  3. $1.0 \times 10^{22}$
  4. $2.1 \times 10^{22}$

Solution

$\begin{aligned} & \mathrm{d}=\frac{\mathrm{z}}{\mathrm{v}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}} \\ & 9.4=\frac{1 \times \mathrm{M}}{\left(3.36 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}}\end{aligned}$ $\mathrm{M}=214.65 \mathrm{~g} / \mathrm{mol}$ No. of atoms $=\frac{5}{214.65} \times 6.02 \times 10^{23}$ $=1.4 \times 10^{22}$

Asked in: MHT CET 2022 (10 Aug Shift 2)

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