An element crystallises in a bcc lattice with cell edge of $500 \mathrm{pm}$. The density of the element is…

An element crystallises in a bcc lattice with cell edge of $500 \mathrm{pm}$. The density of the element is $7.5 \mathrm{~g} \mathrm{~cm}^{-3}$. How many atoms are present in $300 \mathrm{~g}$ of metal ?
  1. $12 \cdot 8 \times 10^{23}$ atoms
  2. $6 \cdot 4 \times 10^{23}$ atoms
  3. $3 \cdot 2 \times 10^{23}$ atoms
  4. $1 \cdot 6 \times 10^{23}$ atoms

Solution

$a=500 \mathrm{pm}=5 \times 10^{-8} \mathrm{~cm}$ $\therefore \mathrm{a}^{3}=\left(5 \times 10^{-8} \mathrm{~cm}\right)^{3}=125 \times 10^{-24} \mathrm{~cm}^{3}$ $\rho=7.5 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{~m}=300 \mathrm{~g}, \mathrm{n}=2($ for bcc cell $)$ $\rho=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \quad \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}$ $\therefore M=\frac{7.5 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2}=282.3 \mathrm{~g} \mathrm{~mol}^{-1}$ $282.3 \mathrm{~g}$ of metal contains $6.022 \times 10^{23}$ atoms $\therefore 300 \mathrm{~g}$ of metal contains $\frac{6.022 \times 10^{23} \times 300}{282.3}=6.399 \times 10^{23}$ atoms $\approx 6.4 \times 10^{23}$ atoms

Asked in: MHT CET 2020 (16 Oct Shift 2)

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