An element crystallises in a fcc lattice with cell edge $250 \mathrm{pm}$. Calculate the density of an…

An element crystallises in a fcc lattice with cell edge $250 \mathrm{pm}$. Calculate the density of an element (at.mass $=90 \cdot 3$ )
  1. $23 \cdot 12 \mathrm{~g} \mathrm{~cm}^{-3}$
  2. $19 \cdot 20 \mathrm{~g} \mathrm{~cm}^{-3}$
  3. $48 \cdot 40 \mathrm{~g} \mathrm{~cm}^{-3}$
  4. $38.40 \mathrm{~g} \mathrm{~cm}^{-3}$

Solution

$\mathrm{a}=250 \mathrm{pm}=2.5 \times 10^{-8} \mathrm{~cm}$ $\mathrm{M}=90.3 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{n}=4($ for fcc cell $)$ $\rho=\frac{n \times M}{a^{3} \times N_{A}}=\frac{4 \times 90.3}{\left(2.5 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}$ $\therefore \rho=38.40 \mathrm{~g} \mathrm{~cm}^{-3}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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