An element crystallising in fcc lattice has a density of $8.92 \mathrm{~g} \mathrm{~cm}^{-3}$ and edge…

An element crystallising in fcc lattice has a density of $8.92 \mathrm{~g} \mathrm{~cm}^{-3}$ and edge length of $3.61 \times 10^{-8} \mathrm{~cm}$. What is the atomic weight of element? $\left(N=6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)$
  1. $126.356 u$
  2. $63.178 u$
  3. $31.589 u$
  4. $47.383 u$

Solution

Given, $d=8.92 \mathrm{~g} \mathrm{~cm}^{-3}$ (Edge length) $a=3.61 \times 10^{-8} \mathrm{~cm}$ $M=$ ? For fcc, $Z=4$ $d=\frac{Z \times M}{a^3 \times N}$ $8.92=\frac{4 \times M}{\left(3.61 \times 10^{-8}\right)^3 \times 6.022 \times 10^{23}}$ $\Rightarrow \quad M=\frac{8.92 \times 47.045 \times 10^{-24} \times 6.022 \times 10^{-23}}{4}$ $=63.178 \mathrm{u}$

Asked in: AP EAMCET 2022 (05 Jul Shift 1)

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