Calculate the volume of unit cell of element with density $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$ that forms bcc…

Calculate the volume of unit cell of element with density $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$ that forms bcc structure. (288 gram of this element $3.35 \times 10^{24}$ atoms)
  1. $3.038 \times 10^{23} \mathrm{~cm}^3$
  2. $4.18 \times 10^{23} \mathrm{~cm}^3$
  3. $6.136 \times 10^{23} \mathrm{~cm}^3$
  4. $3.912 \times 10^{23} \mathrm{~cm}^3$

Solution

$\mathrm{d}=\frac{\mathrm{Z}}{\mathrm{V}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}$ $7.2=\frac{2}{V} \times \frac{288}{3.35 \times 10^{24}}$ $\mathrm{V}=4.18 \times 10^{22} \mathrm{~cm}^3$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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