An element has a bcc structure with cell edge of $288 \mathrm{pm}$. The density of element is $7.2…

An element has a bcc structure with cell edge of $288 \mathrm{pm}$. The density of element is $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$. What is the atomic mass of an element?
  1. $51.78$
  2. $25.89$
  3. $62.43$
  4. $77.68$

Solution

$\begin{array}{l} \mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm} \\ \therefore \mathrm{a}^{3}=\left(2.88 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=2.39 \times 10^{-23} \mathrm{~cm}^{3} \\ \rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{n}=2(\text { for bcc cell }) \\ \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{7.2 \times 2.39 \times 10^{-23} \times 6.022 \times 10^{23}}{2}=51.81 \mathrm{~g} \end{array}$

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

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