Calculate the volume of unit cell when metal having density $1 \mathrm{~g} \mathrm{~cm}^{-3}$ and molar mass…

Calculate the volume of unit cell when metal having density $1 \mathrm{~g} \mathrm{~cm}^{-3}$ and molar mass $23 \mathrm{~g} \mathrm{~mol}^{-1}$ crystallises to form bce structure.
  1. $6.0 \times 10^{-23} \mathrm{~cm}^3$
  2. $8.6 \times 10^{-23} \mathrm{~cm}^3$
  3. $\quad 9.5 \times 10^{-23} \mathrm{~cm}^3$
  4. $7.6 \times 10^{-23} \mathrm{~cm}^3$

Solution

For $b c c$ unit cell, $\mathrm{n}=2$. $\operatorname{Density}(\rho)=\frac{M \times n}{a^3 \times N_A}$ $\begin{aligned} & \text {Volume of unit cell }=\mathrm{a}^3=\frac{\mathrm{M} \times \mathrm{n}}{\rho \times \mathrm{N}_{\mathrm{A}}} \\ & \qquad \begin{aligned} & \\ & \qquad \frac{23 \mathrm{~g} \mathrm{~mol}^{-1} \times 2}{1 \mathrm{~g} \mathrm{~cm}^{-3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}} \\ &=7.6 \times 10^{-23} \mathrm{~cm}^3 \end{aligned} \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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