Silver forms CCP structure and it's density is $10.5 \mathrm{~g} / \mathrm{cm}^3$. What is the edge length…

Silver forms CCP structure and it's density is $10.5 \mathrm{~g} / \mathrm{cm}^3$. What is the edge length of unit cell. (molar mass of silver is $107.9 \mathrm{~g} / \mathrm{mol}$ )
  1. $\sqrt[3]{0.68} Å$
  2. $\sqrt[3]{48} Å$
  3. $\sqrt[3]{68.1} Å$
  4. $\sqrt[3]{680} Å$

Solution

$\mathrm{d}=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{a}^3} \Rightarrow \mathrm{a}^3=\frac{\mathrm{ZM}}{\mathrm{N}_{\mathrm{A}} \mathrm{d}}=\frac{(4)(107.9)}{\left(6.022 \times 10^{23}\right)(10.5)}$ $\begin{aligned} & =6.825 \times 10^{-23} \mathrm{~cm}^3 \\ & =6.825 \times 10^{-23}\left(10^8 Å\right)^3 \\ & =6.825 \times 10^{-23} \times 10^{24} Å \\ & =68.25 Å \Rightarrow \mathrm{a}=\sqrt[3]{68.25} Å\end{aligned}$

Asked in: AP EAMCET 2023 (18 May Shift 1)

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