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Calculate the number of unit cells in 3 grams metal that crystallises to simple cubic unit cell having edge length $336 \mathrm{pm}$. (density of metal $=9.4 \mathrm{~g} \mathrm{~cm}^{-3}$ )
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The correct answer is:
$8.41 \times 10^{21}$
$\mathrm{d}=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{V} \times \mathrm{N}_{\mathrm{A}}}$
$\begin{aligned} & 9.4=\frac{1 \times M}{\left(3.36 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}} \\ & M=214.6 \mathrm{~g} / \mathrm{mol}\end{aligned}$
$\begin{aligned} \text { Number of moles atoms } & =\frac{3}{214.6} \\ & =0.01397 \\ \text { Number of atoms } & =0.01397 \times 6.02 \times 2 \times 10^{23} \\ & =8.41 \times 10^{21}\end{aligned}$
$\begin{aligned} & 9.4=\frac{1 \times M}{\left(3.36 \times 10^{-8}\right)^3 \times 6.02 \times 10^{23}} \\ & M=214.6 \mathrm{~g} / \mathrm{mol}\end{aligned}$
$\begin{aligned} \text { Number of moles atoms } & =\frac{3}{214.6} \\ & =0.01397 \\ \text { Number of atoms } & =0.01397 \times 6.02 \times 2 \times 10^{23} \\ & =8.41 \times 10^{21}\end{aligned}$
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