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Question: Answered & Verified by Expert
Calculate the number of unit cells in $38.6 \mathrm{~g}$ of noble metal haven density $19 \cdot 3 \mathrm{~g} \mathrm{~cm}^{-3}$ and volume of one unit cell is $6 \cdot 18 \times 10^{-23} \mathrm{~cm}^{3}$ ?
ChemistrySolid StateMHT CETMHT CET 2020 (12 Oct Shift 1)
Options:
  • A $3 \cdot 236 \times 10^{22}$
  • B $6 \cdot 180 \times 10^{23}$
  • C $6 \cdot 236 \times 10^{20}$
  • D $3 \cdot 236 \times 10^{23}$
Solution:
1286 Upvotes Verified Answer
The correct answer is: $3 \cdot 236 \times 10^{22}$
$\begin{aligned} \text { Vol. of metal }=\frac{\text { Mass }}{\text { Density }}=\frac{38.6 \mathrm{~g}}{19.3 \mathrm{~g} \mathrm{~cm}^{-3}} &=2 \mathrm{~cm}^{3} \\ \text { No. of unit cells in } 38.6 \mathrm{~g} \text { of noble metal } &=\frac{\text { Total vol. of metal }}{\text { Vol. of one unit cell }} \\ &=\frac{2 \mathrm{~cm}^{3}}{6.18 \times 10^{-23} \mathrm{~cm}^{3}}=3.236 \times 10^{22} \end{aligned}$

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