MHT CET · Chemistry · Solid State
An element crystallises in fcc type of unit cell. The volume of one unit cell is \(24.99 \times 10^{-24} \mathrm{~cm}^{3}\) and density of the element \(7 \cdot 2 \mathrm{~g} \mathrm{~cm}^{-3} .\) Calculate the number of unit cells in \(36 \mathrm{~g}\) of pure sample of element?
- A \(2 \cdot 0 \times 10^{23}\)
- B \(2 \cdot 0 \times 10^{21}\)
- C \(2 \cdot 0 \times 10^{24}\)
- D \(1 \cdot 25 \times 10^{21}\)
Answer & Solution
Correct Answer
(A) \(2 \cdot 0 \times 10^{23}\)
Step-by-step Solution
Detailed explanation
(C)
Vol. of element \(=\frac{\text { Mass }}{\text { Density }}=\frac{36 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=5 \mathrm{~cm}^{3}\)
No. of unit cells in \(36 \mathrm{~g}\) of pure sample of element \(=\frac{\text { Total Vol. of element }}{\text { Vol. of one unit cell }}\)
\(=\frac{5 \mathrm{~cm}^{3}}{24.99 \times 10^{-24} \mathrm{~cm}^{3}}=2.0 \times 10^{23}\)
Vol. of element \(=\frac{\text { Mass }}{\text { Density }}=\frac{36 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=5 \mathrm{~cm}^{3}\)
No. of unit cells in \(36 \mathrm{~g}\) of pure sample of element \(=\frac{\text { Total Vol. of element }}{\text { Vol. of one unit cell }}\)
\(=\frac{5 \mathrm{~cm}^{3}}{24.99 \times 10^{-24} \mathrm{~cm}^{3}}=2.0 \times 10^{23}\)
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