MHT CET · Chemistry · Solid State
How many number of unit cells are present in \(100 \mathrm{~g}\) of an element with fcc crystal having density \(10 \mathrm{~g} / \mathrm{cm}^{3}\) and edge length \(100 \mathrm{pm}\) ?
- A \(3 \times 10^{25}\)
- B \(2 \times 10^{25}\)
- C \(4 \times 10^{25}\)
- D \(1 \times 10^{25}\)
Answer & Solution
Correct Answer
(D) \(1 \times 10^{25}\)
Step-by-step Solution
Detailed explanation
\(\text {Vol. of unit cell} =(100 \mathrm{pm})^{3}=\) \(\left(100 \times 10^{-10} \mathrm{~cm}\right)^{3}\)
\(=10^{-24} \mathrm{~cm}^{3}\)
Vol. of \(100 \mathrm{~g}\) of an element \(=\frac{\text { Mass }}{\text { Density }}=\frac{100 \mathrm{~g}}{10 \mathrm{~g} \mathrm{~cm}^{-3}}=10 \mathrm{~cm}^{3}\)
No. of unit cells in \(100 \mathrm{~g}\) of an element \(=\frac{\text { Total volume }}{\text { Volume of one unit cell }}\)
\(=\frac{10 \mathrm{~cm}^{3}}{10^{-24} \mathrm{~cm}^{3}}=1 \times 10^{25}\) unit cells.
\(=10^{-24} \mathrm{~cm}^{3}\)
Vol. of \(100 \mathrm{~g}\) of an element \(=\frac{\text { Mass }}{\text { Density }}=\frac{100 \mathrm{~g}}{10 \mathrm{~g} \mathrm{~cm}^{-3}}=10 \mathrm{~cm}^{3}\)
No. of unit cells in \(100 \mathrm{~g}\) of an element \(=\frac{\text { Total volume }}{\text { Volume of one unit cell }}\)
\(=\frac{10 \mathrm{~cm}^{3}}{10^{-24} \mathrm{~cm}^{3}}=1 \times 10^{25}\) unit cells.
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