MHT CET · Chemistry · Solid State
Calculate the density of an element having molar mass \(63 \mathrm{~g} \mathrm{~mol}^{-1}\) that forms FCC structure \(\left[\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}=28 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]\)
- A \(6.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
- B \(\quad 9.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
- C \(5.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
- D \(7.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Answer & Solution
Correct Answer
(B) \(\quad 9.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
Step-by-step Solution
Detailed explanation
For FCC lattice, \(n=4\)
\(\rho=\frac{\mathrm{M} \times \mathrm{n}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}=\frac{63 \mathrm{~g} \mathrm{~mol}^{-1} \times 4}{28 \mathrm{~cm}^3 \mathrm{~mol}^{-1}}=9.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
\(\rho=\frac{\mathrm{M} \times \mathrm{n}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}=\frac{63 \mathrm{~g} \mathrm{~mol}^{-1} \times 4}{28 \mathrm{~cm}^3 \mathrm{~mol}^{-1}}=9.0 \mathrm{~g} \mathrm{~cm}^{-3}\)
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